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문제 은행 · 미적분 · 대학 미적분 · 벡터 미적분

선적분

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

CLP-4 · 1.6절 · 2번대학 수준우리말 번역

(a) 극좌표로 r=r(θ)r=r(\theta), θ1≤θ≤θ2\theta_1\le \theta\le\theta_2 와 같이 주어진 곡선 C\mathcal{C} 를 따라 적분한 ∫Cf(x,y) ds\int_\mathcal{C} f(x,y)\,ds 가

∫θ1θ2f(r(θ)cos⁡θ,r(θ)sin⁡θ)r(θ)2+(drdθ(θ))2 dθ\int_{\theta_1}^{\theta_2}f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta 임을 보이세요. (b) r=1+cos⁡θ, 0≤θ≤2πr=1+\cos\theta,\ 0\le \theta\le 2\pi 의 곡선의 길이를 구하세요. 계산을 간단히 하기 위해 다음 공식을 사용해도 됩니다.

1+cos⁡θ=2cos⁡2θ21+\cos\theta=2\cos^2\frac{\theta}{2}

영어 원문 보기

(a) Show that the integral ∫Cf(x,y) ds\int_\mathcal{C} f(x,y)\,ds along the curve C\mathcal{C} given in polar coordinates by r=r(θ)r=r(\theta), θ1≤θ≤θ2\theta_1\le \theta\le\theta_2, is

∫θ1θ2f(r(θ)cos⁡θ,r(θ)sin⁡θ)r(θ)2+(drdθ(θ))2 dθ\int_{\theta_1}^{\theta_2}f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta (b) Compute the arc length of r=1+cos⁡θ, 0≤θ≤2πr=1+\cos\theta,\ 0\le \theta\le 2\pi. You may use the formula

1+cos⁡θ=2cos⁡2θ21+\cos\theta=2\cos^2\frac{\theta}{2}

to simplify the computation.

정답·풀이 보기

(a) 곡선은 r(θ)=x(θ) ı^+y(θ) ȷ^\mathbf{r}(\theta) = x(\theta)\,\hat{\boldsymbol{\imath}}+ y(\theta)\,\hat{\boldsymbol{\jmath}} 이고 x(θ)=r(θ)cos⁡θx(\theta)=r(\theta)\cos\theta, y(θ)=r(θ)sin⁡θy(\theta)=r(\theta)\sin\theta, θ1≤θ≤θ2\theta_1\le \theta\le \theta_2 이다. 이 곡선 위에서

v(θ)=drdθ(θ)=x′(θ)ı^+y′(θ)ȷ^=[r′(θ)cos⁡θ−r(θ)sin⁡θ]ı^+[r′(θ)sin⁡θ+r(θ)cos⁡θ]ȷ^  ⟹  dsdθ(θ)=[r′(θ)cos⁡θ−r(θ)sin⁡θ]2+[r′(θ)sin⁡θ+r(θ)cos⁡θ]2=r′(θ)2+r(θ)2\begin{aligned} \mathbf{v}(\theta) =\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}(\theta)&= x'(\theta)\hat{\boldsymbol{\imath}}+ y'(\theta)\hat{\boldsymbol{\jmath}} =\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]\hat{\boldsymbol{\imath}}+ \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]\hat{\boldsymbol{\jmath}}\\ \implies \frac{\mathrm{d}s}{\mathrm{d}\theta}(\theta) &=\sqrt{\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]^2 + \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]^2}\\ &=\sqrt{r'(\theta)^2+r(\theta)^2} \end{aligned}

따라서

∫Cf(x,y) ds=∫θ1θ2 ⁣ ⁣ ⁣f(x(θ),y(θ))dsdθ dθ=∫θ1θ2 ⁣ ⁣ ⁣f(r(θ)cos⁡θ,r(θ)sin⁡θ)r(θ)2+(drdθ(θ))2 dθ\begin{aligned} \int_\mathcal{C} f(x,y)\,ds &= \int_{\theta_1}^{\theta_2}\!\!\!f\big(x(\theta), y(\theta)\big) \frac{\mathrm{d}s}{\mathrm{d}\theta}\,\mathrm{d}\theta \\ &=\int_{\theta_1}^{\theta_2}\!\!\!f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2 +\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta \end{aligned}

이다.

(b) 이 경우 f(x,y)=1f(x,y)=1, r(θ)=1+cos⁡θr(\theta)=1+\cos\theta, θ1=0\theta_1=0, θ2=2π\theta_2=2\pi 이므로

∫C ds=∫02π[1+cos⁡θ]2+[−sin⁡θ]2 dθ=∫02π2(1+cos⁡θ) dθ=∫02π4cos⁡2θ2 dθ=2∫02π∣cos⁡θ2∣ dθ=4∫0πcos⁡θ2 dθ=8sin⁡θ2∣0π=8\begin{aligned} \int_C \,ds &=\int_0^{2\pi} \sqrt{[1+\cos\theta]^2+[-\sin\theta]^2}\,\mathrm{d}\theta =\int_0^{2\pi} \sqrt{2(1+\cos\theta)}\,\mathrm{d}\theta\\ &=\int_0^{2\pi} \sqrt{4\cos^2\frac{\theta}{2}}\,\mathrm{d}\theta =2\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|\,\mathrm{d}\theta =4\int_0^{\pi} \cos\frac{\theta}{2}\ \mathrm{d}\theta =8 \sin\frac{\theta}{2}\bigg|_0^{\pi} =8 \end{aligned}

이다.

원본 교재 풀이(영어)

(a) The curve is r(θ)=x(θ) ı^+y(θ) ȷ^\mathbf{r}(\theta) = x(\theta)\,\hat{\boldsymbol{\imath}}+ y(\theta)\,\hat{\boldsymbol{\jmath}} with x(θ)=r(θ)cos⁡θx(\theta)=r(\theta)\cos\theta, y(θ)=r(θ)sin⁡θy(\theta)=r(\theta)\sin\theta and θ1≤θ≤θ2\theta_1\le \theta\le \theta_2. On this curve

v(θ)=drdθ(θ)=x′(θ)ı^+y′(θ)ȷ^=[r′(θ)cos⁡θ−r(θ)sin⁡θ]ı^+[r′(θ)sin⁡θ+r(θ)cos⁡θ]ȷ^  ⟹  dsdθ(θ)=[r′(θ)cos⁡θ−r(θ)sin⁡θ]2+[r′(θ)sin⁡θ+r(θ)cos⁡θ]2=r′(θ)2+r(θ)2\begin{aligned} \mathbf{v}(\theta) =\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}(\theta)&= x'(\theta)\hat{\boldsymbol{\imath}}+ y'(\theta)\hat{\boldsymbol{\jmath}} =\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]\hat{\boldsymbol{\imath}}+ \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]\hat{\boldsymbol{\jmath}}\\ \implies \frac{\mathrm{d}s}{\mathrm{d}\theta}(\theta) &=\sqrt{\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]^2 + \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]^2}\\ &=\sqrt{r'(\theta)^2+r(\theta)^2} \end{aligned}

Hence

∫Cf(x,y) ds=∫θ1θ2 ⁣ ⁣ ⁣f(x(θ),y(θ))dsdθ dθ=∫θ1θ2 ⁣ ⁣ ⁣f(r(θ)cos⁡θ,r(θ)sin⁡θ)r(θ)2+(drdθ(θ))2 dθ\begin{aligned} \int_\mathcal{C} f(x,y)\,ds &= \int_{\theta_1}^{\theta_2}\!\!\!f\big(x(\theta), y(\theta)\big) \frac{\mathrm{d}s}{\mathrm{d}\theta}\,\mathrm{d}\theta \\ &=\int_{\theta_1}^{\theta_2}\!\!\!f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2 +\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta \end{aligned}

(b) In this case f(x,y)=1f(x,y)=1, r(θ)=1+cos⁡θr(\theta)=1+\cos\theta, θ1=0\theta_1=0 and θ2=2π\theta_2=2\pi,

∫C ds=∫02π[1+cos⁡θ]2+[−sin⁡θ]2 dθ=∫02π2(1+cos⁡θ) dθ=∫02π4cos⁡2θ2 dθ=2∫02π∣cos⁡θ2∣ dθ=4∫0πcos⁡θ2 dθ=8sin⁡θ2∣0π=8\begin{aligned} \int_C \,ds &=\int_0^{2\pi} \sqrt{[1+\cos\theta]^2+[-\sin\theta]^2}\,\mathrm{d}\theta =\int_0^{2\pi} \sqrt{2(1+\cos\theta)}\,\mathrm{d}\theta\\ &=\int_0^{2\pi} \sqrt{4\cos^2\frac{\theta}{2}}\,\mathrm{d}\theta =2\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|\,\mathrm{d}\theta =4\int_0^{\pi} \cos\frac{\theta}{2}\ \mathrm{d}\theta =8 \sin\frac{\theta}{2}\bigg|_0^{\pi} =8 \end{aligned}

한 번 더 — 비슷한 문제CLP-4 · 1.6절 · 6번우리말 번역

다음에 대해 경로적분 ∫Cf(x,y,z) ds\int_\mathcal{C} f(x,y,z)\,\mathrm{d}s 의 값을 구하세요.

(a) f(x,y,z)=xcos⁡zf(x,y,z)=x\cos z, C:r(t)=tı^+t2ȷ^\mathcal{C}:\mathbf{r}(t)=t\hat{\boldsymbol{\imath}}+t^2\hat{\boldsymbol{\jmath}}, 0≤t≤10\le t\le 1. (b) f(x,y,z)=x+yy+zf(x,y,z)=\frac{x+y}{y+z}, C:r(t)=(t,23t3/2,t)\mathcal{C}:\mathbf{r}(t)= \big(t,\frac{2}{3}t^{3/2},t\big), 1≤t≤21\le t\le 2.

답 쓰고 채점
(a)
(b)
다 풀었으면 정답 확인

(a) 이 경우 r(t)=tı^+t2ȷ^\mathbf{r}(t)=t\hat{\boldsymbol{\imath}}+t^2\hat{\boldsymbol{\jmath}} 이므로 v(t)=drdt(t)=ı^+2tȷ^\mathbf{v}(t)=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)=\hat{\boldsymbol{\imath}}+2t\hat{\boldsymbol{\jmath}} 이고 dsdt=1+4t2\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{1+4t^2} 이다. 따라서

∫Cf(x,y,z) ds=∫01x(t)cos⁡z(t)dsdt dt=∫01t(cos⁡0)1+4t2 dt=18(1+4t2)3/23/2∣01=53/2−112\begin{aligned} \int_\mathcal{C} f(x,y,z)\,\mathrm{d}s &=\int_0^1 x(t)\cos z(t) \frac{\mathrm{d}s}{\mathrm{d}t}\,\mathrm{d}t =\int_0^1 t(\cos 0) \sqrt{1+4t^2}\,\mathrm{d}t =\frac{1}{8}\frac{ {(1+4t^2)}^{3/2}}{3/2}\bigg|_0^1\\ &=\frac{5^{3/2}-1}{12} \end{aligned}

(b) 이 경우 r(t)=(t,23t3/2,t)\mathbf{r}(t)=\big(t,\frac{2}{3}t^{3/2},t\big) 이므로 v(t)=drdt(t)=(1,t1/2,1)\mathbf{v}(t)=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)=\big(1,t^{1/2},1\big) 이고 dsdt=2+t\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{2+t} 이다. 따라서

∫Cf(x,y,z) ds=∫12x(t)+y(t)y(t)+z(t) dsdt dt=∫12t+23t3/223t3/2+t2+t dt=(2+t)3/23/2∣12=8−33/23/2\begin{aligned} \int_\mathcal{C} f(x,y,z)\,\mathrm{d}s &=\int_1^2 \frac{x(t)+y(t)}{y(t)+z(t)} \, \frac{\mathrm{d}s}{\mathrm{d}t}\,\mathrm{d}t =\int_1^2 \frac{t+{2\over 3}t^{3/2}}{{2\over3}t^{3/2}+t} \sqrt{2+t}\,\mathrm{d}t =\frac{(2+t)^{3/2}}{3/2}\bigg|_1^2 \\ &=\frac{8-3^{3/2}}{3/2} \end{aligned}

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CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)

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