Tutorial 11: Multiple Integrals in Polar Coordinate & Its Engineering Application

Tutorial Question

Tutorial Solution

In the following exercises, change the cartesian integral into an equivalent polar coordinate integral. Then solve the integral in polar coordinate:

a. $\displaystyle\int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} d y d x$

Solution
The region is the upper half of the unit disk $x^{2}+y^{2}\le 1$ , $y\ge 0$ . In polar coordinates ( $x=r\cos\theta$ , $y=r\sin\theta$ , $dA=r\,dr\,d\theta$ ) this is $0\le r\le 1$ , $0\le\theta\le\pi$ , so the equivalent polar integral is
$\int_{0}^{\pi} \int_{0}^{1} r\, d r\, d \theta$
Evaluating,
$\begin{aligned} \int_{0}^{\pi} \int_{0}^{1} r\, d r\, d \theta &=\int_{0}^{\pi}\left[\frac{r^{2}}{2}\right]_{0}^{1} d \theta =\int_{0}^{\pi} \frac{1}{2}\, d \theta =\left.\frac{\theta}{2}\right|_{0}^{\pi} =\frac{\pi}{2} \end{aligned}$
(Equivalently, in Cartesian form $\displaystyle\int_{-1}^{1}\sqrt{1-x^{2}}\,dx=\left[\tfrac{1}{2}\sin^{-1}x+\tfrac{1}{2}x\sqrt{1-x^{2}}\right]_{-1}^{1}=\frac{\pi}{2}$ , which is the area of a half unit disk.)

b. $\displaystyle\int_{0}^{2} \int_{0}^{x} y d y d x$

Solution
The region is the triangle $0\le x\le 2$ , $0\le y\le x$ . The line $y=x$ makes an angle $\theta=\frac{\pi}{4}$ with the $x$ -axis, and the edge $x=2$ is $r\cos\theta=2$ , i.e. $r=2\sec\theta$ . So the equivalent polar integral is
$\int_{0}^{\pi/4} \int_{0}^{2\sec\theta} (r\sin\theta)\, r\, d r\, d \theta$
Evaluating,
$\begin{aligned} \int_{0}^{\pi/4} \int_{0}^{2\sec\theta} r^{2}\sin\theta\, d r\, d \theta &=\int_{0}^{\pi/4}\sin\theta\left[\frac{r^{3}}{3}\right]_{0}^{2\sec\theta} d \theta =\frac{8}{3}\int_{0}^{\pi/4}\frac{\sin\theta}{\cos^{3}\theta}\, d \theta \\ &=\frac{8}{3}\left[\frac{1}{2}\tan^{2}\theta\right]_{0}^{\pi/4} =\frac{4}{3}(1-0)=\frac{4}{3} \end{aligned}$
(Check in Cartesian: $\displaystyle\int_0^2\frac{x^2}{2}dx=\left.\frac{x^3}{6}\right|_0^2=\frac{4}{3}$ .)

c. $\displaystyle\int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} \frac{2}{\left(1+x^{2}+y^{2}\right)^{2}} d y d x$

Solution
$\begin{aligned} I &= \int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\frac{2}{(1+x^{2}+y^{2})^{2}} dy dx \quad\text{(region: unit disk }x^{2}+y^{2}\le 1\text{)} \\ &= \int_{0}^{2\pi}\int_{0}^{1}\frac{2}{(1+r^{2})^{2}}\,r\,dr\,d\theta \qquad\text{(polar coordinates }x=r\cos\theta,\;y=r\sin\theta\text{)} \\ &= \int_{0}^{2\pi}\left(\int_{0}^{1}\frac{2r}{(1+r^{2})^{2}}\,dr\right)d\theta \end{aligned}$
Use the substitution $u=1+r^{2}$ , so $du=2r\,dr$ .
When $r=0 \Rightarrow u=1$ , and when $r=1 \Rightarrow u=2$ . Thus,
$\begin{aligned} \int_{0}^{1}\frac{2r}{(1+r^{2})^{2}}dr &=\int_{1}^{2}\frac{1}{u^{2}}du \\ &=\left[-\frac{1}{u}\right]_{1}^{2} \\ &=-\frac{1}{2}+1=\frac{1}{2}. \end{aligned}$
Therefore,
$I=\int_{0}^{2\pi}\frac{1}{2}d\theta=\frac{1}{2}\cdot 2\pi=\pi$

d. $\displaystyle\int_{0}^{\ln 2} \int_{0}^{\sqrt{(\ln 2)^{2}-y^{2}}} e \sqrt{x^{2}+y^{2}} d x d y$
Solution
$\int_{0}^{\ln 2} \int_{0}^{\sqrt{(\ln 2)^{2}-y^{2}}} e^{\sqrt{x^{2}+y^{2}}} \,dx\,dy$
The limits of integration are given by $0 \le y \le \ln 2$ and $0 \le x \le \sqrt{(\ln 2)^{2}-y^{2}}$ . The upper limit for $x$ can be rewritten as $x^2 \le (\ln 2)^2 - y^2$ , which simplifies to $x^2 + y^2 \le (\ln 2)^2$ . This inequality describes the area inside and on a circle centered at the origin with a radius of $R = \ln 2$ .
Given that $x \ge 0$ and $y \ge 0$ , the region of integration is the portion of this circle in the first quadrant.
The circular nature of the domain and the form of the integrand ( $x^2+y^2$ ) make it ideal to convert to polar coordinates. The conversion formulas are:
$\begin{align*} x &= r \cos \theta \\ y &= r \sin \theta \\ x^2 + y^2 &= r^2 \\ dx\,dy &= r\,dr\,d\theta \end{align*}$
The limits for the region in polar coordinates become:
- Radius: $0 \le r \le \ln 2$
- Angle: $0 \le \theta \le \frac{\pi}{2}$
The integrand $e^{\sqrt{x^2+y^2}}$ becomes $e^{\sqrt{r^2}} = e^r$ .
Thus, the integral in polar coordinates is:
$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\ln 2} e^r \cdot r \,dr\,d\theta$
Next, we evaluate the integral with respect to $r$ using integration by parts, where $\int u \,dv = uv - \int v \,du$ .
Let $u = r$ and $dv = e^r \,dr$ . Then $du = dr$ and $v = e^r$ .
$\begin{align*} \int_{0}^{\ln 2} r e^r \,dr &= \left[r e^r - \int e^r \,dr\right]_{0}^{\ln 2} \\ &= \left[r e^r - e^r\right]_{0}^{\ln 2} \\ &= \left[(r-1)e^r\right]_{0}^{\ln 2} \\ &= ((\ln 2 - 1)e^{\ln 2}) - ((0-1)e^0) \\ &= ((\ln 2 - 1) \cdot 2) - (-1 \cdot 1) \\ &= 2\ln 2 - 2 + 1 \\ &= 2\ln 2 - 1 = \ln(4) - 1 \end{align*}$
Finally, we substitute the result from the inner integral into the outer integral and evaluate with respect to $\theta$ :
$\begin{align*} \int_{0}^{\frac{\pi}{2}} (\ln 4 - 1) \,d\theta &= (\ln 4 - 1) \int_{0}^{\frac{\pi}{2}} d\theta \\ &= (\ln 4 - 1) \left[\theta\right]_{0}^{\frac{\pi}{2}} \\ &= (\ln 4 - 1) \left(\frac{\pi}{2} - 0\right) \\ &= \frac{\pi}{2}(\ln 4 - 1) \end{align*}$
Evaluate the $\displaystyle\iint_{R} 1-x^{2}-y^{2} d A$ using polar coordinates, where $R$ is the unit disk $x^{2}+y^{2}\le 1$ .

Solution
The region $R$ is a unit circle, so we can describe it as $R=\{(r, \theta) \mid 0 \leq r \leq 1,0 \leq \theta \leq 2 \pi\}$ .
Using the conversion $x=r \cos \theta, y=r \sin \theta$ , and $d A=r d r d \theta$ , we have
$\begin{equation*} \begin{aligned} \iint_R\left(1-x^2-y^2\right) d A & =\int_0^{2 \pi} \int_0^1\left(1-r^2\right) r d r d \theta \\ & =\int_0^{2 \pi} \int_0^1\left(r-r^3\right) d r d \theta \\ & =\int_0^{2 \pi}\left[\frac{r^2}{2}-\frac{r^4}{4}\right]_0^1 d \theta \\ & =\int_0^{2 \pi} \frac{1}{4} d \theta=\frac{\pi}{2} \end{aligned} \end{equation*}$
Find the volume below $\displaystyle z=\frac{y^{2}}{x^{2}+y^{2}}$ , above $x y$ -plane and between cylinder $x^{2}+y^{2}=1$ and $x^{2}+y^{2}=2$

Solution
$\begin{equation*} \begin{aligned} \int_{\theta=0}^{2\pi} \int_{r=1}^{\sqrt{2}} z\, r\,dr\,d\theta &= \int_{\theta=0}^{2\pi} \int_{r=1}^{\sqrt{2}}\frac{y^{2}}{x^{2}+y^{2}}r\,dr\,d\theta \\ &= \int_{\theta=0}^{2\pi} \int_{r=1}^{\sqrt{2}}\frac{y^{2}}{r^{2}}r\,dr\,d\theta \\ &= \int_{\theta=0}^{2\pi} \int_{r=1}^{\sqrt{2}}\frac{r^{2}\sin^{2}\theta}{r^{2}} r\,dr\,d\theta \\ &= \int_{\theta=0}^{2\pi} \int_{r=1}^{\sqrt{2}} \sin^{2}\theta \, r\,dr\,d\theta \\ &= \int_{\theta=0}^{2\pi} \left. \left( \frac{r^{2}}{2} \sin^{2}\theta \right) \right\vert_{r=1}^{r=\sqrt{2}} d\theta \\ &= \int_{\theta=0}^{2\pi} \frac{\sin^{2}\theta}{2} \, d\theta \\ &= \frac{1}{2} \int_{\theta=0}^{2\pi} \frac{1}{2}\left(1 - \cos 2\theta\right)\, d\theta \\ &= \frac{1}{4} \int_{\theta=0}^{2\pi} (1 - \cos 2\theta)\, d\theta \\ &= \frac{1}{4} \left( \theta - \frac{\sin 2\theta}{2} \right) \Big|_{\theta=0}^{\theta=2\pi} \\ &= \frac{1}{4}(2\pi) = \frac{\pi}{2} \end{aligned} \end{equation*}$
Find the volume between the sphere $x^{2}+y^{2}+z^{2}=1$ and the cone $z=\sqrt{x^{2}+y^{2}}$

Solution
We seek the volume inside the sphere $x^{2}+y^{2}+z^{2}=1$ and above the cone $z=\sqrt{x^{2}+y^{2}}$ .
In spherical coordinates $( \rho,\phi,\theta )$ , $x=\rho\sin\phi\cos\theta,\; y=\rho\sin\phi\sin\theta,\; z=\rho\cos\phi$ , the volume element is
$dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$ .
The cone $z=\sqrt{x^{2}+y^{2}}$ becomes $\rho\cos\phi=\rho\sin\phi\quad\Rightarrow\quad \cos\phi=\sin\phi \quad\Rightarrow\quad \phi=\frac{\pi}{4}$ .
Thus, the region is $0\le\theta\le2\pi,\quad 0\le\phi\le\frac{\pi}{4},\quad 0\le\rho\le1$ .
$\begin{equation*} \begin{aligned} V &= \int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{1} \rho^{2}\sin\phi \; d\rho\,d\phi\,d\theta\\ &= \left(\int_{0}^{2\pi} d\theta\right) \left(\int_{0}^{\pi/4}\sin\phi\,d\phi\right) \left(\int_{0}^{1}\rho^{2}\,d\rho\right)\\ &= (2\pi)\left[-\cos\phi\right]_{0}^{\pi/4}\left[\frac{\rho^{3}}{3}\right]_{0}^{1}\\ &= 2\pi\left(1-\frac{\sqrt{2}}{2}\right)\cdot\frac{1}{3}\\ &= \frac{2\pi}{3}\!\left(1-\frac{\sqrt{2}}{2}\right) = \frac{2\pi}{3}-\frac{\pi\sqrt{2}}{3} = \frac{\pi}{3}\,(2-\sqrt{2}) \end{aligned} \end{equation*}$
Volume is equal to area only if the height ( $z$ ) is equal to 1 . Find the area of $R$ where $R$ is the region bound by $r=3 \cos \theta$ .

Solution
Region bound by $r=3\cos\theta$

How to sketch the region $R$ ?
$\begin{equation*} \begin{aligned} r \cdot r &= 3\cos\theta \cdot r \\ r^{2} &= 3r\cos\theta \\ x^{2} + y^{2} &= 3r\cos\theta \\ x^{2} + y^{2} &= 3x \\ x^{2} - 3x + y^{2} &= 0 \\ \left(x - \tfrac{3}{2}\right)^{2} + y^{2} &= \left(\tfrac{3}{2}\right)^{2} = \tfrac{9}{4} \\ \end{aligned} \end{equation*}$ $x = \tfrac{3}{2}, \quad y = 0$
Volume = area when $z = 1$ . Hence,
$\begin{equation*} \begin{aligned} A &= \iint dA \\ &=2 \int_{\theta=0}^{\frac{\pi}{2}} \int_{r=0}^{r=3\cos\theta} r\,dr\,d\theta\\ &= \int_{\theta=0}^{\pi} \int_{r=0}^{r=3\cos\theta} r\,dr\,d\theta \quad \Leftarrow \small \text{if we set up like this, we will get } \cos\theta = -1 \text{ and } \cos 0 = 1. \text{ The total will be 0.} \\ &= 2 \int_{\theta=0}^{\pi/2} \left. \frac{r^{2}}{2} \right\vert_{r=0}^{r=3\cos\theta} d\theta \\ &= \int_{\theta=0}^{\pi/2} 9\cos^{2}\theta \, d\theta \\ &= \int_{\theta=0}^{\pi/2} \frac{9}{2}(1+\cos 2\theta)\, d\theta = \frac{9\pi}{4} \end{aligned} \end{equation*}$
$\iiint y d V$ , a solid is bound by $z=4-x^{2}-y^{2}$ in the first octant $(x=0, y=0, z=$ $0)$

Solution
$\begin{equation*} \begin{aligned} z &= 4 - x^2 - y^2 \\ 0 &= 4 - x^2 - y^2 \\ 2^2 &= x^2 + y^2, \quad r = 2 \end{aligned} \end{equation*}$ $\begin{equation*} \begin{aligned} \iiint y d V &= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{4-x^2-y^2} y \, dz \, r dr \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \left[ yz \right]_{z=0}^{z=4-x^2-y^2} r \, dr \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (4 - x^2 - y^2)y \, r \, dr \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (4 - r^2)r\sin\theta \, r \, dr \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \sin\theta \cdot r^2(4 - r^2) \, dr \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (4r^2\sin\theta - r^4\sin\theta) \, dr \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \left[ \frac{4r^3}{3}\sin\theta - \frac{r^5}{5}\sin\theta \right]_{r=0}^{r=2} \, d\theta \\ &= \int_{0}^{\frac{\pi}{2}} \left( \frac{32}{3}\sin\theta - \frac{32}{5}\sin\theta \right) \, d\theta \\ &= \left[ -\frac{32}{3}\cos\theta + \frac{32}{5}\cos\theta \right]_{0}^{\frac{\pi}{2}} \\ &= \left( -\frac{32}{3}\cos\frac{\pi}{2} + \frac{32}{5}\cos\frac{\pi}{2} \right) - \left( -\frac{32}{3}\cos 0 + \frac{32}{5}\cos 0 \right) \\ &= (0) - \left( -\frac{32}{3} + \frac{32}{5} \right) \\ &= \frac{32}{3} - \frac{32}{5} \\ &= \frac{160 - 96}{15} \\ &= \frac{64}{15} \end{aligned} \end{equation*}$
Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder $\left(x-2\right)^{2}+y^{2}=4$ by the planes $z=0$ and $z=-y$ .

Solution
First, sketch the integration region.
$(x-2)^2 + y^2 = 4$ is a circle, since $x^2 + y^2 = 4x \Leftrightarrow r^2 = 4r \cos(\theta)$
$r=4\cos(\theta)$
Since $0 \le z \le -y$ , the integration region is on the $y \le 0$ part of the $z = 0$ plane.
$\begin{equation*} \begin{aligned} V &= \int_{3\pi/2}^{2\pi} \int_{0}^{4\cos(\theta)} \int_{0}^{-r\sin(\theta)} r \, dz \, dr \, d\theta. \\ \\ V &= \int_{3\pi/2}^{2\pi} \int_{0}^{4\cos(\theta)} \left[-r\sin(\theta) - 0\right] r \, dr \, d\theta \\ \\ V &= - \int_{3\pi/2}^{2\pi} \left( \left[ \frac{r^3}{3} \right]_{0}^{4\cos(\theta)} \right) \sin(\theta) \, d\theta. \\ \\ V &= - \int_{3\pi/2}^{2\pi} \frac{4^3}{3} \cos^3(\theta) \sin(\theta) \, d\theta. \end{aligned} \end{equation*}$
Introduce the substitution: $u = \cos(\theta)$ , $du = -\sin(\theta) \, d\theta$ ;
$V = \frac{4^3}{3} \int_{0}^{1} u^3 \, du = \frac{4^3}{3} \left( \left. \frac{u^4}{4} \right|_{0}^{1} \right) = \frac{4^3}{3} \cdot \frac{1}{4} = \frac{16}{3}$
Consider the region $E$ inside the right circular cylinder with equation $r=2 \sin \theta$ , bounded below by the $r \theta$ -plane and bounded above by the sphere with radius 4 centered at the origin. Set up a triple integral over this region with a function $f(r, \theta, z)$ in cylindrical coordinates.

Solution
Consider the region $E$ inside the right circular cylinder with equation $r=2\sin\theta$ , bounded below by the $r\theta$ -plane and bounded above by the sphere with radius 4 centered at the origin. Set up a triple integral over this region for a function $f(r, \theta, z)$ in cylindrical coordinates.
The triple integral in cylindrical coordinates for a function $f(r, \theta, z)$ is given by:
$\iiint_E f(r, \theta, z) \, dV = \int_{\theta_{min}}^{\theta_{max}} \int_{r_{min}(\theta)}^{r_{max}(\theta)} \int_{z_{min}(r, \theta)}^{z_{max}(r, \theta)} f(r, \theta, z) \, r \, dz \, dr \, d\theta$
We need to find the bounds for $z$ , $r$ , and $\theta$ that describe the region $E$ .
The region is bounded below by the $r\theta$ -plane, which corresponds to the plane $z=0$ . The region is bounded above by the sphere of radius 4 centered at the origin. The equation of this sphere in Cartesian coordinates is $x^2 + y^2 + z^2 = 16$ . In cylindrical coordinates, since $r^2 = x^2 + y^2$ , this becomes $r^2 + z^2 = 16$ .
Solving for $z$ , we get $z = \sqrt{16 - r^2}$ (we take the positive root for the upper hemisphere). Therefore, the limits for $z$ are:
$0 \le z \le \sqrt{16 - r^2}$
The projection of the volume onto the $r\theta$ -plane (the xy-plane) is determined by the cylinder $r=2\sin\theta$ . For any given angle $\theta$ , the radius $r$ extends from the origin ( $r=0$ ) to the edge of the cylinder ( $r=2\sin\theta$ ). Therefore, the limits for $r$ are:
$0 \le r \le 2\sin\theta$
To find the limits for $\theta$ , we analyze the base of the cylinder $r = 2\sin\theta$ in the xy-plane. To ensure the radius $r$ is non-negative, we must have $\sin\theta \ge 0$ . This condition is met for angles in the first and second quadrants.
The curve starts at the origin when $\theta = 0$ (since $r=2\sin(0)=0$ ), traces a full circle, and returns to the origin when $\theta = \pi$ (since $r=2\sin(\pi)=0$ ). Therefore, the limits for $\theta$ are:
$0 \le \theta \le \pi$
Combining these limits, the triple integral is set up as follows:
$\int_{0}^{\pi} \int_{0}^{2\sin\theta} \int_{0}^{\sqrt{16-r^2}} f(r, \theta, z) \, r \, dz \, dr \, d\theta$
Find the volume of solid bound by $z=2$ and $z=\sqrt{x^{2}+y^{2}}$
Solution
We want to find the volume of the solid region $E$ bounded above by the plane $z=2$ and bounded below by the cone $z = \sqrt{x^2+y^2}$ .
The equations involve $x^2+y^2$ , and the solid has an axis of symmetry along the z-axis. This makes cylindrical coordinates $(r, \theta, z)$ the most convenient choice. The coordinate transformations are:
$\begin{align*} x &= r\cos(\theta) \\ y &= r\sin(\theta) \\ x^2 + y^2 &= r^2 \end{align*}$
The volume element in cylindrical coordinates is $dV = r \, dz \, dr \, d\theta$ . The equations of the bounding surfaces become:
- Cone: $z = \sqrt{r^2} \implies z = r$
- Plane: $z = 2$
The volume $V$ is given by the triple integral over the region E:
$V = \iiint_E dV$
We need to determine the limits of integration for $z$ , $r$ , and $\theta$ .
Limits for $z$ : The solid is bounded below by the cone ( $z=r$ ) and above by the plane ( $z=2$ ). Thus, the limits for $z$ are:
$r \le z \le 2$
Limits for $r$ and $\theta$ : The limits for $r$ and $\theta$ are determined by the projection of the solid onto the xy-plane. This projection is the region where the cone intersects the plane. We find this by setting the z-values equal:
$r=2$
This describes a circle of radius 2 centered at the origin. Therefore, the radius $r$ ranges from $0$ to $2$ , and the angle $\theta$ ranges over a full circle from $0$ to $2\pi$ .
$\begin{align*} 0 \le r \le 2 \\ 0 \le \theta \le 2\pi \end{align*}$
The Integral: Combining these limits, the volume integral is:
$V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{2} r \, dz \, dr \, d\theta$
We evaluate the integral from the inside out.
Step 1: Integrate with respect to $z$
$\int_{r}^{2} r \, dz = r \left[ z \right]_{r}^{2} = r(2 - r) = 2r - r^2$
Step 2: Integrate with respect to $r$
Substitute the result from Step 1 into the next integral:
$\int_{0}^{2} (2r - r^2) \, dr = \left[ r^2 - \frac{r^3}{3} \right]_{0}^{2} = \left( 2^2 - \frac{2^3}{3} \right) - (0) = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$
Step 3: Integrate with respect to $\theta$
Substitute the result from Step 2 into the final integral:
$\int_{0}^{2\pi} \frac{4}{3} \, d\theta = \frac{4}{3} \left[ \theta \right]_{0}^{2\pi} = \frac{4}{3} (2\pi - 0) = \frac{8\pi}{3}$
The volume of the solid bounded by the plane $z=2$ and the cone $z=\sqrt{x^2+y^2}$ is:
$V = \frac{8\pi}{3}$
Use spherical coordinates to find the volume of the region outside the sphere $\rho=2$ $\cos (\phi)$ and inside the sphere $\rho=2$ with $\phi \in[0, \pi / 2]$ .

Solution
$\begin{align*} V &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{2\cos(\phi)}^{2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta. \\ V &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \left( \frac{\rho^3}{3} \right) \Bigg|_{2\cos(\phi)}^{2} \sin(\phi) \, d\phi \, d\theta \\ V &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \left[ \frac{2^3}{3} - \frac{(2\cos(\phi))^3}{3} \right] \sin(\phi) \, d\phi \, d\theta \\ V &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \frac{1}{3} \left[ 8 - 8\cos^3(\phi) \right] \sin(\phi) \, d\phi \, d\theta \\ V &= \frac{8}{3} \int_{0}^{2\pi} \int_{0}^{\pi/2} \left[ 1 - \cos^3(\phi) \right] \sin(\phi) \, d\phi \, d\theta \\ V &= \frac{8}{3} (2\pi) \int_{0}^{\pi/2} \left[ \sin(\phi) - \cos^3(\phi)\sin(\phi) \right] \, d\phi \\ V &= \frac{16\pi}{3} \left[ \left( -\cos(\phi) \right) \Bigg|_{0}^{\pi/2} - \int_{0}^{\pi/2} \cos^3(\phi)\sin(\phi) \, d\phi \right]. \end{align*}$
Introduce the substitution for the integral: $u = \cos(\phi)$ , $du = -\sin(\phi) \, d\phi$ .
When $\phi = 0$ , $u = \cos(0) = 1$ .
When $\phi = \pi/2$ , $u = \cos(\pi/2) = 0$ .
Now substitute into the integral:
$\int_{0}^{\pi/2} \cos^3(\phi)\sin(\phi) \, d\phi = \int_{1}^{0} u^3 (-du) = - \int_{1}^{0} u^3 \, du = \int_{0}^{1} u^3 \, du.$
So, the entire expression for $V$ becomes:
$\begin{align*} V &= \frac{16\pi}{3} \left[ 1 - \int_{0}^{1} u^3 \, du \right] \\ V &= \frac{16\pi}{3} \left[ 1 - \left( \frac{u^4}{4} \right) \Bigg|_{0}^{1} \right] \\ V &= \frac{16\pi}{3} \left[ 1 - \left( \frac{1^4}{4} - \frac{0^4}{4} \right) \right] \\ V &= \frac{16\pi}{3} \left[ 1 - \frac{1}{4} \right] \\ V &= \frac{16\pi}{3} \left( \frac{3}{4} \right) \\ V &= \frac{16\pi \cdot 3}{3 \cdot 4} \\ V &= 4\pi. \end{align*}$
Given a solid bound by $z=2$ and $z=\sqrt{x^{2}+y^{2}}$ , find the mass density if the mass density is directly proportional to the square of the distance from origin.

Solution
The problem states that the mass density, let's call it $\delta(x,y,z)$ , is directly proportional to the square of the distance from the origin. The square of the distance from the origin to a point $(x,y,z)$ is $x^2+y^2+z^2$ .
Therefore, the density function is:
$\delta(x,y,z) = k(x^2+y^2+z^2)$
where $k$ is the constant of proportionality.
Given the geometry of the solid (bounded by a cone), it is best to use spherical coordinates $(\rho, \phi, \theta)$ . In spherical coordinates, the square of the distance from the origin is simply $\rho^2$ . So, the density function becomes:
$\delta(\rho, \phi, \theta) = k\rho^2$
The solid is bounded by the plane $z=2$ and the cone $z=\sqrt{x^2+y^2}$ . We must convert these boundaries into spherical coordinates.
The Cone: In cylindrical coordinates, the cone is $z=r$ . Using the transformations $z=\rho\cos(\phi)$ and $r=\rho\sin(\phi)$ , we get:
$\rho\cos(\phi) = \rho\sin(\phi) \implies \tan(\phi) = 1 \implies \phi = \frac{\pi}{4}$
This means the cone forms a constant angle with the positive z-axis. The solid is above this cone, so the angle $\phi$ ranges from the z-axis ( $\phi=0$ ) to the cone itself.
$0 \le \phi \le \frac{\pi}{4}$
The Plane: The upper bound is the plane $z=2$ . Using $z=\rho\cos(\phi)$ :
$\rho\cos(\phi) = 2 \implies \rho = \frac{2}{\cos(\phi)} \implies \rho = 2\sec(\phi)$
The distance $\rho$ starts from the origin ( $\rho=0$ ) and extends to this plane.
$0 \le \rho \le 2\sec(\phi)$
The Angle $\theta$ : Since the solid is symmetric about the z-axis, the angle $\theta$ completes a full revolution.
$0 \le \theta \le 2\pi$
The total mass $M$ is the triple integral of the density function over the volume of the solid. The volume element in spherical coordinates is $dV = \rho^2\sin(\phi) \, d\rho \, d\phi \, d\theta$ .
$M = \iiint_E \delta \, dV = \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{2\sec(\phi)} (k\rho^2) (\rho^2\sin(\phi)) \, d\rho \, d\phi \, d\theta$ $M = k \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{2\sec(\phi)} \rho^4 \sin(\phi) \, d\rho \, d\phi \, d\theta$
Step 1: Integrate with respect to $\rho$
$\int_{0}^{2\sec(\phi)} \rho^4 \sin(\phi) \, d\rho = \sin(\phi) \left[ \frac{\rho^5}{5} \right]_{0}^{2\sec(\phi)} = \sin(\phi) \frac{(2\sec(\phi))^5}{5} = \frac{32}{5} \frac{\sin(\phi)}{\cos^5(\phi)} = \frac{32}{5} \tan(\phi)\sec^4(\phi)$
Step 2: Integrate with respect to $\phi$
$\int_{0}^{\pi/4} \frac{32}{5} \tan(\phi)\sec^4(\phi) \, d\phi$
We use the substitution $u = \tan(\phi)$ , so $du = \sec^2(\phi) \, d\phi$ . We rewrite $\sec^4(\phi) = \sec^2(\phi)\sec^2(\phi) = (1+\tan^2(\phi))\sec^2(\phi) = (1+u^2)\sec^2(\phi)$ .
The limits change: when $\phi=0, u=0$ ; when $\phi=\pi/4, u=1$ .
$\begin{align*} \frac{32}{5} \int_{0}^{1} u(1+u^2) \, du &= \frac{32}{5} \int_{0}^{1} (u+u^3) \, du \\ &= \frac{32}{5} \left[ \frac{u^2}{2} + \frac{u^4}{4} \right]_{0}^{1} \\ &= \frac{32}{5} \left( \frac{1}{2} + \frac{1}{4} - 0 \right) = \frac{32}{5} \left( \frac{3}{4} \right) = \frac{24}{5} \end{align*}$
Step 3: Integrate with respect to $\theta$
$\int_{0}^{2\pi} k \left( \frac{24}{5} \right) \, d\theta = \frac{24k}{5} \left[ \theta \right]_{0}^{2\pi} = \frac{24k}{5}(2\pi) = \frac{48k\pi}{5}$
The total mass of the solid is:
$M = \frac{48k\pi}{5}$
where $k$ is the constant of proportionality.
Find the mass of $T$ , $\rho(x, y, z)=y$ , where $T$ is region bound by $y=x^{2}+z^{2}$ and $y=4$ .
Solution
We need to find the mass $M$ of a solid region $T$ . The density is given by the function $\rho(x, y, z) = y$ . The solid $T$ is bounded by the surfaces:
- A paraboloid opening along the y-axis: $y = x^2 + z^2$
- A plane perpendicular to the y-axis: $y = 4$
The formula for mass is given by the triple integral of the density function over the region $T$ :
$M = \iiint_T \rho(x, y, z) \, dV$
The solid has a circular cross-section in the xz-plane and is symmetric about the y-axis. This suggests using a modified version of cylindrical coordinates. Let's define our coordinates as follows:
$\begin{align*} x &= r\cos(\theta) \\ z &= r\sin(\theta) \\ y &= y \end{align*}$
In this system, $x^2 + z^2 = r^2$ . The volume element is $dV = r \, dy \, dr \, d\theta$ .
The bounding surfaces and the density function become:
- Paraboloid: $y = r^2$
- Plane: $y = 4$
- Density: $\rho(r, y, \theta) = y$
We set up the integral in the order $dy \, dr \, d\theta$ .
Limits for $y$ : For any point $(r, \theta)$ in the xz-plane, the vertical extent of the solid goes from the paraboloid surface up to the plane.
$r^2 \le y \le 4$
Limits for $r$ and $\theta$ : The projection of the solid onto the xz-plane is a disk. The radius of this disk is found at the intersection of the two surfaces:
$r^2 = 4 \implies r = 2$
Thus, the radius $r$ ranges from the y-axis ( $r=0$ ) to the edge of the disk ( $r=2$ ). Since the disk is complete, the angle $\theta$ ranges over a full circle.
$\begin{align*} 0 \le r \le 2 \\ 0 \le \theta \le 2\pi \end{align*}$
Now we substitute the density, volume element, and limits into the mass formula.
$M = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} (y) \cdot (r \, dy \, dr \, d\theta)$
We evaluate the integral from the inside out.
Step 1: Integrate with respect to $y$
$\int_{r^2}^{4} ry \, dy = r \left[ \frac{y^2}{2} \right]_{r^2}^{4} = r \left( \frac{4^2}{2} - \frac{(r^2)^2}{2} \right) = r \left( \frac{16}{2} - \frac{r^4}{2} \right) = 8r - \frac{1}{2}r^5$
Step 2: Integrate with respect to $r$
$\int_{0}^{2} \left( 8r - \frac{1}{2}r^5 \right) \, dr = \left[ 4r^2 - \frac{r^6}{12} \right]_{0}^{2} = \left( 4(2^2) - \frac{2^6}{12} \right) - (0) = 16 - \frac{64}{12} = 16 - \frac{16}{3} = \frac{32}{3}$
Step 3: Integrate with respect to $\theta$
$\int_{0}^{2\pi} \frac{32}{3} \, d\theta = \frac{32}{3} \left[ \theta \right]_{0}^{2\pi} = \frac{32}{3}(2\pi - 0) = \frac{64\pi}{3}$
The total mass of the solid is:
$M = \frac{64\pi}{3}$