극좌표 이중적분

$D=\{(r,\theta )|4\le r\le 5,\frac{\pi }{2}\le \theta \le \pi \}$

$\frac{3\pi }{4}$

$\frac{1}{3}\left(4\pi -3\sqrt{3}\right)$

$\frac{\pi }{18}$

a. $\frac{2\pi }{3};$ b. $\frac{\pi }{3};$ c. $\frac{\pi }{3}$

$\frac{256\pi }{3}\,{\text{cm}}^{3}$

$\frac{3\pi }{32}$

$4\pi$

$$\begin{aligned} \iint_R f(x,y)\ dA &= \int_0^{2\pi}\int_0^a \left(h-h\sqrt{\frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{a^2}}\right)r\ dr\ d\theta \\ &= \int_0^{2\pi}\int_0^a \left(hr-h\frac{r^2}{a}\right)\ dr\ d\theta \\ &= \int_0^{2\pi}\left.\left(\frac12hr^2-\frac{h}{3a}r^3\right)\right|_0^a d\theta \\ &= \int_0^{2\pi} \left(\frac16a^2h\right)\ d\theta\\ &= \frac13\pi a^2h. \end{aligned}$$

(a)

• The equation $x^2 + y^2 \le 2y$ is equivalent to the equation $x^2 + (y-1)^2=1$, which is the equation of the cylinder whose $z=z_0$ cross–section is the horizontal circle of radius $1$, centred on $x=0$, $y=1$, $z=z_0$. The part of this cylinder in the first octant is sketched in the figure on the left below.
• $z \le \sqrt{x^2 + y^2}$ is the equation of the cone with vertex $(0,0,0)$, and axis the positive $z$–axis, whose radius at height $z=2$ is $2$. The part of this cone in the first octant is sketched in the figure on the right below.

The region $E$ is the part of the cylinder that is above the $xy$–plane (since $z\ge 0$) outside the cone (since $z\le\sqrt{x^2+y^2}$). The part of $E$ that is in the first octant is outlined in red in the figure below. Both $x^2 + y^2 \le 2y$ and $0 \le z \le \sqrt{x^2 + y^2}$ are invariant under $x\rightarrow -x$. So $E$ is also invariant under $x\rightarrow -x$. That is, $E$ is symmetric about the $yz$–plane and contains, in the octant $x\le 0$, $y\ge 0$, $z\ge 0$, a mirror image of the first octant part of $E$.

(b) In polar coordinates, $x^2 + y^2 \le 2y$ becomes

$$r^2\le 2r\sin\theta \iff r\le 2\sin\theta$$

Let us denote by $D$ the base region of the part of $E$ in the first octant (i.e. the shaded region in the figure above). Think of $D$ as being part of the $xy$–plane. In polar coordinates, on $D$

• $\theta$ runs from $0$ to $\frac{\pi}{2}$. (Recall that $D$ is contained in the first quadrant.)
• For each $\theta$ in that range, $r$ runs from $0$ to $2\sin\theta$.

Because

• in polar coordinates $\mathrm{d}A = r\,\mathrm{d}r\,\mathrm{d}\theta$, and
• the height of $E$ above each point $(x,y)$ in $D$ is $\sqrt{x^2+y^2}$, or, in polar coordinates, $r$, and
• the volume of $E$ is twice the volume of the part of $E$ in the first octant,

we have

$$\begin{aligned} \text{Volume}(E) &= 2\int_0^{\frac{\pi}{2}}\mathrm{d}\theta \int_0^{2\sin\theta}\mathrm{d}r\ r^2 \\ &=\frac{16}{3} \int_0^{\frac{\pi}{2}}\mathrm{d}\theta\ \sin^3\theta =\frac{16}{3} \int_0^{\frac{\pi}{2}}\mathrm{d}\theta\ \sin\theta\ \big(1-\cos^2\theta\big) \\ &=-\frac{16}{3} \int_1^0\mathrm{d}u\ \big(1-u^2\big)\qquad \text{with }u=\cos\theta,\ \mathrm{d}u=-\sin\theta\ \mathrm{d}\theta \\ &=\frac{16}{3}\left[1-\frac{1}{3}\right] \\ &=\frac{32}{9} \end{aligned}$$

The paraboloid hits the $xy$–plane at $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$.

$$\begin{aligned} \text{Volume} &=\int_0^a \mathrm{d}x\int_0^{b\sqrt{1-{x^2\over a^2}}} \mathrm{d}y\ \left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right) \\ &=b\int_0^a \mathrm{d}x\int_0^{\sqrt{1-{x^2\over a^2}}} \mathrm{d}v\ \left(1-\frac{x^2}{a^2}-v^2\right) \qquad\text{ where $y=bv$}\\ \end{aligned}$$

Think of this integral as being of the form

$$\displaystyle b\int_0^a \mathrm{d}x\ g(x)\quad \text{with}\quad \displaystyle g(x)=\int_0^{\sqrt{1-{x^2\over a^2}}} \mathrm{d}v\ \left(1-\frac{x^2}{a^2}-v^2\right)$$

Then, substituting $x=au$,

$$\begin{aligned} \text{Volume} &=ab\int_0^1 \mathrm{d}u\int_0^{\sqrt{1-u^2}} \mathrm{d}v\ \big(1-u^2-v^2\big) \\ &=ab\iint _{u^2+v^2\le 1\atop u,v\ge 0} \mathrm{d}u \mathrm{d}v\ \big(1-u^2-v^2\big) \end{aligned}$$

Now switch to polar coordinates using $u=r\cos\theta$, $v=r\sin\theta$.

$$\begin{aligned} \text{Volume} &=ab\int_0^1 \mathrm{d}r\int_0^{\pi\over 2} \mathrm{d}\theta\ r\big(1-r^2\big) =ab\ \frac{\pi}{2}\ \left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1 =\frac{\pi}{8}ab \end{aligned}$$