빌드라인 스쿨

문제 은행 · 미적분 · 대학 미적분 · 다중적분

다중적분 전체 문제

이 단원의 모든 연습문제입니다. 소단원으로 좁혀 보려면 아래 꼬리표를 누르세요.

유형전체개념 확인계산그래프·표 해석활용증명·논술
난이도전체기본보통도전
풀이전체우리말 풀이 있는 문제만

유형·난이도는 우리말 풀이가 붙은 문제는 선생님이 정했고, 나머지는 자동 분류라 다를 수 있어요. 난이도가 정해지지 않은 문제는 난이도로 거를 때 빠집니다.

691문제 · 35 / 35쪽

681.CLP-3 · 3.7절 · 20번대학 수준우리말 번역

어떤 입체 VV 는 직원기둥이다. 밑면은 xyxy평면에서 원점을 중심으로 하는 반지름 22 인 원판이다. 높이는 22 이고 밀도는 x2+y2\sqrt{x^2 + y^2} 이다.

VV 의 윗면을 밑면으로 하고 점 (0,0,0)(0, 0, 0) 을 꼭짓점으로 하는 거꾸로 놓인 원뿔을 제거하여, 더 작은 입체 UU 를 얻는다.

(a) 원기둥좌표를 사용하여 UU 의 질량을 나타내는 적분을 세우세요. (b) 구면좌표를 사용하여 UU 의 질량을 나타내는 적분을 세우세요. (c) 그 질량을 구하세요.

영어 원문 보기

A certain solid VV is a right–circular cylinder. Its base is the disk of radius 22 centred at the origin in the xyxy–plane. It has height 22 and density x2+y2\sqrt{x^2 + y^2}.

A smaller solid UU is obtained by removing the inverted cone, whose base is the top surface of VV and whose vertex is the point (0,0,0)(0, 0, 0).

(a) Use cylindrical coordinates to set up an integral giving the mass of UU. (b) Use spherical coordinates to set up an integral giving the mass of UU. (c) Find that mass.

정답·풀이 보기

xyxy평면에서 원점을 중심으로 하는 반지름 22 인 원판은 x2+y2≤4x^2+y^2\le 4 이다. 따라서

V={ (x,y,z) ∣ x2+y2≤4, 0≤z≤2 }\begin{aligned} V&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2\le 4,\ 0\le z\le 2\ \big\} \end{aligned}

UU 의 윗 가장자리 x+y2=4x+y^2=4, z=2z=2 를 포함하면서 꼭짓점이 원점인 원뿔은 x2+y2=z2x^2+y^2=z^2 이다. 따라서

U={ (x,y,z) ∣ x2+y2≤4, 0≤z≤2, x2+y2≥z2 }\begin{aligned} U&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2\le 4,\ 0\le z\le 2,\ x^2+y^2\ge z^2\ \big\} \end{aligned}

다음은 VV (왼쪽)와 UU (오른쪽)의 y=0y=0 단면을 그린 그림이다.

그림

그림

(a) 원기둥좌표에서 x2+y2≤4x^2+y^2\le 4 는 r≤2r\le 2 가 되고, x2+y2≥z2x^2+y^2\ge z^2 는 r≥∣z∣r\ge |z| 이며, 밀도는 x2+y2=r\sqrt{x^2+y^2}=r 이다. 따라서

U={ (rcos⁡θ,rsin⁡θ,z) ∣ r≤2, 0≤z≤2, r≥z }\begin{aligned} U&=\big\{ \ (r\cos\theta,r\sin\theta,z)\ \big|\ r\le 2,\ 0\le z\le 2,\ r\ge z\ \big\} \end{aligned}

아래 왼쪽 그림을 보면, UU 에서

  • zz 는 00 에서 22 까지 변하고,
  • 그 범위의 각 zz 에 대해 rr 는 zz 에서 22 까지, θ\theta 는 00 에서 2π2\pi 까지 변한다.
  • dV=r dr dθ dz\mathrm{d}V =r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z

따라서

질량=∫02dz∫02πdθ∫z2dr r r⏞밀도=∫02dz∫02πdθ∫z2dr r2\begin{aligned} \text{질량}&= \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r\ \overbrace{r}^{\text{밀도}} = \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r^2 \end{aligned}

그림

그림

(b) 구면좌표에서

x=ρsin⁡φcos⁡θy=ρsin⁡φsin⁡θz=ρcos⁡φx2+y2=ρ2sin⁡2φ\begin{aligned} x&=\rho\sin\varphi\cos\theta \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ x^2+y^2 &=\rho^2\sin^2\varphi \end{aligned}

이므로 x2+y2≤4x^2+y^2\le 4 는 ρsin⁡φ≤2\rho\sin\varphi\le 2 가 되고, x2+y2≥z2x^2+y^2\ge z^2 는 다음과 같이 된다.

ρsin⁡φ≥ρcos⁡φ  ⟺  tan⁡φ≥1  ⟺  φ≥π4\begin{aligned} \rho\sin\varphi \ge \rho\cos\varphi \iff \tan\varphi\ge 1 \iff \varphi\ge\frac{\pi}{4} \end{aligned}

그리고 밀도는 x2+y2=ρsin⁡φ\sqrt{x^2+y^2}=\rho\sin\varphi 이다. 따라서

U={ (ρsin⁡φcos⁡θ,ρsin⁡φsin⁡θ,ρcos⁡φ) ∣ π/4≤φ≤π/2, 0≤θ≤2π, ρsin⁡φ≤2 }\begin{aligned} U&=\big\{ \ (\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\ \big|\ {\pi}/{4}\le\varphi\le{\pi}/{2},\ 0\le \theta\le 2\pi,\ \rho\sin\varphi\le 2\ \big\} \end{aligned}

위의 오른쪽 그림을 보면, UU 에서

  • φ\varphi 는 π4\frac{\pi}{4} (원뿔 위)에서 π2\frac{\pi}{2} (xyxy평면 위)까지 변하고,
  • 그 범위의 각 φ\varphi 에 대해 ρ\rho 는 00 에서 2sin⁡φ\frac{2}{\sin\varphi} 까지, θ\theta 는 00 에서 2π2\pi 까지 변한다.
  • dV=ρ2sin⁡φ dρ dθ dφ\mathrm{d}V =\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi

따라서

질량=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ2sin⁡φ ρsin⁡φ⏞밀도=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ3sin⁡2φ\begin{aligned} \text{질량}&= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{2/\sin\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi}^{\text{밀도}} \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{2/\sin\varphi}\mathrm{d}\rho\ \rho^3\sin^2\varphi \end{aligned}

(c) 원기둥좌표로 나타낸 식을 사용한다.

질량=∫02dz∫02πdθ∫z2dr r2=2π∫02dz 8−z33=2π3 [16−244]=8π\begin{aligned} \text{질량} &= \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r^2 \\ &=2\pi \int_0^2\mathrm{d}z\ \frac{8-z^3}{3} \\ &=\frac{2\pi}{3}\ \left[16-\frac{2^4}{4}\right] \\ &=8\pi \end{aligned}

원본 교재 풀이(영어)

The disk of radius 22 centred at the origin in the xyxy–plane is x2+y2≤4x^2+y^2\le 4. So

V={ (x,y,z) ∣ x2+y2≤4, 0≤z≤2 }\begin{aligned} V&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2\le 4,\ 0\le z\le 2\ \big\} \end{aligned}

The cone with vertex at the origin that contains the top edge, x+y2=4x+y^2=4, z=2z=2, of UU is x2+y2=z2x^2+y^2=z^2. So

U={ (x,y,z) ∣ x2+y2≤4, 0≤z≤2, x2+y2≥z2 }\begin{aligned} U&=\big\{ \ (x,y,z)\ \big|\ x^2+y^2\le 4,\ 0\le z\le 2,\ x^2+y^2\ge z^2\ \big\} \end{aligned}

Here are sketches of the y=0y=0 cross–section of VV, on the left, and UU, on the right.

figure

figure

(a) In cylindrical coordinates, x2+y2≤4x^2+y^2\le 4 becomes r≤2r\le 2 and x2+y2≥z2x^2+y^2\ge z^2 is r≥∣z∣r\ge |z|, and the density is x2+y2=r\sqrt{x^2+y^2}=r. So

U={ (rcos⁡θ,rsin⁡θ,z) ∣ r≤2, 0≤z≤2, r≥z }\begin{aligned} U&=\big\{ \ (r\cos\theta,r\sin\theta,z)\ \big|\ r\le 2,\ 0\le z\le 2,\ r\ge z\ \big\} \end{aligned}

Looking at the figure on the left below, we see that, on UU

  • zz runs from 00 to 22, and
  • for each zz is that range, rr runs from zz to 22 and θ\theta runs from 00 to 2π2\pi.
  • dV=r dr dθ dz\mathrm{d}V =r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z

So

Mass=∫02dz∫02πdθ∫z2dr r r⏞density=∫02dz∫02πdθ∫z2dr r2\begin{aligned} \text{Mass}&= \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r\ \overbrace{r}^{\text{density}} = \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r^2 \end{aligned}

figure

figure

(b) Recall that in spherical coordinates,

x=ρsin⁡φcos⁡θy=ρsin⁡φsin⁡θz=ρcos⁡φx2+y2=ρ2sin⁡2φ\begin{aligned} x&=\rho\sin\varphi\cos\theta \\ y&=\rho\sin\varphi\sin\theta \\ z&=\rho\cos\varphi \\ x^2+y^2 &=\rho^2\sin^2\varphi \end{aligned}

so that x2+y2≤4x^2+y^2\le 4 becomes ρsin⁡φ≤2\rho\sin\varphi\le 2, and x2+y2≥z2x^2+y^2\ge z^2 becomes

ρsin⁡φ≥ρcos⁡φ  ⟺  tan⁡φ≥1  ⟺  φ≥π4\begin{aligned} \rho\sin\varphi \ge \rho\cos\varphi \iff \tan\varphi\ge 1 \iff \varphi\ge\frac{\pi}{4} \end{aligned}

and the density x2+y2=ρsin⁡φ\sqrt{x^2+y^2}=\rho\sin\varphi. So

U={ (ρsin⁡φcos⁡θ,ρsin⁡φsin⁡θ,ρcos⁡φ) ∣ π/4≤φ≤π/2, 0≤θ≤2π, ρsin⁡φ≤2 }\begin{aligned} U&=\big\{ \ (\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\ \big|\ {\pi}/{4}\le\varphi\le{\pi}/{2},\ 0\le \theta\le 2\pi,\ \rho\sin\varphi\le 2\ \big\} \end{aligned}

Looking at the figure on the right above, we see that, on UU

  • φ\varphi runs from π4\frac{\pi}{4} (on the cone) to π2\frac{\pi}{2} (on the xyxy–plane), and
  • for each φ\varphi is that range, ρ\rho runs from 00 to 2sin⁡φ\frac{2}{\sin\varphi} and θ\theta runs from 00 to 2π2\pi.
  • dV=ρ2sin⁡φ dρ dθ dφ\mathrm{d}V =\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi

So

Mass=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ2sin⁡φ ρsin⁡φ⏞density=∫π/4π/2dφ∫02πdθ∫02/sin⁡φdρ ρ3sin⁡2φ\begin{aligned} \text{Mass}&= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{2/\sin\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi\ \overbrace{\rho\sin\varphi}^{\text{density}} \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{2/\sin\varphi}\mathrm{d}\rho\ \rho^3\sin^2\varphi \end{aligned}

(c) We'll use the cylindrical form.

Mass=∫02dz∫02πdθ∫z2dr r2=2π∫02dz 8−z33=2π3 [16−244]=8π\begin{aligned} \text{Mass} &= \int_0^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\int_z^2\mathrm{d}r\ r^2 \\ &=2\pi \int_0^2\mathrm{d}z\ \frac{8-z^3}{3} \\ &=\frac{2\pi}{3}\ \left[16-\frac{2^4}{4}\right] \\ &=8\pi \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 21번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
682.CLP-3 · 3.7절 · 21번대학 수준우리말 번역

어떤 입체가 아래로는 원뿔 z ⁣= ⁣x2 ⁣+ ⁣y2 z\!=\!\sqrt{x^2\!+\!y^2}\ 으로, 위로는 구 x2+y2+z2=2x^2+y^2+z^2 = 2 로 유계이다. 이 입체의 밀도는 δ(x,y,z)=x2+y2\delta (x,y,z) = x^2 + y^2 이다.

(a) 이 입체의 질량 MM 을 원기둥좌표계에서 적분 구간과 함께 삼중적분으로 나타내세요. (b) (a)와 같이 하되 구면좌표계에서 나타내세요. (c) MM 의 값을 구하세요.

영어 원문 보기

A solid is bounded below by the cone z ⁣= ⁣x2 ⁣+ ⁣y2 z\!=\!\sqrt{x^2\!+\!y^2}\ and above by the sphere x2+y2+z2=2x^2+y^2+z^2 = 2. It has density δ(x,y,z)=x2+y2\delta (x,y,z) = x^2 + y^2.

(a) Express the mass MM of the solid as a triple integral, with limits, in cylindrical coordinates. (b) Same as (a) but in spherical coordinates. (c) Evaluate MM.

정답·풀이 보기

(a) 이 입체를 VV 라 하자. 원기둥좌표계에서

  • x2+y2+z2≤2x^2+y^2+z^2\le 2 는 r2+z2≤2r^2+z^2\le 2 이고
  • x2+y2≤z\sqrt{x^2+y^2}\le z 는 r≤zr\le z 이고
  • 밀도는 δ=r2\delta =r^2 이고
  • dV\mathrm{d}V 는 r dr dθ dzr\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z 이다.

r2+z2=2r^2+z^2= 2 와 r=zr= z 는 2r2=22r^2=2 일 때 만나므로 r=z=1r=z=1 임을 알 수 있다. 다음은 EE 의 y=0y=0 단면의 그림이다.

그림

따라서

V={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1, 0≤θ≤2π, r≤z≤2−r2 }V =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le 1,\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{2-r^2}\ \big\}

이고

M=∭Vρ(x,y,z) dV=∫01dr∫02πdθ∫r2−r2dz r(r2)⏞δ=∫01dr∫02πdθ∫r2−r2dz r3\begin{aligned} M &=\iiint _V \rho(x,y,z)\ \mathrm{d}V =\int_0^1\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{2-r^2}}\mathrm{d}z\ r\overbrace{(r^2)}^{\delta } =\int_0^1\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{2-r^2}}\mathrm{d}z\ r^3 \end{aligned}

이다.

(b)

구면좌표계에서

  • x2+y2+z2≤2x^2+y^2+z^2\le 2 는 ρ≤2\rho\le\sqrt{2} 이고,
  • x2+y2≤z\sqrt{x^2+y^2}\le z 는 ρsin⁡φ≤ρcos⁡φ\rho\sin\varphi\le \rho\cos\varphi, 즉 tan⁡φ≤1\tan\varphi\le 1 또는 φ≤π4\varphi\le\frac{\pi}{4} 이고,
  • 밀도는 x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi 이고,
  • dV\mathrm{d}V 는 ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi 이다.

따라서

V={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤2, 0≤θ≤2π, 0≤φ≤π/4 }V =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le \sqrt{2},\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{4}\ \big\}

이고, 피적분함수가 x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi 이므로

M=∭V(x2+y2) dV=∫02dρ∫02πdθ∫0π/4dφ ρ2sin⁡φ ρ2sin⁡2φ=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ\begin{aligned} M &=\iiint _V \big(x^2+y^2\big)\ \mathrm{d}V =\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \rho^2\sin^2\varphi \\ &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \end{aligned}

이다.

(c) 구면좌표계로 나타낸 식을 이용한다.

M=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡φ[1−cos⁡2φ]=2π∫02dρ ρ4[−cos⁡φ+cos⁡3φ3]0π/4=2π[23−562]∫02dρ ρ4=2π425[23−562]=π[16215−43]≈0.5503\begin{aligned} M &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \\ &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^{\sqrt{2}}\mathrm{d}\rho\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/4} =2\pi \left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] \int_0^{\sqrt{2}}\mathrm{d}\rho\ \rho^4 \\ &=2\pi \frac{4\sqrt{2}}{5}\left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] =\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right] \approx 0.5503 \end{aligned}

원본 교재 풀이(영어)

(a) Call the solid VV. In cylindrical coordinates

  • x2+y2+z2≤2x^2+y^2+z^2\le 2 is r2+z2≤2r^2+z^2\le 2 and
  • x2+y2≤z\sqrt{x^2+y^2}\le z is r≤zr\le z and
  • the density δ=r2\delta =r^2, and
  • dV\mathrm{d}V is r dr dθ dzr\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z

Observe that r2+z2=2r^2+z^2= 2 and r=zr= z intersect when 2r2=22r^2=2 so that r=z=1r=z=1. Here is a sketch of the y=0y=0 cross–section of EE.

figure

So

V={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1, 0≤θ≤2π, r≤z≤2−r2 }V =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le 1,\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{2-r^2}\ \big\}

and

M=∭Vρ(x,y,z) dV=∫01dr∫02πdθ∫r2−r2dz r(r2)⏞δ=∫01dr∫02πdθ∫r2−r2dz r3\begin{aligned} M &=\iiint _V \rho(x,y,z)\ \mathrm{d}V =\int_0^1\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{2-r^2}}\mathrm{d}z\ r\overbrace{(r^2)}^{\delta } =\int_0^1\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{2-r^2}}\mathrm{d}z\ r^3 \end{aligned}

(b)

In spherical coordinates

  • x2+y2+z2≤2x^2+y^2+z^2\le 2 is ρ≤2\rho\le\sqrt{2}, and
  • x2+y2≤z\sqrt{x^2+y^2}\le z is ρsin⁡φ≤ρcos⁡φ\rho\sin\varphi\le \rho\cos\varphi, or tan⁡φ≤1\tan\varphi\le 1 or φ≤π4\varphi\le\frac{\pi}{4}, and
  • the density x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi, and
  • dV\mathrm{d}V is ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi

So

V={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤2, 0≤θ≤2π, 0≤φ≤π/4 }V =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le \sqrt{2},\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{4}\ \big\}

and, since the integrand x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi,

M=∭V(x2+y2) dV=∫02dρ∫02πdθ∫0π/4dφ ρ2sin⁡φ ρ2sin⁡2φ=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ\begin{aligned} M &=\iiint _V \big(x^2+y^2\big)\ \mathrm{d}V =\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \rho^2\sin^2\varphi \\ &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \end{aligned}

(c) We'll use the spherical coordinate form.

M=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡3φ=∫02dρ∫02πdθ∫0π/4dφ ρ4sin⁡φ[1−cos⁡2φ]=2π∫02dρ ρ4[−cos⁡φ+cos⁡3φ3]0π/4=2π[23−562]∫02dρ ρ4=2π425[23−562]=π[16215−43]≈0.5503\begin{aligned} M &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \\ &=\int_0^{\sqrt{2}}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^{\sqrt{2}}\mathrm{d}\rho\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/4} =2\pi \left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] \int_0^{\sqrt{2}}\mathrm{d}\rho\ \rho^4 \\ &=2\pi \frac{4\sqrt{2}}{5}\left[\frac{2}{3} - \frac{5}{6\sqrt{2}}\right] =\pi \left[\frac{16\sqrt{2}}{15} - \frac{4}{3}\right] \approx 0.5503 \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 22번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
683.CLP-3 · 3.7절 · 22번대학 수준우리말 번역

I=∭Exz dV\begin{aligned} I = \iiint _E xz\ \mathrm{d}V \end{aligned}

라 하자. 여기서 EE 는 구 x2+y2+z2≤1x^2+y^2+z^2\le 1 의 8분의 1 부분으로, x,y,z≥0x,y,z\ge 0 이다.

(a) II 를 구면좌표계에서의 삼중적분으로 나타내세요. (b) II 를 원기둥좌표계에서의 삼중적분으로 나타내세요. (c) 어떤 방법으로든 II 의 값을 구하세요.

영어 원문 보기

Let

I=∭Exz dV\begin{aligned} I = \iiint _E xz\ \mathrm{d}V \end{aligned}

where EE is the eighth of the sphere x2+y2+z2≤1x^2+y^2+z^2\le 1 with x,y,z≥0x,y,z\ge 0.

(a) Express II as a triple integral in spherical coordinates. (b) Express II as a triple integral in cylindrical coordinates. (c) Evaluate II by any method.

정답·풀이 보기

(a) EE 위에서

  • 구면좌표 φ\varphi 는 00 (양의 zz축)에서 π/2{\pi}/{2} (xyxy평면)까지 움직이고,
  • 그 범위의 각 고정된 φ\varphi 에 대해 θ\theta 는 00 에서 π/2{\pi}/{2} 까지 움직이며,
  • 각 고정된 φ\varphi 와 θ\theta 에 대해 구면좌표 ρ\rho 는 00 에서 11 까지 움직인다.
  • 구면좌표에서 dV=ρ2 sin⁡φ dρ dθ dφ\mathrm{d}V=\rho^2\,\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\, \mathrm{d}\varphi 이고

xz=(ρsin⁡φcos⁡θ)(ρcos⁡φ)=ρ2sin⁡φ cos⁡φ cos⁡θxz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta

이다. 따라서

I=∫0π/2dφ∫0π/2dθ∫01dρ ρ4sin⁡2φcos⁡φ cos⁡θ\begin{aligned} I=\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{\pi/2}\mathrm{d}\theta\int_0^1\mathrm{d}\rho\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{aligned}

(b) 원기둥좌표에서 조건 x2+y2+z2≤1x^2+y^2+z^2\le 1 은 r2+z2≤1r^2+z^2\le 1 이 된다. 따라서 EE 위에서

  • 원기둥좌표 zz 는 00 (xyxy평면 위)에서 11 ((0,0,1)(0,0,1) 에서)까지 움직이고,
  • 그 범위의 각 고정된 zz 에 대해 θ\theta 는 00 에서 π/2\pi/2 까지 움직이며,
  • 각 고정된 zz 와 θ\theta 에 대해 원기둥좌표 rr 은 00 에서 1−z2\sqrt{1-z^2} 까지 움직인다 (r2+z2≤1r^2+z^2\le 1 임을 떠올리자).
  • 원기둥좌표에서 dV=r dr dθ dz\mathrm{d}V=r\,\mathrm{d}r\,\mathrm{d}\theta\, \mathrm{d}z 이고

xz=(rcos⁡θ)(z)=r z cos⁡θxz=\big(r\cos\theta\big)\big(z\big) =r\,z\, \cos\theta

이다. 따라서

I=∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ\begin{aligned} I=\int_0^1\mathrm{d}z\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\sqrt{1-z^2}}\mathrm{d}r \ r^2\,z\,\cos\theta \end{aligned}

(c) 구면좌표와 원기둥좌표의 적분 모두 계산이 간단하다. 두 가지를 모두 계산해 보자. 먼저 구면좌표에서는

I=[∫0π/2dφ sin⁡2φcos⁡φ][∫0π/2dθ cos⁡θ][∫01dρ ρ4]=[sin⁡3φ3]0π/2[sin⁡θ]0π/2[ρ55]01=115\begin{aligned} I&= \left[\int_0^{\pi/2}\mathrm{d}\varphi\ \sin^2\varphi\cos\varphi\right] \left[\int_0^{\pi/2}\mathrm{d}\theta\ \cos\theta\right] \left[\int_0^1\mathrm{d}\rho\ \rho^4\right] \\ &=\left[\frac{\sin^3\varphi}{3}\right]_0^{\pi/2} \left[\sin\theta\right]_0^{\pi/2} \left[\frac{\rho^5}{5}\right]_0^1 \\ &=\frac{1}{15} \end{aligned}

이다. 이제 원기둥좌표에서는

I=∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ=13∫01dz∫0π/2dθ z(1−z2)3/2 cos⁡θ=13∫01dz z(1−z2)3/2=13[−12 (1−z2)5/25/2]01=115\begin{aligned} I&=\int_0^1\mathrm{d}z\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\sqrt{1-z^2}}\mathrm{d}r \ r^2\,z\,\cos\theta \\ &=\frac{1}{3}\int_0^1\mathrm{d}z\int_0^{\pi/2}\mathrm{d}\theta\ z{(1-z^2\big)}^{3/2}\, \cos\theta \\ &=\frac{1}{3}\int_0^1\mathrm{d}z\ z{(1-z^2\big)}^{3/2} \\ &=\frac{1}{3}\left[-\frac{1}{2}\ \frac{(1-z^2)^{5/2}}{5/2}\right]_0^1\\ &=\frac{1}{15} \end{aligned}

이다.

원본 교재 풀이(영어)

(a) On EE,

  • the spherical coordinate φ\varphi runs from 00 (the positive zz–xis) to π/2{\pi}/{2} (the xyxy–plane), and
  • for each fixed φ\varphi in that range, θ\theta runs from 00 to π/2{\pi}/{2}, and
  • for each fixed φ\varphi and θ\theta, the spherical coordinate ρ\rho runs from 00 to 11.
  • In spherical coordinates dV=ρ2 sin⁡φ dρ dθ dφ\mathrm{d}V=\rho^2\,\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\, \mathrm{d}\varphi and

xz=(ρsin⁡φcos⁡θ)(ρcos⁡φ)=ρ2sin⁡φ cos⁡φ cos⁡θxz=\big(\rho\sin\varphi\cos\theta\big)\big(\rho\cos\varphi\big) =\rho^2 \sin\varphi\ \cos\varphi\ \cos\theta

So

I=∫0π/2dφ∫0π/2dθ∫01dρ ρ4sin⁡2φcos⁡φ cos⁡θ\begin{aligned} I=\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{\pi/2}\mathrm{d}\theta\int_0^1\mathrm{d}\rho\ \rho^4\sin^2\varphi\cos\varphi\ \cos\theta \end{aligned}

(b) In cylindrical coordinates, the condition x2+y2+z2≤1x^2+y^2+z^2\le 1 becomes r2+z2≤1r^2+z^2\le 1. So, on EE

  • the cylindrical coordinate zz runs from 00 (in the xyxy–plane) to 11 (at (0,0,1)(0,0,1)) and
  • for each fixed zz in that range, θ\theta runs from 00 to π/2\pi/2 and
  • for each such fixed zz and θ\theta, the cylindrical coordinate rr runs from 00 to 1−z2\sqrt{1-z^2} (recall that r2+z2≤1r^2+z^2\le 1).
  • In cylindrical coordinates dV=r dr dθ dz\mathrm{d}V=r\,\mathrm{d}r\,\mathrm{d}\theta\, \mathrm{d}z and

xz=(rcos⁡θ)(z)=r z cos⁡θxz=\big(r\cos\theta\big)\big(z\big) =r\,z\, \cos\theta

So

I=∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ\begin{aligned} I=\int_0^1\mathrm{d}z\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\sqrt{1-z^2}}\mathrm{d}r \ r^2\,z\,\cos\theta \end{aligned}

(c) Both spherical and cylindrical integrals are straight forward to evaluate. Here are both. First, in spherical coordinates,

I=[∫0π/2dφ sin⁡2φcos⁡φ][∫0π/2dθ cos⁡θ][∫01dρ ρ4]=[sin⁡3φ3]0π/2[sin⁡θ]0π/2[ρ55]01=115\begin{aligned} I&= \left[\int_0^{\pi/2}\mathrm{d}\varphi\ \sin^2\varphi\cos\varphi\right] \left[\int_0^{\pi/2}\mathrm{d}\theta\ \cos\theta\right] \left[\int_0^1\mathrm{d}\rho\ \rho^4\right] \\ &=\left[\frac{\sin^3\varphi}{3}\right]_0^{\pi/2} \left[\sin\theta\right]_0^{\pi/2} \left[\frac{\rho^5}{5}\right]_0^1 \\ &=\frac{1}{15} \end{aligned}

Now in cylindrical coordinates

I=∫01dz∫0π/2dθ∫01−z2dr r2 z cos⁡θ=13∫01dz∫0π/2dθ z(1−z2)3/2 cos⁡θ=13∫01dz z(1−z2)3/2=13[−12 (1−z2)5/25/2]01=115\begin{aligned} I&=\int_0^1\mathrm{d}z\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\sqrt{1-z^2}}\mathrm{d}r \ r^2\,z\,\cos\theta \\ &=\frac{1}{3}\int_0^1\mathrm{d}z\int_0^{\pi/2}\mathrm{d}\theta\ z{(1-z^2\big)}^{3/2}\, \cos\theta \\ &=\frac{1}{3}\int_0^1\mathrm{d}z\ z{(1-z^2\big)}^{3/2} \\ &=\frac{1}{3}\left[-\frac{1}{2}\ \frac{(1-z^2)^{5/2}}{5/2}\right]_0^1\\ &=\frac{1}{15} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 23번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
684.CLP-3 · 3.7절 · 23번대학 수준우리말 번역

I=∭T(x2+y2) dVI = \iiint _T (x^2+y^2)\ \mathrm{d}V

라 하자. 여기서 TT 는 아래로는 원뿔 z=3x2+3y2z =\sqrt{3x^2+3y^2} 로, 위로는 구 x2+y2+z2=9x^2 + y^2 + z^2 = 9 로 유계인 입체 영역이다.

(a) II 를 구면좌표계의 삼중적분으로 나타내세요. (b) II 를 원기둥좌표계의 삼중적분으로 나타내세요. (c) 어떤 방법으로든 II 의 값을 구하세요.

영어 원문 보기

Let

I=∭T(x2+y2) dVI = \iiint _T (x^2+y^2)\ \mathrm{d}V

where TT is the solid region bounded below by the cone z=3x2+3y2z =\sqrt{3x^2+3y^2} and above by the sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9.

(a) Express II as a triple integral in spherical coordinates. (b) Express II as a triple integral in cylindrical coordinates. (c) Evaluate II by any method.

정답·풀이 보기

(a) 구면좌표계에서

x=ρ sin⁡φ cos⁡θy=ρ sin⁡φ sin⁡θz=ρ cos⁡φ\begin{aligned} x=\rho\,\sin\varphi\,\cos\theta\qquad y=\rho\,\sin\varphi\,\sin\theta\qquad z=\rho\,\cos\varphi \end{aligned}

이므로

  • x2+y2+z2≤9x^2+y^2+z^2\le 9 는 ρ≤3\rho\le 3 이고,
  • 3x2+3y2≤z\sqrt{3x^2+3y^2}\le z 는 3ρsin⁡φ≤ρcos⁡φ\sqrt{3}\rho\sin\varphi\le \rho\cos\varphi, 즉 tan⁡φ≤13\tan\varphi\le \frac{1}{\sqrt{3}} 또는 φ≤π6\varphi\le\frac{\pi}{6} 이고,
  • 피적분함수 x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi 이고,
  • dV\mathrm{d}V 는 ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi 이다.

따라서

T={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤3, 0≤θ≤2π, 0≤φ≤π/6 }T =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 3,\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{6}\ \big\}

이고,

I=∭T(x2+y2) dV=∫03dρ∫02πdθ∫0π/6dφ ρ2sin⁡φ ρ2sin⁡2φ⏞x2+y2=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ\begin{aligned} I &=\iiint _T \big(x^2+y^2\big)\ \mathrm{d}V =\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \overbrace{\rho^2\sin^2\varphi}^{x^2+y^2} \\ &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \end{aligned}

(b) 원기둥좌표계에서

  • x2+y2+z2≤9x^2+y^2+z^2\le 9 는 r2+z2≤9r^2+z^2\le 9 이고,
  • 3x2+3y2≤z\sqrt{3 x^2+3 y^2}\le z 는 3 r≤z\sqrt{3}\,r\le z 이고,
  • 피적분함수 x2+y2=r2x^2+y^2=r^2 이고,
  • dV\mathrm{d}V 는 r dr dθ dzr\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z 이다.

r2+z2=9r^2+z^2= 9 와 3 r=z\sqrt{3}\,r= z 는 r2+3r2=9r^2+3r^2=9 일 때 만나므로 r=32r=\frac{3}{2}, z=332z=\frac{3\sqrt{3}}{2} 임을 알 수 있다. 다음은 TT 의 y=0y=0 단면의 그림이다.

그림

따라서

T={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤32, 0≤θ≤2π, 3 r≤z≤9−r2 }T =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le \tfrac{3}{2},\ 0\le\theta\le2\pi,\ \sqrt{3}\,r\le z\le \sqrt{9-r^2}\ \big\}

이고

I=∭V(x2+y2) dV=∫03/2dr∫02πdθ∫3 r9−r2dz r(r2)⏞x2+y2=∫03/2dr∫02πdθ∫3 r9−r2dz  r3\begin{aligned} I &=\iiint _V (x^2+y^2)\ \mathrm{d}V =\int_0^{3/2}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\mathrm{d}z\ r\overbrace{(r^2)}^{x^2+y^2} =\int_0^{3/2}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\mathrm{d}z\ \ r^3 \end{aligned}

(c) 구면좌표로 나타낸 식을 이용한다.

I=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡φ[1−cos⁡2φ]=2π∫03dρ ρ4[−cos⁡φ+cos⁡3φ3]0π/6=2π[−32+38+1−13]∫03dρ ρ4=2π355[23−338]=81⏞34π[45−9320]≈5.24\begin{aligned} I &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \\ &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^3\mathrm{d}\rho\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/6} =2\pi \left[-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8}+1-\frac{1}{3}\right] \int_0^3\mathrm{d}\rho\ \rho^4 \\ &=2\pi \frac{3^5}{5}\left[\frac{2}{3} - \frac{3\sqrt{3}}{8}\right] =\overbrace{81}^{3^4}\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right] \approx 5.24 \end{aligned}

원본 교재 풀이(영어)

(a) Recall that in spherical coordinates

x=ρ sin⁡φ cos⁡θy=ρ sin⁡φ sin⁡θz=ρ cos⁡φ\begin{aligned} x=\rho\,\sin\varphi\,\cos\theta\qquad y=\rho\,\sin\varphi\,\sin\theta\qquad z=\rho\,\cos\varphi \end{aligned}

so that

  • x2+y2+z2≤9x^2+y^2+z^2\le 9 is ρ≤3\rho\le 3, and
  • 3x2+3y2≤z\sqrt{3x^2+3y^2}\le z is 3ρsin⁡φ≤ρcos⁡φ\sqrt{3}\rho\sin\varphi\le \rho\cos\varphi, or tan⁡φ≤13\tan\varphi\le \frac{1}{\sqrt{3}} or φ≤π6\varphi\le\frac{\pi}{6}, and
  • the integrand x2+y2=ρ2sin⁡2φx^2+y^2=\rho^2\sin^2\varphi, and
  • dV\mathrm{d}V is ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi

So

T={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤3, 0≤θ≤2π, 0≤φ≤π/6 }T =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 3,\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{6}\ \big\}

and,

I=∭T(x2+y2) dV=∫03dρ∫02πdθ∫0π/6dφ ρ2sin⁡φ ρ2sin⁡2φ⏞x2+y2=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ\begin{aligned} I &=\iiint _T \big(x^2+y^2\big)\ \mathrm{d}V =\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \overbrace{\rho^2\sin^2\varphi}^{x^2+y^2} \\ &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \end{aligned}

(b) In cylindrical coordinates

  • x2+y2+z2≤9x^2+y^2+z^2\le 9 is r2+z2≤9r^2+z^2\le 9 and
  • 3x2+3y2≤z\sqrt{3 x^2+3 y^2}\le z is 3 r≤z\sqrt{3}\,r\le z and
  • the integand x2+y2=r2x^2+y^2=r^2, and
  • dV\mathrm{d}V is r dr dθ dzr\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z

Observe that r2+z2=9r^2+z^2= 9 and 3 r=z\sqrt{3}\,r= z intersect when r2+3r2=9r^2+3r^2=9 so that r=32r=\frac{3}{2} and z=332z=\frac{3\sqrt{3}}{2}. Here is a sketch of the y=0y=0 cross–section of TT.

figure

So

T={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤32, 0≤θ≤2π, 3 r≤z≤9−r2 }T =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le \tfrac{3}{2},\ 0\le\theta\le2\pi,\ \sqrt{3}\,r\le z\le \sqrt{9-r^2}\ \big\}

and

I=∭V(x2+y2) dV=∫03/2dr∫02πdθ∫3 r9−r2dz r(r2)⏞x2+y2=∫03/2dr∫02πdθ∫3 r9−r2dz  r3\begin{aligned} I &=\iiint _V (x^2+y^2)\ \mathrm{d}V =\int_0^{3/2}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\mathrm{d}z\ r\overbrace{(r^2)}^{x^2+y^2} =\int_0^{3/2}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}\mathrm{d}z\ \ r^3 \end{aligned}

(c) We'll use the spherical coordinate form.

I=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡3φ=∫03dρ∫02πdθ∫0π/6dφ ρ4sin⁡φ[1−cos⁡2φ]=2π∫03dρ ρ4[−cos⁡φ+cos⁡3φ3]0π/6=2π[−32+38+1−13]∫03dρ ρ4=2π355[23−338]=81⏞34π[45−9320]≈5.24\begin{aligned} I &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin^3\varphi \\ &=\int_0^{3}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/6}\mathrm{d}\varphi\ \rho^4\sin\varphi \big[1-\cos^2\varphi\big] \\ &=2\pi \int_0^3\mathrm{d}\rho\ \rho^4 \left[-\cos\varphi +\frac{\cos^3\varphi}{3}\right]_0^{\pi/6} =2\pi \left[-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8}+1-\frac{1}{3}\right] \int_0^3\mathrm{d}\rho\ \rho^4 \\ &=2\pi \frac{3^5}{5}\left[\frac{2}{3} - \frac{3\sqrt{3}}{8}\right] =\overbrace{81}^{3^4}\pi \left[\frac{4}{5} - \frac{9\sqrt{3}}{20}\right] \approx 5.24 \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 8번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
685.CLP-3 · 3.7절 · 24번대학 수준우리말 번역

EE 를 “아이스크림콘” x2+y2+z2≤1x^2 + y^2 + z^2 \le 1, x2+y2≤z2x^2 + y^2 \le z^2, z≥0z \ge 0 이라 하자. 다음을 생각하세요.

J=∭Ex2+y2+z2 dVJ =\iiint _E \sqrt{x^2+y^2+z^2}\ \mathrm{d}V

(a) JJ 를 원통좌표계에서 적분 구간을 포함한 반복적분으로 나타내세요. (b) JJ 를 구면좌표계에서 적분 구간을 포함한 반복적분으로 나타내세요. (c) JJ 의 값을 구하세요.

영어 원문 보기

Let EE be the “ice cream cone” x2+y2+z2≤1x^2 + y^2 + z^2 \le 1, x2+y2≤z2x^2 + y^2 \le z^2 , z≥0z \ge 0. Consider

J=∭Ex2+y2+z2 dVJ =\iiint _E \sqrt{x^2+y^2+z^2}\ \mathrm{d}V

(a) Write JJ as an iterated integral, with limits, in cylindrical coordinates. (b) Write JJ as an iterated integral, with limits, in spherical coordinates. (c) Evaluate JJ.

정답·풀이 보기

(a) 원통좌표계에서

  • x2+y2+z2≤1x^2+y^2+z^2\le 1 은 r2+z2≤1r^2+z^2\le 1 이고
  • x2+y2≤z2x^2+y^2\le z^2 은 r2≤z2r^2\le z^2 이고
  • dV\mathrm{d}V 는 r dr dθ dzr\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z 이다.

r2+z2=1r^2+z^2= 1 과 r2=z2r^2= z^2 은 r2=z2=12r^2=z^2=\frac{1}{2} 일 때 만난다. 다음은 EE 의 y=0y=0 단면의 그림이다.

그림

따라서

E={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1/2, 0≤θ≤2π, r≤z≤1−r2 }E =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le{1}/{\sqrt{2}},\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{1-r^2}\ \big\}

이고

J=∭Ex2+y2+z2 dV=∫01/2dr∫02πdθ∫r1−r2dz rr2+z2\begin{aligned} J &=\iiint _E \sqrt{x^2+y^2+z^2}\ \mathrm{d}V =\int_0^{1/\sqrt{2}}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{1-r^2}}\mathrm{d}z\ r\sqrt{r^2+z^2} \end{aligned}

이다.

(b) 구면좌표계에서

  • x2+y2+z2≤1x^2+y^2+z^2\le 1 은 ρ≤1\rho\le 1 이고
  • x2+y2≤z2x^2+y^2\le z^2 은 ρ2sin⁡2φ≤ρ2cos⁡2φ\rho^2\sin^2\varphi\le \rho^2\cos^2\varphi, 즉 tan⁡φ≤1\tan\varphi\le 1 또는 φ≤π4\varphi\le\frac{\pi}{4} 이고,
  • dV\mathrm{d}V 는 ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi 이다.

따라서

E={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤1, 0≤θ≤2π, 0≤φ≤π/4 }E =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 1,\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{4}\ \big\}

이고, 피적분함수 x2+y2+z2=ρ\sqrt{x^2+y^2+z^2}=\rho 이므로

J=∭Ex2+y2+z2 dV=∫01dρ∫02πdθ∫0π/4dφ ρ2sin⁡φ ρ=∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ\begin{aligned} J &=\iiint _E \sqrt{x^2+y^2+z^2}\ \mathrm{d}V =\int_0^{1}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \rho \\ &=\int_0^{1}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^3\sin\varphi \end{aligned}

이다.

(c) 구면좌표계로 나타낸 식을 이용하여 값을 구한다.

J=∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ=2π∫01dρ ρ3[−cos⁡φ]0π/4=2π 14 [1−12]=π2 [1−12]\begin{aligned} J &=\int_0^{1}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^3\sin\varphi \\ &=2\pi \int_0^{1}\mathrm{d}\rho\ \rho^3 \Big[-\cos\varphi\Big]_0^{\pi/4} =2\pi\ \frac{1}{4}\ \left[1-\frac{1}{\sqrt{2}}\right] \\ &=\frac{\pi}{2}\ \left[1-\frac{1}{\sqrt{2}}\right] \end{aligned}

원본 교재 풀이(영어)

(a) In cylindrical coordinates

  • x2+y2+z2≤1x^2+y^2+z^2\le 1 is r2+z2≤1r^2+z^2\le 1 and
  • x2+y2≤z2x^2+y^2\le z^2 is r2≤z2r^2\le z^2 and
  • dV\mathrm{d}V is r dr dθ dzr\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z

Observe that r2+z2=1r^2+z^2= 1 and r2=z2r^2= z^2 intersect when r2=z2=12r^2=z^2=\frac{1}{2}. Here is a sketch of the y=0y=0 cross–section of EE.

figure

So

E={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1/2, 0≤θ≤2π, r≤z≤1−r2 }E =\big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le{1}/{\sqrt{2}},\ 0\le\theta\le2\pi,\ r\le z\le \sqrt{1-r^2}\ \big\}

and

J=∭Ex2+y2+z2 dV=∫01/2dr∫02πdθ∫r1−r2dz rr2+z2\begin{aligned} J &=\iiint _E \sqrt{x^2+y^2+z^2}\ \mathrm{d}V =\int_0^{1/\sqrt{2}}\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta \int_r^{\sqrt{1-r^2}}\mathrm{d}z\ r\sqrt{r^2+z^2} \end{aligned}

(b) In spherical coordinates

  • x2+y2+z2≤1x^2+y^2+z^2\le 1 is ρ≤1\rho\le 1 and
  • x2+y2≤z2x^2+y^2\le z^2 is ρ2sin⁡2φ≤ρ2cos⁡2φ\rho^2\sin^2\varphi\le \rho^2\cos^2\varphi, or tan⁡φ≤1\tan\varphi\le 1 or φ≤π4\varphi\le\frac{\pi}{4}, and
  • dV\mathrm{d}V is ρ2sin⁡φ dρ dθ dφ\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi

So

E={ (ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ 0≤ρ≤1, 0≤θ≤2π, 0≤φ≤π/4 }E =\big\{ \ (\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \big|\ 0\le\rho\le 1,\ 0\le\theta\le2\pi,\ 0\le \varphi\le {\pi}/{4}\ \big\}

and, since the integrand x2+y2+z2=ρ\sqrt{x^2+y^2+z^2}=\rho,

J=∭Ex2+y2+z2 dV=∫01dρ∫02πdθ∫0π/4dφ ρ2sin⁡φ ρ=∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ\begin{aligned} J &=\iiint _E \sqrt{x^2+y^2+z^2}\ \mathrm{d}V =\int_0^{1}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^2\sin\varphi\ \rho \\ &=\int_0^{1}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^3\sin\varphi \end{aligned}

(c) We'll use the spherical coordinate form to evaluate

J=∫01dρ∫02πdθ∫0π/4dφ ρ3sin⁡φ=2π∫01dρ ρ3[−cos⁡φ]0π/4=2π 14 [1−12]=π2 [1−12]\begin{aligned} J &=\int_0^{1}\mathrm{d}\rho\int_0^{2\pi}\mathrm{d}\theta \int_0^{\pi/4}\mathrm{d}\varphi\ \rho^3\sin\varphi \\ &=2\pi \int_0^{1}\mathrm{d}\rho\ \rho^3 \Big[-\cos\varphi\Big]_0^{\pi/4} =2\pi\ \frac{1}{4}\ \left[1-\frac{1}{\sqrt{2}}\right] \\ &=\frac{\pi}{2}\ \left[1-\frac{1}{\sqrt{2}}\right] \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 26번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
686.CLP-3 · 3.7절 · 26번대학 수준우리말 번역

(a) 구면좌표계에서 방정식 ρ=8sin⁡(φ)\rho = 8 \sin(\varphi) 로 정의된 곡면의 내부에 있는 입체의 부피를 구하세요.

다음을 이용해도 됩니다.

∫sin⁡4(φ)=132(12φ−8sin⁡(2φ)+sin⁡(4φ))+C\int \sin^4(\varphi) =\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big) +C (b) 이 입체의 그림을 그리거나 어떤 모양인지 설명하세요.

영어 원문 보기

(a) Find the volume of the solid inside the surface defined by the equation ρ=8sin⁡(φ)\rho = 8 \sin(\varphi) in spherical coordinates.

You may use that

∫sin⁡4(φ)=132(12φ−8sin⁡(2φ)+sin⁡(4φ))+C\int \sin^4(\varphi) =\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big) +C (b) Sketch this solid or describe what it looks like.

정답·풀이 보기

(a) 구면좌표계에서 φ\varphi 는 00 (양의 zz축)부터 π\pi (음의 zz축)까지 변하고, θ\theta 는 00 부터 2π2\pi 까지 변하며 (θ\theta 는 일반적인 극좌표 또는 원기둥좌표이다) dV=ρ2 sin⁡φ dρ dθ dφ\mathrm{d}V = \rho^2\ \sin\varphi\ \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi 임을 떠올리자. 따라서

부피=∫0πdφ∫02πdθ∫08sin⁡φdρ ρ2 sin⁡φ=∫0πdφ∫02πdθ (8sin⁡φ)33sin⁡φ=2(83) π3∫0πdφ sin⁡4φ=2(83) π3[132(12φ−8sin⁡(2φ)+sin⁡(4φ))]0π=2(83) π3 12 π32=128π2\begin{aligned} \text{부피} & = \int_0^\pi \mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{8\sin\varphi}\mathrm{d}\rho\ \rho^2\ \sin\varphi \\ & = \int_0^\pi \mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta\ \frac{(8\sin\varphi)^3}{3} \sin\varphi \\ & = \frac{2(8^3)\,\pi}{3} \int_0^\pi \mathrm{d}\varphi\ \sin^4\varphi \\ &= \frac{2(8^3)\,\pi}{3} \left[\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big)\right]_0^\pi \\ &=\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32} =128\pi^2 \end{aligned}

(b) 00 과 π\pi 사이의 임의의 φ\varphi 를 고정하자. ρ=8sin⁡φ\rho=8\sin\varphi 일 때, θ\theta 가 00 부터 2π2\pi 까지 변하면

(x,y,z)=(ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ)=(8sin⁡2φcos⁡θ , 8sin⁡2φsin⁡θ , 8sin⁡φcos⁡φ)=(Rcos⁡θ , Rsin⁡θ , Z)여기서 R=8sin⁡2φ, Z=8sin⁡φcos⁡φ\begin{aligned} (x,y,z) &= \big(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi\big) \\ &= \big(8\sin^2\varphi\cos\theta\,,\,8\sin^2\varphi\sin\theta\,,\, 8\sin\varphi\cos\varphi\big) \\ &= \big(R\cos\theta\,,\,R\sin\theta\,,\,Z\big)\qquad \text{여기서 }R=8\sin^2\varphi,\ Z= 8\sin\varphi\cos\varphi \end{aligned}

는 평면 z=Z=8sin⁡φcos⁡φz=Z=8\sin\varphi\cos\varphi 위에 있고 (0,0,Z=8sin⁡φcos⁡φ)\big(0,0,Z=8\sin\varphi\cos\varphi\big) 을 중심으로 하는 반지름 R=8sin⁡2φR=8\sin^2\varphi 인 원을 그린다. 따라서 이 곡면은 원들을 차곡차곡 쌓아 올린 것이며, 회전면이다. 이 곡면은 다음과 같이 그릴 수 있다.

  • 먼저 곡면의 θ=0\theta=0 단면 (곡면 중 xzxz평면의 오른쪽 절반에 있는 부분)을 그린다.
  • 그다음 그 결과를 zz축을 중심으로 회전시킨다.

곡면의 θ=0\theta=0 부분은

{ (x,y,z) ∣ x=8sin⁡2φ, y=0, z=8sin⁡φcos⁡φ, 0≤φ≤π }={ (x,y,z) ∣ x=4−4cos⁡(2φ), y=0, z=4sin⁡(2φ), 0≤φ≤π }\begin{aligned} &\big\{ \ (x,y,z)\ \big|\ x=8\sin^2\varphi,\ y=0,\ z=8\sin\varphi\cos\varphi,\ 0\le\varphi\le \pi\ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ x=4-4\cos(2\varphi),\ y=0,\ z=4\sin(2\varphi),\ 0\le\varphi\le \pi\ \big\} \end{aligned}

이다. 이것은 xzxz평면 (즉 y=0y=0) 위에 있고 (4,0,0)(4,0,0) 을 중심으로 하는 반지름 44 인 원이다! 아래 왼쪽 그림은 이 원의 위쪽 절반을 그린 것이다. 이 원을 zz축을 중심으로 회전시키면 토러스(도넛 모양)가 되는데, 가운데 구멍이 한 점으로 줄어든 모양이다. 아래 오른쪽 그림은 토러스 중 제1팔분공간에 있는 부분을 그린 것이다.

그림

그림

원본 교재 풀이(영어)

(a) Recall that, in spherical coordinates, φ\varphi runs from 00 (that's the positive zz–axis) to π\pi (that's the negative zz–axis), θ\theta runs from 00 to 2π2\pi (θ\theta is the regular polar or cylindrical coordinate) and dV=ρ2 sin⁡φ dρ dθ dφ\mathrm{d}V = \rho^2\ \sin\varphi\ \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi. So

Volume=∫0πdφ∫02πdθ∫08sin⁡φdρ ρ2 sin⁡φ=∫0πdφ∫02πdθ (8sin⁡φ)33sin⁡φ=2(83) π3∫0πdφ sin⁡4φ=2(83) π3[132(12φ−8sin⁡(2φ)+sin⁡(4φ))]0π=2(83) π3 12 π32=128π2\begin{aligned} \text{Volume} & = \int_0^\pi \mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta \int_0^{8\sin\varphi}\mathrm{d}\rho\ \rho^2\ \sin\varphi \\ & = \int_0^\pi \mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta\ \frac{(8\sin\varphi)^3}{3} \sin\varphi \\ & = \frac{2(8^3)\,\pi}{3} \int_0^\pi \mathrm{d}\varphi\ \sin^4\varphi \\ &= \frac{2(8^3)\,\pi}{3} \left[\frac{1}{32}\big(12\varphi -8\sin(2\varphi) +\sin(4\varphi)\big)\right]_0^\pi \\ &=\frac{2(8^3)\,\pi}{3}\ \frac{12\,\pi}{32} =128\pi^2 \end{aligned}

(b) Fix any φ\varphi between 00 and π\pi. If ρ=8sin⁡φ\rho=8\sin\varphi, then as θ\theta runs from 00 to 2π2\pi,

(x,y,z)=(ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ)=(8sin⁡2φcos⁡θ , 8sin⁡2φsin⁡θ , 8sin⁡φcos⁡φ)=(Rcos⁡θ , Rsin⁡θ , Z)with R=8sin⁡2φ, Z=8sin⁡φcos⁡φ\begin{aligned} (x,y,z) &= \big(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi\big) \\ &= \big(8\sin^2\varphi\cos\theta\,,\,8\sin^2\varphi\sin\theta\,,\, 8\sin\varphi\cos\varphi\big) \\ &= \big(R\cos\theta\,,\,R\sin\theta\,,\,Z\big)\qquad \text{with }R=8\sin^2\varphi,\ Z= 8\sin\varphi\cos\varphi \end{aligned}

sweeps out a circle of radius R=8sin⁡2φR=8\sin^2\varphi contained in the plane z=Z=8sin⁡φcos⁡φz=Z=8\sin\varphi\cos\varphi and centred on (0,0,Z=8sin⁡φcos⁡φ)\big(0,0,Z=8\sin\varphi\cos\varphi\big). So the surface is a bunch of circles stacked one on top of the other. It is a surface of revolution. We can sketch it by

  • first sketching the θ=0\theta=0 section of the surface (that's the part of the surface in the right half of the xzxz–plane)
  • and then rotate the result about the zz–axis.

The θ=0\theta=0 part of the surface is

{ (x,y,z) ∣ x=8sin⁡2φ, y=0, z=8sin⁡φcos⁡φ, 0≤φ≤π }={ (x,y,z) ∣ x=4−4cos⁡(2φ), y=0, z=4sin⁡(2φ), 0≤φ≤π }\begin{aligned} &\big\{ \ (x,y,z)\ \big|\ x=8\sin^2\varphi,\ y=0,\ z=8\sin\varphi\cos\varphi,\ 0\le\varphi\le \pi\ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ x=4-4\cos(2\varphi),\ y=0,\ z=4\sin(2\varphi),\ 0\le\varphi\le \pi\ \big\} \end{aligned}

It's a circle of radius 44, contained in the xzxz–plane (i.e. y=0y=0) and centred on (4,0,0)(4,0,0)! The figure on the left below is a sketch of the top half of the circle. When we rotate the circle about the zz–axis we get a torus (a donut) but with the hole in the centre shrunk to a point. The figure on the right below is a sketch of the part of the torus in the first octant.

figure

figure

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 27번
# 원기둥·구면좌표 삼중적분그래프·표 해석도전자동 분류
687.CLP-3 · 3.7절 · 27번대학 수준우리말 번역

EE 를 다음과 같은 입체라 하자.

0≤z≤x2+y2,x2+y2≤1,0 \le z \le \sqrt{x^2 + y^2},\qquad x^2 + y^2 \le 1,

그리고 다음 적분을 생각하자.

I=∭Ezx2+y2+z2 dV.I = \iiint _E z \sqrt{x^2 + y^2 + z^2}\ \mathrm{d}V.

(a) 적분 II 를 원기둥좌표계로 나타내세요. (b) 적분 II 를 구면좌표계로 나타내세요. (c) 두 형태 중 하나를 사용하여 적분 II 의 값을 구하세요.

영어 원문 보기

Let EE be the solid

0≤z≤x2+y2,x2+y2≤1,0 \le z \le \sqrt{x^2 + y^2},\qquad x^2 + y^2 \le 1,

and consider the integral

I=∭Ezx2+y2+z2 dV.I = \iiint _E z \sqrt{x^2 + y^2 + z^2}\ \mathrm{d}V.

(a) Write the integral II in cylindrical coordinates. (b) Write the integral II in spherical coordinates. (c) Evaluate the integral II using either form.

정답·풀이 보기

(a) 원기둥좌표계에서 0≤z≤x2+y20\le z\le \sqrt{x^2+y^2} 은 0≤z≤r0\le z\le r 이 되고, x2+y2≤1x^2+y^2\le 1 은 0≤r≤10\le r\le 1 이 된다. 따라서

E={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1, 0≤θ≤2π, 0≤z≤r }E = \big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le 1,\ 0\le\theta\le 2\pi, \ 0\le z\le r\ \big\}

이고, dV=r dr dθ dz\mathrm{d}V = r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z 이므로

I=∭Ezx2+y2+z2 dV=∫01dr∫02πdθ∫0rdz r z r2+z2⏟x2+y2+z2\begin{aligned} I &= \iiint _E z \sqrt{x^2 + y^2 + z^2}\ \mathrm{d}V = \int_0^1\mathrm{d}r \int_0^{2\pi}\mathrm{d}\theta \int_0^r\mathrm{d}z\ r\ z\ \sqrt{\underbrace{r^2+z^2}_{x^2+y^2+z^2}} \end{aligned}

(b) 다음은 EE 를 θ\theta 가 상수인 평면으로 자른 단면의 그림이다.

그림

구면좌표 φ\varphi 는 zz축과 반지름 벡터 사이의 각임을 기억하자. 따라서 구면좌표계에서 z=rz=r (이는 zz축과 π4\frac{\pi}{4} 의 각을 이룬다)은 φ=π4\varphi=\frac{\pi}{4} 가 되고, 평면 z=0z=0, 즉 xyxy평면은 φ=π2\varphi=\frac{\pi}{2} 가 되며, r=1r=1 은 ρsin⁡φ=1\rho\sin\varphi = 1 이 된다. 따라서

E={(ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ π4≤φ≤π2, 0≤θ≤2π, 0≤ρ≤1sin⁡φ }E = \left\{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \left|\ \frac{\pi}{4}\le \varphi\le \frac{\pi}{2}, \ 0\le\theta\le 2\pi, \ 0\le \rho\le \frac{1}{\sin\varphi}\ \right.\right\}

이고, dV=ρ2sin⁡φ dρ dθ dφ\mathrm{d}V = \rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi 이므로

I=∭Ezx2+y2+z2 dV=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ2sin⁡φ ρcos⁡φ⏞z ρ=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φ\begin{aligned} I &= \iiint _E z \sqrt{x^2 + y^2 + z^2}\ \mathrm{d}V = \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{1/\sin\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z}\ \rho \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{1/\sin\varphi}\mathrm{d}\rho\ \rho^4\sin\varphi \ \cos\varphi \end{aligned}

(c) 구면좌표계 형태를 사용하여 적분하자.

I=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φ=∫π/4π/2dφ∫02πdθ 15sin⁡5φsin⁡φ cos⁡φ=2π5∫π/4π/2dφ cos⁡φsin⁡4φ=2π5∫1/21duu4단, u=sin⁡φ, du=cos⁡φ dφ=2π5[u−3−3]1/21=2(22−1)π15\begin{aligned} I &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{1/\sin\varphi}\mathrm{d}\rho\ \rho^4\sin\varphi \ \cos\varphi \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta\ \frac{1}{5\sin^5\varphi}\sin\varphi \ \cos\varphi \\ &= \frac{2\pi}{5}\int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\ \frac{\cos\varphi}{\sin^4\varphi} \\ &= \frac{2\pi}{5}\int_{1/\sqrt{2}}^{1} \frac{\mathrm{d}u}{u^4}\qquad\text{단, } u=\sin\varphi,\ \mathrm{d}u=\cos\varphi\,\mathrm{d}\varphi \\ &= \frac{2\pi}{5} \left[\frac{u^{-3}}{-3}\right]_{1/\sqrt{2}}^{1} \\ &= \frac{2(2\sqrt{2}-1)\pi}{15} \end{aligned}

원본 교재 풀이(영어)

(a) In cylindrical coordinates 0≤z≤x2+y20\le z\le \sqrt{x^2+y^2} becomes 0≤z≤r0\le z\le r, and x2+y2≤1x^2+y^2\le 1 becomes 0≤r≤10\le r\le 1. So

E={ (rcos⁡θ , rsin⁡θ , z) ∣ 0≤r≤1, 0≤θ≤2π, 0≤z≤r }E = \big\{ \ (r\cos\theta\,,\,r\sin\theta\,,\,z)\ \big|\ 0\le r\le 1,\ 0\le\theta\le 2\pi, \ 0\le z\le r\ \big\}

and, since dV=r dr dθ dz\mathrm{d}V = r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z,

I=∭Ezx2+y2+z2 dV=∫01dr∫02πdθ∫0rdz r z r2+z2⏟x2+y2+z2\begin{aligned} I &= \iiint _E z \sqrt{x^2 + y^2 + z^2}\ \mathrm{d}V = \int_0^1\mathrm{d}r \int_0^{2\pi}\mathrm{d}\theta \int_0^r\mathrm{d}z\ r\ z\ \sqrt{\underbrace{r^2+z^2}_{x^2+y^2+z^2}} \end{aligned}

(b) Here is a sketch of a constant θ\theta section of EE.

figure

Recall that the spherical coordinate φ\varphi is the angle between the zz–axis and the radius vector. So, in spherical coordinates z=rz=r (which makes an angle π4\frac{\pi}{4} with the zz axis) becomes φ=π4\varphi=\frac{\pi}{4}, and the plane z=0z=0, i.e. the xyxy–plane, becomes φ=π2\varphi=\frac{\pi}{2}, and r=1r=1 becomes ρsin⁡φ=1\rho\sin\varphi = 1. So

E={(ρsin⁡φcos⁡θ , ρsin⁡φsin⁡θ , ρcos⁡φ) ∣ π4≤φ≤π2, 0≤θ≤2π, 0≤ρ≤1sin⁡φ }E = \left\{(\rho\sin\varphi\cos\theta\,,\,\rho\sin\varphi\sin\theta\,,\, \rho\cos\varphi)\ \left|\ \frac{\pi}{4}\le \varphi\le \frac{\pi}{2}, \ 0\le\theta\le 2\pi, \ 0\le \rho\le \frac{1}{\sin\varphi}\ \right.\right\}

and, since dV=ρ2sin⁡φ dρ dθ dφ\mathrm{d}V = \rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi,

I=∭Ezx2+y2+z2 dV=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ2sin⁡φ ρcos⁡φ⏞z ρ=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φ\begin{aligned} I &= \iiint _E z \sqrt{x^2 + y^2 + z^2}\ \mathrm{d}V = \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{1/\sin\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z}\ \rho \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{1/\sin\varphi}\mathrm{d}\rho\ \rho^4\sin\varphi \ \cos\varphi \end{aligned}

(c) We'll integrate using the spherical coordinate version.

I=∫π/4π/2dφ∫02πdθ∫01/sin⁡φdρ ρ4sin⁡φ cos⁡φ=∫π/4π/2dφ∫02πdθ 15sin⁡5φsin⁡φ cos⁡φ=2π5∫π/4π/2dφ cos⁡φsin⁡4φ=2π5∫1/21duu4with u=sin⁡φ, du=cos⁡φ dφ=2π5[u−3−3]1/21=2(22−1)π15\begin{aligned} I &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{1/\sin\varphi}\mathrm{d}\rho\ \rho^4\sin\varphi \ \cos\varphi \\ &= \int_{\pi/4}^{\pi/2}\mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta\ \frac{1}{5\sin^5\varphi}\sin\varphi \ \cos\varphi \\ &= \frac{2\pi}{5}\int_{\pi/4}^{\pi/2}\mathrm{d}\varphi\ \frac{\cos\varphi}{\sin^4\varphi} \\ &= \frac{2\pi}{5}\int_{1/\sqrt{2}}^{1} \frac{\mathrm{d}u}{u^4}\qquad\text{with } u=\sin\varphi,\ \mathrm{d}u=\cos\varphi\,\mathrm{d}\varphi \\ &= \frac{2\pi}{5} \left[\frac{u^{-3}}{-3}\right]_{1/\sqrt{2}}^{1} \\ &= \frac{2(2\sqrt{2}-1)\pi}{15} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 20번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
688.CLP-3 · 3.7절 · 28번대학 수준우리말 번역

다음 반복적분을 생각하세요.

I=∫−a0∫−a2−x20∫0a2−x2−y2(x2+y2+z2)2014 dz dy dxI =\int_{-a}^0\int_{-\sqrt{a^2-x^2}}^0 \int_0^{\sqrt{a^2-x^2-y^2}} \big(x^2+y^2+z^2\big)^{2014}\ \mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x

여기서 aa 는 양의 상수입니다.

(a) II 를 원기둥좌표계에서의 반복적분으로 나타내세요. (b) II 를 구면좌표계에서의 반복적분으로 나타내세요. (c) 원하는 방법을 사용하여 II 의 값을 구하세요.

영어 원문 보기

Consider the iterated integral

I=∫−a0∫−a2−x20∫0a2−x2−y2(x2+y2+z2)2014 dz dy dxI =\int_{-a}^0\int_{-\sqrt{a^2-x^2}}^0 \int_0^{\sqrt{a^2-x^2-y^2}} \big(x^2+y^2+z^2\big)^{2014}\ \mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x

where aa is a positive constant.

(a) Write II as an iterated integral in cylindrical coordinates. (b) Write II as an iterated integral in spherical coordinates. (c) Evaluate I using whatever method you prefer.

정답·풀이 보기

핵심은 적분 영역이 어떤 모양인지 알아내는 것이다.

  • 가장 바깥쪽 적분은 xx 가 −a-a 에서 00 까지 움직인다는 뜻이다.
  • 가운데 적분은 그 범위의 각 xx 에 대하여 yy 가 −a2−x2-\sqrt{a^2-x^2} 에서 00 까지 움직인다는 뜻이다. y=−a2−x2y=-\sqrt{a^2-x^2} 을 더 익숙한 꼴인 x2+y2=a2x^2+y^2=a^2, y≤0y\le 0 으로 고쳐 쓸 수 있다. 따라서 (x,y)(x,y) 는 원점을 중심으로 하는 반지름 aa 인 원판 중 제3사분면에 있는 부분 위를 움직인다.

그림

  • 마지막으로 가장 안쪽 적분은 사분원판 안의 각 (x,y)(x,y) 에 대하여 zz 가 00 에서 a2−x2−y2\sqrt{a^2-x^2-y^2} 까지 움직인다는 뜻이다. 여기서도 z=a2−x2−y2z=\sqrt{a^2-x^2-y^2} 을 더 익숙한 꼴인 x2+y2+z2=a2x^2+y^2+z^2=a^2, z≥0z\ge 0 으로 고쳐 쓸 수 있다.

따라서 적분 영역은 원점을 중심으로 하는 반지름 aa 인 구의 내부 중 팔분공간 x≤0x\le 0, y≤0y\le 0, z≥0z\ge 0 에 놓인 부분이다.

V={ (x,y,z) ∣ −a≤x≤0, −a2−x2≤y≤0, 0≤z≤a2−x2−y2 }={ (x,y,z) ∣ x2+y2+z2≤a2, x≤0, y≤0, z≥0 }\begin{aligned} V&=\big\{ \ (x,y,z)\ \big|\ -a\le x\le 0,\ -\sqrt{a^2-x^2}\le y\le 0,\ 0\le z\le\sqrt{a^2-x^2-y^2} \ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ x^2+y^2+z^2\le a^2,\ x\le 0,\ y\le 0,\ z\ge 0 \ \big\} \end{aligned}

그림

(a) VV 에서 (x,y)(x,y) 는 제3사분면에 제한되며, 이는 원기둥좌표계에서 π≤θ≤3π2\pi\le\theta\le\frac{3\pi}{2} 이다. 따라서 원기둥좌표계에서

V={(rcos⁡θ,rsin⁡θ,z) ∣ r2+z2≤a2, π≤θ≤3π2, z≥0}={(rcos⁡θ,rsin⁡θ,z) ∣ 0≤z≤a, π≤θ≤3π2, 0≤r≤a2−z2}\begin{aligned} V&=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ r^2+z^2\le a^2,\ \pi\le\theta\le\frac{3\pi}{2},\ z\ge 0 \Big\} \\ &=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ 0\le z\le a,\ \pi\le\theta\le\frac{3\pi}{2},\ 0\le r\le\sqrt{a^2-z^2} \Big\} \end{aligned}

이고

I=∭V(x2+y2+z2)2014 dV=∭V(r2+z2)2014 r dr dθ dz=∫0adz∫π3π/2dθ∫0a2−z2dr r(r2+z2)2014\begin{aligned} I&= \iiint _V \big(x^2+y^2+z^2\big)^{2014}\ \mathrm{d}V = \iiint _V \big(r^2+z^2\big)^{2014}\ r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z \\ &=\int_0^a \mathrm{d}z\int_{\pi}^{3\pi/2}\mathrm{d}\theta \int_0^{\sqrt{a^2-z^2}}\mathrm{d}r\ r\big(r^2+z^2\big)^{2014} \end{aligned}

(b) 구면좌표 φ\varphi 는 00 (반지름 벡터가 양의 zz축 방향일 때)에서 π/2{\pi}/{2} (반지름 벡터가 xyxy평면 위에 있을 때)까지 움직이므로

I=∭V(x2+y2+z2)2014 dV=∭Vρ2×2014 ρ2sin⁡φ dρ dθ dφ=∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ\begin{aligned} I&= \iiint _V \big(x^2+y^2+z^2\big)^{2014}\ \mathrm{d}V = \iiint _V \rho^{2\times 2014}\ \rho^2\sin\varphi\ \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi \\ &=\int_0^{\pi/2} \mathrm{d}\varphi\int_{\pi}^{3\pi/2}\mathrm{d}\theta \int_0^a\mathrm{d}\rho\ \rho^{4030}\sin\varphi \end{aligned}

(c) 구면좌표로 나타낸 식을 이용하면

I=∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ=a40314031∫0π/2dφ∫π3π/2dθ sin⁡φ=a4031π8062∫0π/2dφ sin⁡φ=a4031π8062\begin{aligned} I&=\int_0^{\pi/2} \mathrm{d}\varphi\int_{\pi}^{3\pi/2}\mathrm{d}\theta \int_0^a\mathrm{d}\rho\ \rho^{4030}\sin\varphi \\ &=\frac{a^{4031}}{4031} \int_0^{\pi/2} \mathrm{d}\varphi\int_{\pi}^{3\pi/2}\mathrm{d}\theta \ \sin\varphi \\ &=\frac{a^{4031}\pi}{8062} \int_0^{\pi/2} \mathrm{d}\varphi\ \sin\varphi \\ &=\frac{a^{4031}\pi}{8062} \end{aligned}

원본 교재 풀이(영어)

The main step is to figure out what the domain of integration looks like.

  • The outside integral says that xx runs from −a-a to 00.
  • The middle integrals says that, for each xx in that range, yy runs from −a2−x2-\sqrt{a^2-x^2} to 00. We can rewrite y=−a2−x2y=-\sqrt{a^2-x^2} in the more familiar form x2+y2=a2x^2+y^2=a^2, y≤0y\le 0. So (x,y)(x,y) runs over the third quadrant part of the disk of radius aa, centred on the origin.

figure

  • Finally, the inside integral says that, for each (x,y)(x,y) in the quarter disk, zz runs from 00 to a2−x2−y2\sqrt{a^2-x^2-y^2} . We can also rewrite z=a2−x2−y2z=\sqrt{a^2-x^2-y^2} in the more familiar form x2+y2+z2=a2x^2+y^2+z^2=a^2, z≥0z\ge 0.

So the domain of integration is the part of the interior of the sphere of radius aa, centred on the origin, that lies in the octant x≤0x\le 0, y≤0y\le 0, z≥0z\ge 0.

V={ (x,y,z) ∣ −a≤x≤0, −a2−x2≤y≤0, 0≤z≤a2−x2−y2 }={ (x,y,z) ∣ x2+y2+z2≤a2, x≤0, y≤0, z≥0 }\begin{aligned} V&=\big\{ \ (x,y,z)\ \big|\ -a\le x\le 0,\ -\sqrt{a^2-x^2}\le y\le 0,\ 0\le z\le\sqrt{a^2-x^2-y^2} \ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ x^2+y^2+z^2\le a^2,\ x\le 0,\ y\le 0,\ z\ge 0 \ \big\} \end{aligned}

figure

(a) Note that, in VV, (x,y)(x,y) is restricted to the third quadrant, which in cylindrical coordinates is π≤θ≤3π2\pi\le\theta\le\frac{3\pi}{2}. So, in cylindrical coordinates,

V={(rcos⁡θ,rsin⁡θ,z) ∣ r2+z2≤a2, π≤θ≤3π2, z≥0}={(rcos⁡θ,rsin⁡θ,z) ∣ 0≤z≤a, π≤θ≤3π2, 0≤r≤a2−z2}\begin{aligned} V&=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ r^2+z^2\le a^2,\ \pi\le\theta\le\frac{3\pi}{2},\ z\ge 0 \Big\} \\ &=\Big\{(r\cos\theta,r\sin\theta,z)\ \Big|\ 0\le z\le a,\ \pi\le\theta\le\frac{3\pi}{2},\ 0\le r\le\sqrt{a^2-z^2} \Big\} \end{aligned}

and

I=∭V(x2+y2+z2)2014 dV=∭V(r2+z2)2014 r dr dθ dz=∫0adz∫π3π/2dθ∫0a2−z2dr r(r2+z2)2014\begin{aligned} I&= \iiint _V \big(x^2+y^2+z^2\big)^{2014}\ \mathrm{d}V = \iiint _V \big(r^2+z^2\big)^{2014}\ r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z \\ &=\int_0^a \mathrm{d}z\int_{\pi}^{3\pi/2}\mathrm{d}\theta \int_0^{\sqrt{a^2-z^2}}\mathrm{d}r\ r\big(r^2+z^2\big)^{2014} \end{aligned}

(b) The spherical coordinate φ\varphi runs from 00 (when the radius vector is along the positive zz–axis) to π/2{\pi}/{2} (when the radius vector lies in the xyxy–plane) so that

I=∭V(x2+y2+z2)2014 dV=∭Vρ2×2014 ρ2sin⁡φ dρ dθ dφ=∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ\begin{aligned} I&= \iiint _V \big(x^2+y^2+z^2\big)^{2014}\ \mathrm{d}V = \iiint _V \rho^{2\times 2014}\ \rho^2\sin\varphi\ \mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi \\ &=\int_0^{\pi/2} \mathrm{d}\varphi\int_{\pi}^{3\pi/2}\mathrm{d}\theta \int_0^a\mathrm{d}\rho\ \rho^{4030}\sin\varphi \end{aligned}

(c) Using the spherical coordinate version

I=∫0π/2dφ∫π3π/2dθ∫0adρ ρ4030sin⁡φ=a40314031∫0π/2dφ∫π3π/2dθ sin⁡φ=a4031π8062∫0π/2dφ sin⁡φ=a4031π8062\begin{aligned} I&=\int_0^{\pi/2} \mathrm{d}\varphi\int_{\pi}^{3\pi/2}\mathrm{d}\theta \int_0^a\mathrm{d}\rho\ \rho^{4030}\sin\varphi \\ &=\frac{a^{4031}}{4031} \int_0^{\pi/2} \mathrm{d}\varphi\int_{\pi}^{3\pi/2}\mathrm{d}\theta \ \sin\varphi \\ &=\frac{a^{4031}\pi}{8062} \int_0^{\pi/2} \mathrm{d}\varphi\ \sin\varphi \\ &=\frac{a^{4031}\pi}{8062} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 29번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
689.CLP-3 · 3.7절 · 29번대학 수준우리말 번역

입체 EE 는 아래로는 포물면 z=x2+y2z = x^2 + y^2 으로, 위로는 원뿔 z=x2+y2z=\sqrt{x^2+y^2} 으로 유계이다. 다음과 같이 두자.

I=∭Ez(x2+y2+z2) dVI = \iiint _E z\big(x^2+y^2+z^2\big)\ \mathrm{d}V

(a) II 를 원기둥좌표로 나타내세요. 값을 구하지는 마세요. (b) II 를 구면좌표로 나타내세요. 값을 구하지는 마세요. (c) II 의 값을 구하세요.

영어 원문 보기

The solid EE is bounded below by the paraboloid z=x2+y2z = x^2 + y^2 and above by the cone z=x2+y2z=\sqrt{x^2+y^2}. Let

I=∭Ez(x2+y2+z2) dVI = \iiint _E z\big(x^2+y^2+z^2\big)\ \mathrm{d}V

(a) Write II in terms of cylindrical coordinates. Do not evaluate. (b) Write II in terms of spherical coordinates. Do not evaluate. (c) Calculate II.

정답·풀이 보기

(a) 원기둥좌표에서 포물면은 z=r2z=r^2 이고 원뿔은 z=rz=r 이다. 두 곡면은 r2=rr^2=r 일 때 만난다. 즉, r=0r=0 일 때와 r=1r=1 일 때이다. 따라서 원기둥좌표로 나타내면

I=∫01dr r∫02πdθ∫r2rdz z(r2+z2)\begin{aligned} I = \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta \int_{r^2}^r\mathrm{d}z\ z(r^2+z^2) \end{aligned}

(b) 구면좌표에서 포물면은

ρcos⁡φ=ρ2sin⁡2φ또는ρ=cos⁡φsin⁡2φ\rho\cos\varphi=\rho^2\sin^2\varphi\qquad\text{또는}\qquad \rho=\frac{\cos\varphi}{\sin^2\varphi}

이고 원뿔은

ρcos⁡φ=ρsin⁡φ또는tan⁡φ=1또는φ=π4\rho\cos\varphi=\rho\sin\varphi\quad\text{또는}\quad \tan\varphi=1\quad\text{또는}\quad \varphi=\frac{\pi}{4}

이다. 아래 그림은 θ\theta 가 상수인 EE 의 단면을 보여 준다. 이 그림을 보면 φ\varphi 는 π4\frac{\pi}{4} (즉, 원뿔)에서 π2\frac{\pi}{2} (즉, xyxy평면)까지 변한다.

그림

따라서 구면좌표로 나타내면

I=∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ2sin⁡φ ρcos⁡φ⏞zρ2⏞x2+y2+z2=∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ5sin⁡φcos⁡φ\begin{aligned} I &= \int_{\pi/4}^{\pi/2} \mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{\cos\varphi/\sin^2\varphi} \mathrm{d}\rho\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z} \overbrace{\rho^2}^{x^2+y^2+z^2} \\ &= \int_{\pi/4}^{\pi/2} \mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{\cos\varphi/\sin^2\varphi} \mathrm{d}\rho\ \rho^5\sin\varphi \cos\varphi \end{aligned}

(c) 원기둥좌표 적분이 더 쉬워 보인다.

I=∫01dr r∫02πdθ∫r2rdz z(r2+z2)=∫01dr r∫02πdθ [r2z22+z44]r2r=2π∫01dr r[r2r22+r44−r2r42−r84]=2π[112+124−116−140]=π2[13+16−14−110]=3π40\begin{aligned} I &= \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta \int_{r^2}^r\mathrm{d}z\ z(r^2+z^2) \\ &= \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta\ \left[r^2\frac{z^2}{2}+\frac{z^4}{4}\right]_{r^2}^r \\ &= 2\pi \int_0^1 \mathrm{d}r\ r \left[r^2\frac{r^2}{2}+\frac{r^4}{4} -r^2\frac{r^4}{2}-\frac{r^8}{4}\right] \\ &=2\pi \left[\frac{1}{12}+\frac{1}{24} -\frac{1}{16}-\frac{1}{40}\right] =\frac{\pi}{2} \left[\frac{1}{3}+\frac{1}{6} -\frac{1}{4}-\frac{1}{10}\right] \\ &=\frac{3\pi}{40} \end{aligned}

원본 교재 풀이(영어)

(a) In cylindrical coordinates, the paraboloid is z=r2z=r^2 and the cone is z=rz=r. The two meet when r2=rr^2=r. That is, when r=0r=0 and when r=1r=1. So, in cylindrical coordinates

I=∫01dr r∫02πdθ∫r2rdz z(r2+z2)\begin{aligned} I = \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta \int_{r^2}^r\mathrm{d}z\ z(r^2+z^2) \end{aligned}

(b) In spherical coordinates, the paraboloid is

ρcos⁡φ=ρ2sin⁡2φorρ=cos⁡φsin⁡2φ\rho\cos\varphi=\rho^2\sin^2\varphi\qquad\text{or}\qquad \rho=\frac{\cos\varphi}{\sin^2\varphi}

and the cone is

ρcos⁡φ=ρsin⁡φortan⁡φ=1orφ=π4\rho\cos\varphi=\rho\sin\varphi\quad\text{or}\quad \tan\varphi=1\quad\text{or}\quad \varphi=\frac{\pi}{4}

The figure below shows a constant θ\theta cross–section of EE. Looking at that figure, we see that φ\varphi runs from π4\frac{\pi}{4} (i.e. the cone) to π2\frac{\pi}{2} (i.e. the xyxy–plane).

figure

So, is spherical coordinates,

I=∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ2sin⁡φ ρcos⁡φ⏞zρ2⏞x2+y2+z2=∫π/4π/2dφ∫02πdθ∫0cos⁡φ/sin⁡2φdρ ρ5sin⁡φcos⁡φ\begin{aligned} I &= \int_{\pi/4}^{\pi/2} \mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{\cos\varphi/\sin^2\varphi} \mathrm{d}\rho\ \rho^2\sin\varphi \ \overbrace{\rho\cos\varphi}^{z} \overbrace{\rho^2}^{x^2+y^2+z^2} \\ &= \int_{\pi/4}^{\pi/2} \mathrm{d}\varphi \int_0^{2\pi}\mathrm{d}\theta \int_0^{\cos\varphi/\sin^2\varphi} \mathrm{d}\rho\ \rho^5\sin\varphi \cos\varphi \end{aligned}

(c) The cylindrical coordinates integral looks easier.

I=∫01dr r∫02πdθ∫r2rdz z(r2+z2)=∫01dr r∫02πdθ [r2z22+z44]r2r=2π∫01dr r[r2r22+r44−r2r42−r84]=2π[112+124−116−140]=π2[13+16−14−110]=3π40\begin{aligned} I &= \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta \int_{r^2}^r\mathrm{d}z\ z(r^2+z^2) \\ &= \int_0^1 \mathrm{d}r\ r \int_0^{2\pi}\mathrm{d}\theta\ \left[r^2\frac{z^2}{2}+\frac{z^4}{4}\right]_{r^2}^r \\ &= 2\pi \int_0^1 \mathrm{d}r\ r \left[r^2\frac{r^2}{2}+\frac{r^4}{4} -r^2\frac{r^4}{2}-\frac{r^8}{4}\right] \\ &=2\pi \left[\frac{1}{12}+\frac{1}{24} -\frac{1}{16}-\frac{1}{40}\right] =\frac{\pi}{2} \left[\frac{1}{3}+\frac{1}{6} -\frac{1}{4}-\frac{1}{10}\right] \\ &=\frac{3\pi}{40} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 19번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
690.CLP-3 · 3.7절 · 30번대학 수준우리말 번역

SS 를 제1팔분공간(x,y,z≥0x,y,z\ge 0)에 있는 영역으로서 원뿔 z=x2+y2z=\sqrt{x^2+y^2} 의 위쪽, 구 (z−1)2+x2+y2=1(z-1)^2 +x^2+y^2=1 의 아래쪽에 놓인 부분이라 하자. VV 를 그 부피라 하자.

(a) VV 를 원기둥좌표계에서의 삼중적분으로 나타내세요. (b) VV 를 구면좌표계에서의 삼중적분으로 나타내세요. (c) 위의 두 적분 중 하나를 이용하여 VV 를 계산하세요.

영어 원문 보기

Let SS be the region on the first octant (so that x,y,z≥0x,y,z\ge 0) which lies above the cone z=x2+y2z=\sqrt{x^2+y^2} and below the sphere (z−1)2+x2+y2=1(z-1)^2 +x^2+y^2=1. Let VV be its volume.

(a) Express VV as a triple integral in cylindrical coordinates. (b) Express VV as an triple integral in spherical coordinates. (c) Calculate VV using either of the integrals above.

정답·풀이 보기

구 x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1 과 원뿔 z=x2+y2z=\sqrt{x^2+y^2} 은 모두 zz축을 중심으로 한 회전에 대하여 변하지 않는다. 구 x2+y2+(z−1)2=1x^2 + y^2 + (z-1)^2 = 1 과 원뿔 z=x2+y2z = \sqrt{x^2 + y^2} 은 z=x2+y2z=\sqrt{x^2+y^2} 일 때 만나므로 x2+y2=z2x^2+y^2=z^2 이고,

x2+y2+(z−1)2=z2+(z−1)2=1  ⟺  2z2−2z=0  ⟺  2z(z−1)=0  ⟺  z=0,1\begin{aligned} x^2 + y^2 + (z-1)^2 = z^2+(z-1)^2=1 &\iff 2z^2-2z=0 \iff 2z(z-1)=0 \\ &\iff z=0,1 \end{aligned}

따라서 두 곡면은 원 z=1z=1, x2+y2=1x^2+y^2=1 에서 만나며

S={ (x,y,z) ∣ x,y≥0, x2+y2≤1, x2+y2≤z≤1+1−x2−y2 }S = \big\{ \ (x,y,z)\ \big|\ x,y\ge 0,\ x^2+y^2\le 1,\ \sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2}\ \big\}

이다.

다음은 SS 의 y=0y=0 단면의 개형이다.

그림

(a) 원기둥좌표계에서

  • 조건 x,y≥0x,y\ge 0 은 0≤θ≤π/20\le\theta\le{\pi}/{2} 이고,
  • 조건 x2+y2≤1x^2+y^2\le 1 은 r≤1r\le 1 이며,
  • 조건 x2+y2≤z≤1+1−x2−y2\sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2} 은 r≤z≤1+1−r2r\le z\le 1+\sqrt{1-r^2} 이고,
  • dV=r dr dθ dz\mathrm{d}V = r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z 이다.

따라서

V=∭SdV=∫01dr∫0π/2dθ∫r1+1−r2dz r\begin{aligned} V = \iiint _S\mathrm{d}V = \int_0^1\mathrm{d}r \int_0^{\pi/2}\mathrm{d}\theta\int_r^{1+\sqrt{1-r^2}}\mathrm{d}z \ r \end{aligned}

이다.

(b) 구면좌표계에서

  • 원뿔 z=x2+y2z=\sqrt{x^2+y^2} 은 다음과 같이 된다.

ρ cos⁡φ=ρ2sin⁡2φ cos⁡2θ+ρ2sin⁡2φ sin⁡2θ=ρsin⁡φ  ⟺  tan⁡φ=1  ⟺  φ=π4\begin{aligned} \rho\,\cos\varphi =\sqrt{\rho^2\sin^2\varphi\,\cos^2\theta +\rho^2\sin^2\varphi\,\sin^2\theta} =\rho\sin\varphi \iff \tan\varphi=1 \iff \varphi=\frac{\pi}{4} \end{aligned}

  • 따라서 SS 위에서 구면좌표 φ\varphi 는 φ=0\varphi=0 (zz축의 양의 부분)에서 φ=π/4\varphi={\pi}/{4} (원뿔 z=x2+y2z=\sqrt{x^2+y^2})까지 변하며, 이는 원뿔의 위쪽에 머물도록 한다.
  • 조건 x,y≥0x,y\ge 0 은 0≤θ≤π/20\le\theta\le{\pi}/{2} 이다.
  • 조건 x2+y2+(z−1)2≤1x^2+y^2+(z-1)^2\le 1 (구의 내부에 머물도록 하는 조건)은 다음과 같이 된다.

ρ2sin⁡2φ cos⁡2θ+ρ2sin⁡2φ sin⁡2θ+(ρcos⁡φ−1)2≤1  ⟺  ρ2 sin⁡2φ+ρ2cos⁡2φ−2ρcos⁡φ+1≤1  ⟺  ρ2−2ρcos⁡φ≤0  ⟺  ρ≤2cos⁡φ\begin{aligned} &\rho^2\sin^2\varphi\,\cos^2\theta + \rho^2\sin^2\varphi\,\sin^2\theta +\big(\rho\cos\varphi-1\big)^2\le 1 \\ &\iff \rho^2\,\sin^2\varphi +\rho^2\cos^2\varphi -2\rho\cos\varphi +1 \le 1 \\ &\iff \rho^2-2\rho\cos\varphi\le 0 \\ &\iff \rho\le 2 \cos\varphi \end{aligned}

  • 그리고 dV=ρ2sin⁡φ dρ dθ dφ\mathrm{d}V = \rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi 이다.

따라서

V=∭SdV=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φ\begin{aligned} V = \iiint _S\mathrm{d}V = \int_0^{\pi/4}\mathrm{d}\varphi \int_0^{\pi/2}\mathrm{d}\theta \int_0^{2\cos\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi \end{aligned}

이다.

(c) (b)의 구면좌표계 적분을 이용하여 VV 를 계산한다.

V=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φ=83∫0π/4dφ∫0π/2dθ cos⁡3φ sin⁡φ=83 π2 [−cos⁡4φ4]0π/4=π3[1−1(2)4]=π4\begin{aligned} V &= \int_0^{\pi/4}\mathrm{d}\varphi \int_0^{\pi/2}\mathrm{d}\theta \int_0^{2\cos\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi \\ &=\frac{8}{3} \int_0^{\pi/4}\mathrm{d}\varphi \int_0^{\pi/2}\mathrm{d}\theta\ \cos^3\varphi\ \sin\varphi \\ &=\frac{8}{3}\ \frac{\pi}{2}\ \left[-\frac{\cos^4\varphi}{4}\right]_0^{\pi/4} \\ &=\frac{\pi}{3} \left[1-\frac{1}{(\sqrt{2})^4}\right] \\ &=\frac{\pi}{4} \end{aligned}

원본 교재 풀이(영어)

Note that both the sphere x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1 and the cone z=x2+y2z=\sqrt{x^2+y^2} are invariant under rotations around the zz–axis. The sphere x2+y2+(z−1)2=1x^2 + y^2 + (z-1)^2 = 1 and the cone z=x2+y2z = \sqrt{x^2 + y^2} intersect when z=x2+y2z=\sqrt{x^2+y^2}, so that x2+y2=z2x^2+y^2=z^2, and

x2+y2+(z−1)2=z2+(z−1)2=1  ⟺  2z2−2z=0  ⟺  2z(z−1)=0  ⟺  z=0,1\begin{aligned} x^2 + y^2 + (z-1)^2 = z^2+(z-1)^2=1 &\iff 2z^2-2z=0 \iff 2z(z-1)=0 \\ &\iff z=0,1 \end{aligned}

So the surfaces intersect on the circle z=1z=1, x2+y2=1x^2+y^2=1 and

S={ (x,y,z) ∣ x,y≥0, x2+y2≤1, x2+y2≤z≤1+1−x2−y2 }S = \big\{ \ (x,y,z)\ \big|\ x,y\ge 0,\ x^2+y^2\le 1,\ \sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2}\ \big\}

Here is a sketch of the y=0y=0 cross section of S.

figure

(a) In cylindrical coordinates

  • the condition x,y≥0x,y\ge 0 is 0≤θ≤π/20\le\theta\le{\pi}/{2},
  • the condition x2+y2≤1x^2+y^2\le 1 is r≤1r\le 1, and
  • the conditions x2+y2≤z≤1+1−x2−y2\sqrt{x^2+y^2}\le z\le 1+\sqrt{1-x^2-y^2} are r≤z≤1+1−r2r\le z\le 1+\sqrt{1-r^2}, and
  • dV=r dr dθ dz\mathrm{d}V = r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z.

So

V=∭SdV=∫01dr∫0π/2dθ∫r1+1−r2dz r\begin{aligned} V = \iiint _S\mathrm{d}V = \int_0^1\mathrm{d}r \int_0^{\pi/2}\mathrm{d}\theta\int_r^{1+\sqrt{1-r^2}}\mathrm{d}z \ r \end{aligned}

(b) In spherical coordinates,

  • the cone z=x2+y2z=\sqrt{x^2+y^2} becomes

ρ cos⁡φ=ρ2sin⁡2φ cos⁡2θ+ρ2sin⁡2φ sin⁡2θ=ρsin⁡φ  ⟺  tan⁡φ=1  ⟺  φ=π4\begin{aligned} \rho\,\cos\varphi =\sqrt{\rho^2\sin^2\varphi\,\cos^2\theta +\rho^2\sin^2\varphi\,\sin^2\theta} =\rho\sin\varphi \iff \tan\varphi=1 \iff \varphi=\frac{\pi}{4} \end{aligned}

  • so that, on SS, the spherical coordinate φ\varphi runs from φ=0\varphi=0 (the positive zz –axis) to φ=π/4\varphi={\pi}/{4} (the cone z=x2+y2z=\sqrt{x^2+y^2}), which keeps us above the cone,
  • the condition x,y≥0x,y\ge 0 is 0≤θ≤π/20\le\theta\le{\pi}/{2},
  • the condition x2+y2+(z−1)2≤1x^2+y^2+(z-1)^2\le 1, (which keeps us inside the sphere), becomes

ρ2sin⁡2φ cos⁡2θ+ρ2sin⁡2φ sin⁡2θ+(ρcos⁡φ−1)2≤1  ⟺  ρ2 sin⁡2φ+ρ2cos⁡2φ−2ρcos⁡φ+1≤1  ⟺  ρ2−2ρcos⁡φ≤0  ⟺  ρ≤2cos⁡φ\begin{aligned} &\rho^2\sin^2\varphi\,\cos^2\theta + \rho^2\sin^2\varphi\,\sin^2\theta +\big(\rho\cos\varphi-1\big)^2\le 1 \\ &\iff \rho^2\,\sin^2\varphi +\rho^2\cos^2\varphi -2\rho\cos\varphi +1 \le 1 \\ &\iff \rho^2-2\rho\cos\varphi\le 0 \\ &\iff \rho\le 2 \cos\varphi \end{aligned}

  • and dV=ρ2sin⁡φ dρ dθ dφ\mathrm{d}V = \rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi.

So

V=∭SdV=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φ\begin{aligned} V = \iiint _S\mathrm{d}V = \int_0^{\pi/4}\mathrm{d}\varphi \int_0^{\pi/2}\mathrm{d}\theta \int_0^{2\cos\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi \end{aligned}

(c) We'll evaluate VV using the spherical coordinate integral of part (b).

V=∫0π/4dφ∫0π/2dθ∫02cos⁡φdρ ρ2sin⁡φ=83∫0π/4dφ∫0π/2dθ cos⁡3φ sin⁡φ=83 π2 [−cos⁡4φ4]0π/4=π3[1−1(2)4]=π4\begin{aligned} V &= \int_0^{\pi/4}\mathrm{d}\varphi \int_0^{\pi/2}\mathrm{d}\theta \int_0^{2\cos\varphi}\mathrm{d}\rho\ \rho^2\sin\varphi \\ &=\frac{8}{3} \int_0^{\pi/4}\mathrm{d}\varphi \int_0^{\pi/2}\mathrm{d}\theta\ \cos^3\varphi\ \sin\varphi \\ &=\frac{8}{3}\ \frac{\pi}{2}\ \left[-\frac{\cos^4\varphi}{4}\right]_0^{\pi/4} \\ &=\frac{\pi}{3} \left[1-\frac{1}{(\sqrt{2})^4}\right] \\ &=\frac{\pi}{4} \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 28번
# 원기둥·구면좌표 삼중적분계산도전자동 분류
691.CLP-3 · 3.7절 · 31번대학 수준우리말 번역

입체가 원뿔 z=3x2+3y2z=\sqrt{3x^2+3y^2} 로 아래로 유계이고 구 x2+y2+z2=9x^2+y^2+z^2=9 로 위로 유계이다. 이 입체의 밀도는 δ(x,y,z)=x2+y2\delta (x,y,z)=x^2+y^2 이다.

(a) 이 입체의 질량 mm 을 원기둥좌표계에서의 삼중적분으로 나타내세요. (b) 이 입체의 질량 mm 을 구면좌표계에서의 삼중적분으로 나타내세요. (c) mm 의 값을 구하세요.

영어 원문 보기

A solid is bounded below by the cone z=3x2+3y2z=\sqrt{3x^2+3y^2} and above by the sphere x2+y2+z2=9x^2+y^2+z^2=9. It has density δ(x,y,z)=x2+y2\delta (x,y,z)=x^2+y^2.

(a) Express the mass mm of the solid as a triple integral in cylindrical coordinates. (b) Express the mass mm of the solid as a triple integral in spherical coordinates. (c) Evaluate mm.

정답·풀이 보기

(a) 원기둥좌표계에서 밀도는 δ=x2+y2=r2\delta =x^2+y^2=r^2 이고, 입체의 아랫면은 z=3x2+3y2=3 rz=\sqrt{3x^2+3y^2}=\sqrt{3}\,r 이고 입체의 윗면은 z=9−x2−y2=9−r2z=\sqrt{9-x^2-y^2}=\sqrt{9-r^2} 이다. 윗면과 아랫면은 다음 조건에서 만난다.

3 r=9−r2  ⟺  3r2=9−r2  ⟺  4r2=9  ⟺  r=32\sqrt{3}\,r=\sqrt{9-r^2}\iff 3r^2=9-r^2\iff 4r^2=9\iff r=\frac{3}{2}

질량은 다음과 같다.

m=∫02πdθ∫03/2dr r∫3 r9−r2dz r2⏞δ\begin{aligned} m=\int_0^{2\pi}\mathrm{d}\theta\int_0^{3/2}\mathrm{d}r\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ \overbrace{r^2}^{\delta } \end{aligned}

그림

(b) 구면좌표계에서 밀도는 δ=x2+y2=ρ2sin⁡2φ\delta =x^2+y^2=\rho^2\sin^2\varphi 이고, 입체의 아랫면은

z=3 r  ⟺  ρcos⁡φ=3 ρsin⁡φ  ⟺  tan⁡φ=13  ⟺  φ=π6z=\sqrt{3}\,r\iff \rho\cos\varphi=\sqrt{3}\,\rho\sin\varphi\iff \tan\varphi=\frac{1}{\sqrt{3}}\iff\varphi=\frac{\pi}{6}

에 있고, 입체의 윗면은 x2+y2+z2=ρ2=9x^2+y^2+z^2=\rho^2=9 에 있다. 질량은 다음과 같다.

m=∫02πdθ∫0π/6dφ∫03dρ (ρ2sin⁡φ)(ρ2sin⁡2φ)⏞δm=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/6}\mathrm{d}\varphi\int_0^3\mathrm{d}\rho\ \big(\rho^2\sin\varphi\big) \overbrace{\big(\rho^2\sin^2\varphi\big)}^{\delta }

(c) 풀이 1: (b)의 적분에서 s=cos⁡φs=\cos\varphi, ds=−sin⁡φ dφ\mathrm{d}s=-\sin\varphi\ \mathrm{d}\varphi 로 변수를 바꾸면,

m=∫02πdθ∫0π/6dφ∫03dρ ρ4sin⁡φ(1−cos⁡2φ)=355∫02πdθ∫0π/6dφ sin⁡φ(1−cos⁡2φ)=−355∫02πdθ∫13/2ds (1−s2)=−355∫02πdθ [s−s33]13/2=2π355[1−13−32+38]=2π355[23−338]\begin{aligned} m&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/6}\mathrm{d}\varphi \int_0^3\mathrm{d}\rho\ \rho^4\sin\varphi \big(1-\cos^2\varphi\big) \\ &=\frac{3^5}{5}\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/6}\mathrm{d}\varphi\ \sin\varphi \big(1-\cos^2\varphi\big) \\ &=-\frac{3^5}{5}\int_0^{2\pi}\mathrm{d}\theta\int_1^{\sqrt{3}/2}\mathrm{d}s\ (1-s^2) \\ &=-\frac{3^5}{5}\int_0^{2\pi}\mathrm{d}\theta\ \left[s-\frac{s^3}{3}\right]_1^{\sqrt{3}/2} \\ &=2\pi \frac{3^5}{5} \left[1-\frac{1}{3}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{8}\right] \\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{aligned}

(c) 풀이 2: 다른 풀이로, (a)의 적분을 계산할 수도 있다.

m=∫02πdθ∫03/2dr r∫3 r9−r2dz r2=∫02πdθ∫03/2dr r3(9−r2−3 r)=2π∫03/2dr r3(9−r2−3 r)\begin{aligned} m&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{3/2}\mathrm{d}r\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ r^2\\ &=\int_0^{2\pi}\mathrm{d}\theta\int_0^{3/2}\mathrm{d}r\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big)\\ &=2\pi\int_0^{3/2}\mathrm{d}r\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big) \end{aligned}

둘째 항은

−2π∫03/2dr 3 r4=−2π3 r55∣03/2=−2π3 355×25-2\pi\int_0^{3/2}\mathrm{d}r\ \sqrt{3}\,r^4=-2\pi\sqrt{3}\,\frac{r^5}{5}\bigg|_0^{3/2} =-2\pi\sqrt{3}\,\frac{3^5}{5\times 2^5}

이다. 첫째 항은 s=9−r2s=9-r^2, ds=−2r dr\mathrm{d}s=-2r\,\mathrm{d}r 로 치환한다.

2π∫03/2dr r39−r2=2π∫927/4ds−2⏞r dr(9−s)⏞r2s=−π[6s3/2−25s5/2]927/4=−π[3534−2×34−372453+2355]\begin{aligned} 2\pi\int_0^{3/2}\mathrm{d}r\ r^3\sqrt{9-r^2} &=2\pi\int_9^{27/4}\overbrace{\frac{\mathrm{d}s}{-2}}^{r\,\mathrm{d}r} \overbrace{(9-s)}^{r^2} \sqrt{s} =-\pi \left[6s^{3/2}-\frac{2}{5}s^{5/2}\right]_9^{27/4}\\ &=-\pi \left[\frac{3^5\sqrt{3}}{4}-2\times 3^4 -\frac{3^7}{2^4 5}\sqrt{3}+2\frac{3^5}{5}\right] \end{aligned}

두 항을 더하면,

m=−2π355332−2π355538+2π35553+2π3559332−2π355=2π355[(53−1)−3(132+58−932)]=2π355[23−338]\begin{aligned} m&=-2\pi\frac{3^5}{5} \frac{\sqrt{3}}{32} -2\pi\frac{3^5}{5} \frac{5\sqrt{3}}{8} +2\pi\frac{3^5}{5} \frac{5}{3} +2\pi\frac{3^5}{5} \frac{9\sqrt{3}}{32} -2\pi\frac{3^5}{5} \\ &=2\pi\frac{3^5}{5}\left[\left(\frac{5}{3}-1\right) -\sqrt{3}\left(\frac{1}{32}+\frac{5}{8}-\frac{9}{32}\right)\right] \\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{aligned}

원본 교재 풀이(영어)

(a) In cylindrical coordinates, the density of is δ=x2+y2=r2\delta =x^2+y^2=r^2, the bottom of the solid is at z=3x2+3y2=3 rz=\sqrt{3x^2+3y^2}=\sqrt{3}\,r and the top of the solid is at z=9−x2−y2=9−r2z=\sqrt{9-x^2-y^2}=\sqrt{9-r^2}. The top and bottom meet when

3 r=9−r2  ⟺  3r2=9−r2  ⟺  4r2=9  ⟺  r=32\sqrt{3}\,r=\sqrt{9-r^2}\iff 3r^2=9-r^2\iff 4r^2=9\iff r=\frac{3}{2}

The mass is

m=∫02πdθ∫03/2dr r∫3 r9−r2dz r2⏞δ\begin{aligned} m=\int_0^{2\pi}\mathrm{d}\theta\int_0^{3/2}\mathrm{d}r\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ \overbrace{r^2}^{\delta } \end{aligned}

figure

(b) In spherical coordinates, the density of is δ=x2+y2=ρ2sin⁡2φ\delta =x^2+y^2=\rho^2\sin^2\varphi, the bottom of the solid is at

z=3 r  ⟺  ρcos⁡φ=3 ρsin⁡φ  ⟺  tan⁡φ=13  ⟺  φ=π6z=\sqrt{3}\,r\iff \rho\cos\varphi=\sqrt{3}\,\rho\sin\varphi\iff \tan\varphi=\frac{1}{\sqrt{3}}\iff\varphi=\frac{\pi}{6}

and the top of the solid is at x2+y2+z2=ρ2=9x^2+y^2+z^2=\rho^2=9. The mass is

m=∫02πdθ∫0π/6dφ∫03dρ (ρ2sin⁡φ)(ρ2sin⁡2φ)⏞δm=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/6}\mathrm{d}\varphi\int_0^3\mathrm{d}\rho\ \big(\rho^2\sin\varphi\big) \overbrace{\big(\rho^2\sin^2\varphi\big)}^{\delta }

(c) Solution 1: Making the change of variables s=cos⁡φs=\cos\varphi, ds=−sin⁡φ dφ\mathrm{d}s=-\sin\varphi\ \mathrm{d}\varphi, in the integral of part (b),

m=∫02πdθ∫0π/6dφ∫03dρ ρ4sin⁡φ(1−cos⁡2φ)=355∫02πdθ∫0π/6dφ sin⁡φ(1−cos⁡2φ)=−355∫02πdθ∫13/2ds (1−s2)=−355∫02πdθ [s−s33]13/2=2π355[1−13−32+38]=2π355[23−338]\begin{aligned} m&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/6}\mathrm{d}\varphi \int_0^3\mathrm{d}\rho\ \rho^4\sin\varphi \big(1-\cos^2\varphi\big) \\ &=\frac{3^5}{5}\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/6}\mathrm{d}\varphi\ \sin\varphi \big(1-\cos^2\varphi\big) \\ &=-\frac{3^5}{5}\int_0^{2\pi}\mathrm{d}\theta\int_1^{\sqrt{3}/2}\mathrm{d}s\ (1-s^2) \\ &=-\frac{3^5}{5}\int_0^{2\pi}\mathrm{d}\theta\ \left[s-\frac{s^3}{3}\right]_1^{\sqrt{3}/2} \\ &=2\pi \frac{3^5}{5} \left[1-\frac{1}{3}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{8}\right] \\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{aligned}

(c) Solution 2: As an alternate solution, we can also evaluate the integral of part (a).

m=∫02πdθ∫03/2dr r∫3 r9−r2dz r2=∫02πdθ∫03/2dr r3(9−r2−3 r)=2π∫03/2dr r3(9−r2−3 r)\begin{aligned} m&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{3/2}\mathrm{d}r\ r\int_{\sqrt{3}\,r}^{\sqrt{9-r^2}}dz\ r^2\\ &=\int_0^{2\pi}\mathrm{d}\theta\int_0^{3/2}\mathrm{d}r\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big)\\ &=2\pi\int_0^{3/2}\mathrm{d}r\ r^3\big(\sqrt{9-r^2}-\sqrt{3}\,r\big) \end{aligned}

The second term

−2π∫03/2dr 3 r4=−2π3 r55∣03/2=−2π3 355×25-2\pi\int_0^{3/2}\mathrm{d}r\ \sqrt{3}\,r^4=-2\pi\sqrt{3}\,\frac{r^5}{5}\bigg|_0^{3/2} =-2\pi\sqrt{3}\,\frac{3^5}{5\times 2^5}

For the first term, we substitute s=9−r2s=9-r^2, ds=−2r dr\mathrm{d}s=-2r\,\mathrm{d}r.

2π∫03/2dr r39−r2=2π∫927/4ds−2⏞r dr(9−s)⏞r2s=−π[6s3/2−25s5/2]927/4=−π[3534−2×34−372453+2355]\begin{aligned} 2\pi\int_0^{3/2}\mathrm{d}r\ r^3\sqrt{9-r^2} &=2\pi\int_9^{27/4}\overbrace{\frac{\mathrm{d}s}{-2}}^{r\,\mathrm{d}r} \overbrace{(9-s)}^{r^2} \sqrt{s} =-\pi \left[6s^{3/2}-\frac{2}{5}s^{5/2}\right]_9^{27/4}\\ &=-\pi \left[\frac{3^5\sqrt{3}}{4}-2\times 3^4 -\frac{3^7}{2^4 5}\sqrt{3}+2\frac{3^5}{5}\right] \end{aligned}

Adding the two terms together,

m=−2π355332−2π355538+2π35553+2π3559332−2π355=2π355[(53−1)−3(132+58−932)]=2π355[23−338]\begin{aligned} m&=-2\pi\frac{3^5}{5} \frac{\sqrt{3}}{32} -2\pi\frac{3^5}{5} \frac{5\sqrt{3}}{8} +2\pi\frac{3^5}{5} \frac{5}{3} +2\pi\frac{3^5}{5} \frac{9\sqrt{3}}{32} -2\pi\frac{3^5}{5} \\ &=2\pi\frac{3^5}{5}\left[\left(\frac{5}{3}-1\right) -\sqrt{3}\left(\frac{1}{32}+\frac{5}{8}-\frac{9}{32}\right)\right] \\ &=2\pi \frac{3^5}{5} \left[\frac{2}{3}-\frac{3\sqrt{3}}{8}\right] \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-3 · 3.7절 · 9번
# 원기둥·구면좌표 삼중적분계산도전자동 분류