Week 5

Tutorial 5: Volume Integrals

1. Find a volume bounded by the surface $𝑥 + 𝑧 ≤ 𝑦 ≤ 𝑥 + 𝑧 + 5$, $\frac{x}{2}\leq z \leq \sin x$ and, $0 ≤ 𝑥 ≤ \frac{\pi}{2}$.
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Solution
$$\begin{aligned}\iiint_VdV&=\int_{x=0}^{\frac{\pi}{2}}\int_{z=\frac{x}{2}}^{\sin x}\int_{y=x+z}^{x+z+5}dydzdx\\&=\int_{x=0}^{\frac{\pi}{2}}\int_{z=\frac{x}{2}}^{\sin x}5dzdx\\&=\int_{x=0}^{\frac{\pi}{2}}5\left(\sin x-\frac{x}{2}\right)dx\\&=5\left[-\cos x-\frac{x^2}{4}\right]_0^{\frac{\pi}{2}}\\&=\underline{\underline{5\left(1-\frac{\pi^2}{16}\right)}}\end{aligned}$$

2. Find the total mass of a spherical object with radius of $𝑟 = 3$ and with a density function of $\rho = e^{-h}$, where $ℎ$ is the radial distance from the center of the sphere.
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Solution
$$\begin{aligned}\text{Total mass}&=\iiint_V\rho dzdydx\\&=\iiint_V\rho r^2\sin\theta d\phi d\theta dr\\&=\iiint_Ve^{-r}r^2\sin \theta d\phi d\theta dr\\&=\int_{r=0}^3\int_{\theta=0}^\pi\int_{\phi=0}^{2\pi}e^{-r}r^2\sin\theta d\phi d\theta dr\\&=\int_{r=0}^3e^{-r}r^2dr\int_{\theta=0}^\pi\sin\theta d\theta\int_{\phi=0}^{2\pi}d\phi\\&=\left[-e^{-r}(r^2+2r+2)\right]_{r=0}^3\left[-\cos\theta\right]_{\theta=0}^\pi\left[\phi\right]_{\phi=0}^{2\pi}\\&=(2-17e^{-3})(2)(2\pi)\\&=\underline{\underline{4\pi(2-17e^{-3})}}\end{aligned}$$

3. Find the moment of inertia of the following hollow cylinder with constant density $\rho$ along the $z$-axis.



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Solution
$$\begin{aligned}\text{Moment of inertia}&=\iiint_V\rho(x^2+y^2)dxdydx\\&=\iiint_V\rho(x^2+y^2)rdzd\theta dr\\&=\iiint_V\rho(r^2\cos^2\theta+r^2\sin^2\theta)rdzd\theta dr\\&=\iiint_V\rho r^3 dzd\theta dr\\&=\rho\int_{\theta=0}^{2\pi}\int_{r=r_1}^{r_2}\int_{z=0}^hr^3dzd\theta dr\\&=\rho\int_{\theta=0}^{2\pi}d\theta\int_{r=r_1}^{r_2}r^3dr\int_{z=0}^hdz\\&=\rho\left[\theta\right]_{\theta=0}^{2\pi}\left[\frac{r^4}{4}\right]_{r=r_1}^{r_2}\left[z\right]_{z=0}^h\\&=\rho(2\pi)\left(\frac{r_2^4}{4}-\frac{r_1^4}{4}\right)(h)\\&=\underline{\underline{\frac{1}{2}\pi\rho h(r_2^4-r_1^4)}}\end{aligned}$$

4. Find the volume of the following paraboloid. Hint: Inside of the cone boundary is not empty, but it is used to calculate the height upper boundary of the paraboloid ($z=A$).



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Solution
Radius upper boundary:
$$\begin{aligned}z&=x^2+y^2\\&=r^2\cos^2\theta+r^2\sin^2\theta\\&=r^2\\r&=\sqrt{z}\end{aligned}$$
Height upper boundary:
$$\begin{aligned}z&=2y\\y&=2\\z&=2(2)=4\end{aligned}$$
Hence, the cylindrical coordinate boundary is: $0\le\theta\le2\pi$, $0\le r\le\sqrt{z}$, $0\le z\le4$
$$\begin{aligned}\iiint_VdV&=\int_{\theta=0}^{2\pi}\int_{z=0}^4\int_{r=0}^{\sqrt{z}}rdrdzd\theta\\&=\int_{\theta=0}^{2\pi}\int_{z=0}^4\left[\frac{r^2}{2}\right]_0^{\sqrt{z}}dzd\theta\\&=\int_{\theta=0}^{2\pi}\int_{z=0}^4\frac{z}{2}dzd\theta\\&=\int_{\theta=0}^{2\pi}\left[\frac{z^2}{4}\right]_0^4d\theta\\&=\int_{\theta=0}^{2\pi}4d\theta\\&=\underline{\underline{8\pi}}\end{aligned}$$

5. Evaluate $\iiint_E xdV$ where $𝐸$ is enclosed by $𝑧 = 0$, $𝑧 = 𝑥 + 𝑦 + 5$, $x^2+y^2=4$ and $x^2+y^2=9$.
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Solution
We will use cylindrical coordinates to easily solve this sum. Converting the given outer limits of $E$ we get:
$$z=0,\quad z=r\cos\theta+r\sin\theta+5,\quad r=2\text{ and }r=3$$
Since there is no limitation on the values of $\theta$, we assume it has the values of $0$ to $2\pi$. This is also in accordance with the diagram which is obtained by drawing these surfaces.
Also, we substitute $x=r\cos\theta$ in the argument of the integral.
Thus, the integral is,
$$\begin{aligned}&\int_0^{2\pi}\int_2^3\int_0^{r\cos\theta+r\sin\theta+5}(r\cos\theta)rdzdrd\theta\\&=\int_0^{2\pi}\int_2^3(r^2\cos\theta)(r\cos\theta+r\sin\theta+5)drd\theta\\&=\int_0^{2\pi}\left[\frac{r^4}{4}\cos^2\theta+\frac{r^4}{4}\cos\theta\sin\theta+\frac{5}{3}r^3\cos\theta\right]_{r=2}^{r=3}d\theta\\&=\int_0^{2\pi}\left[\frac{65}{8}(1+\cos 2\theta)+\frac{65}{8}\sin(2\theta)+\frac{95}{3}\cos\theta\right]d\theta\\&=\frac{65}{8}\left(2\pi+0-\frac{1}{2}+0-0+\frac{1}{2}\right)+\frac{95}{3}(0)\\&=\underline{\underline{\frac{65\pi}{4}}}\end{aligned}$$

6. Integrate the function $𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦$ over the volume enclose by the planes $𝑧 = 𝑥 + 𝑦$ and $𝑧 = 0$, and between the surfaces $𝑦 = 𝑥^2$ and $𝑥 = 𝑦^2$.
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Solution
Since $z$ is expressed as a function of $(x,y)$, we should integrate in the $z$ direction first. After this we consider the $xy$-plane.
The two curves meet at $(0,0)$ and $(1,1)$. We can integrate in $x$ or $y$ first. If we choose $y$, we can see that the region begins at $y=x^2$ and ends at $y=\sqrt{x}$, and so $x$ is between 0 and 1.
Thus, the integral is,
$$\begin{aligned}&\int_0^1\int_{x^2}^{\sqrt{x}}\int_0^{x+y}f(x,y,z)dzdydx\\&=\int_0^1\int_{x^2}^{\sqrt{x}}\int_0^{x+y}xydzdydx\\&=\int_0^1\int_{x^2}^{\sqrt{x}}xy\left(\left[z\right]_{z=0}^{z=x+y}\right)dydx\\&=\int_0^1\int_{x^2}^{\sqrt{x}}xy(x+y)dydx\\&=\int_0^1\int_{x^2}^{\sqrt{x}}(x^2y+y^2x)dydx\\&=\int_0^1\left[\frac{x^2y^2}{2}+\frac{y^3x}{3}\right]_{y=x^2}^{y=\sqrt{x}}dx\\&=\frac{1}{2}\int_0^1\left(x^2(\sqrt{x})^2-x^2(x^2)^2\right)dx+\frac{1}{3}\int_0^1\left((\sqrt{x})^3x-(x^2)^3x\right)dx\\&=\frac{1}{2}\int_0^1(x^3-x^6)dx+\frac{1}{3}\int_0^1\left(x^{\frac{5}{2}}-x^7\right)dx\\&=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}\left(\frac{2}{7}-\frac{1}{8}\right)\\&=\underline{\underline{\frac{3}{28}}}\end{aligned}$$

7. Find the volume of the tetrahedron with corners at $(0,0,0)$, $(0,3,0)$, $(2,3,0)$, and $(2,3,5)$.
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Solution
The whole problem comes down to correctly describing the region by inequalities:
$$0\le x\le2,\quad\frac{3x}{2}\le y\le3,\quad0\le z\le\frac{5x}{2}$$
The lower $y$ limit comes from the equation of the line $y = \frac{3x}{2}$ that forms one edge of the tetrahedron in the x-y plane
The upper $z$ limit comes from the equation of the plane $z = \frac{5x}{2}$ that forms the “upper” side of the tetrahedron. Now the volume is
$$\begin{aligned}\int_0^2\int_{\frac{3x}{2}}^3\int_0^{\frac{5x}{2}}dzdydx&=\int_0^2\int_{\frac{3x}{2}}^3\left.z\right|_0^{\frac{5x}{2}}dydx\\&=\int_0^3\int_{\frac{3x}{2}}^3\frac{5x}{2}dydx\\&=\int_0^2\left.\frac{5x}{2}y\right|_{\frac{3x}{2}}^3dx\\&=\int_0^2\frac{15x}{2}-\frac{15x^2}{4}dx\\&=\left.\frac{15x}{4}-\frac{15x^3}{12}\right|_0^2\\&=15-10=\underline{\underline{5}}\end{aligned}$$

8. Find the volume under $𝑧 =\sqrt{4-r^2}$ above the quarter circle inside $x^2+y^2=4$ in the first quadrant.
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Solution
$$\begin{aligned}\int_0^{\frac{\pi}{2}}\int_0^2\int_0^{\sqrt{4-r^2}}rdzdrd\theta&=\int_0^{\frac{\pi}{2}}\int_0^2\sqrt{4-r^2}rdrd\theta\\&=\int_0^{\frac{\pi}{2}}\left.-\frac{1}{3}(4-r^2)^{\frac{3}{2}}\right|_0^2d\theta\\&=\int_0^{\frac{\pi}{2}}\frac{8}{3}d\theta\\&=\underline{\underline{\frac{4\pi}{3}}}\end{aligned}$$

9. An object occupies the space inside both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$ , and has density $𝑥^2$ at $(𝑥, 𝑦, 𝑧)$. Find the total mass.
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Solution
$$\begin{aligned}\int_0^{2\pi}\int_0^1\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}r^3\cos^2(\theta)dzdrd\theta&=\int_0^{2\pi}\int_0^12\sqrt{4-r^2}r^3\cos^2(\theta)drd\theta\\&=\int_0^{2\pi}\left(\frac{128}{15}-\frac{22}{5}\sqrt{3}\right)\cos^2(\theta)d\theta\\&=\underline{\underline{\left(\frac{128}{15}-\frac{22}{5}\sqrt{3}\right)\pi}}\end{aligned}$$

10. Find the volume bounded by the cylinder $x^2+y^2=4$ and the planes $𝑦 + 𝑧 = 4$; $𝑧 = −1$.



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Solution
Place the volume element in general position, and integrate to a rod along the $z$-direction. Then integrate the rod over the circle:
$$\begin{aligned}V&=\iint_{\text{Circle}}\int_{z=-1}^{z=4-y}dzdxdy\\&=\iint_{\text{Circle}}(5-y)dxdy\\&=\int_{\phi=0}^{2\pi}\int_{r=0}^2(5-r\sin\phi)rdrd\phi\\&=\int_{\phi=0}^{2\pi}\left.5r-\frac{r^2}{2}\sin\phi\right|_{r=0}^{r=2}d\phi\\&=\int_{\phi=0}^{2\pi}[10-2\sin\phi]d\phi\\&=\left.10\phi+2\cos\phi\right|_0^{2\pi}\\&=\underline{\underline{20\pi}}\end{aligned}$$