Triple Integrala

Let $T$ be a closed bounded region in $xyz$ space and the projection of $T$ onto $xy$-plane be $\Omega_{xy}$. Then $T$ is expressed as

$$T = \{(x,y,z) : (x,y) \in \Omega_{xy}, \psi_{1}(x,y) \leq z \leq \psi_{2}(x,y) \}$$

Thus,

$$\iiint_{T}f(x,y,z)dxdydz = \iint_{\Omega_{xy}}\big(\int_{\psi_{1}(x,y)}^{\psi_{2}(x,y)}f(x,y,z)dz\big)dxdy$$

Furthermore, if

$$\Omega = \{(x,y) : a \leq x \leq b, \phi_{1}(x) \leq y \leq \phi_{2}(x) \}$$

then

$$\iiint_{T}f(x,y,z)dxdydz = \int_{a}^{b}\left(\int_{\phi_{1}(x)}^{... ..._{2}(x)}\left(\int_{\psi_{1}(x,y)}^{\psi_{2}(x,y)}f(x,y,z)dz\right)dy\right)dx$$

NOTE Consider a small rectangular solid $\Delta x \Delta y \Delta z$. Then as in figure5.11, fill the solid by piling up these small rectangular solid to the direction of $z$-axis.

Figure 5.11: Triple Integral

Then the height of the long rectangular solid can be expressed by $\int_{\psi_1}^{\psi_2}dz = (\psi_{2} - \psi_1)$, the volume of the small long rectangular solid is $(\psi_{2} - \psi_1)\Delta x \Delta y$. Thus

$$\iiint_{T}f(x,y,z)dxdydz = \iint_{\Omega_{xy}}[\int_{\psi_{1}(x,y)}^{\psi_{2}(x,y)}f(x,y,z)dz]dxdy.$$

This way, a triple integral can be reduced a double integral.

Example 5..9   For $${T = \{(x,y,z) : 0 \leq x \leq y \leq z \leq 1 \}}$$. evaluate the following triple integrals.

$$\iiint_{T} dx dydz$$

SOLUTION The projection of $T = \{(x,y,z) : 0 \leq x \leq y \leq z \leq 1\}$ onto the $xy$-plane is the region $\Omega_{xy} = \{(x,y) : 0 \leq x \leq y \leq 1$. Now using vertical simple region, $\Omega_{xy} = \{(x,y) : 0 \leq x \leq 1, x \leq y \leq 1\}.$ Its volume is given by

$$V = \iint_{\Omega}\int_{z=y}^{1}dz dx dy = \int_{x=0}^{1}\int_{x}^{1} [z]_{y}^{1}dy dx$$

$$= \int_{0}^{1}\int_{x}^{1}(1 - y)dy dx = \int_{0}^{1}\left[y - \frac{y^2}{2}\right]_{x}^{1} dx$$

$$= \int_{0}^{1}(1 - \frac{1}{2} - (x - \frac{x^2}{2})) dx = \left[\frac{x}{2} - \frac{x^2}{2} + \frac{x^3}{6}\right]_{0}^{1}$$

$$= \frac{1}{2} - \frac{1}{2} + \frac{1}{6} = \frac{1}{6}$$

Exercise 5..9   For $${T = \{(x,y,z) : 0 \leq x \leq y \leq z \leq 1 \}}$$, evaluate the following triple integral.

$$\iiint_{T} e^{x+y+z} dx dydz$$

SOLUTION The projection of $T = \{(x,y,z) : 0 \leq x \leq y \leq z \leq 1\}$ onto $xy$-plane is the region $\Omega_{xy} = \{(x,y) : 0 \leq x \leq y \leq 1\}$. Using vertical simple region, $\Omega_{xy} = \{(x,y) : 0 \leq x \leq 1, x \leq y \leq 1\}$. Thus

$$V = \iint_{\Omega}\int_{z=y}^{1}e^{x+y+z}dz dx dy = \int_{x=0}^{1}\int_{x}^{1} [e^{x+y+z}]_{y}^{1}dy dx$$

$$= \int_{0}^{1}\int_{x}^{1}(e^{x+y+1} - e^{x+2y})dy dx$$

$$= \int_0^1 \left[e^{x+y+1} - \frac{1}{2}e^{x+2y}\right]_x^1 dx$$

$$= \int_{0}^{1}(e^{x+2} - \frac{1}{2}e^{x+2} - e^{2x+1} + \frac{1}{2}e^{3x})dx$$

$$= \left[\frac{1}{2}e^{x+2} - \frac{1}{2}e^{2x+1} + \frac{1}{6}e^{3x}\right]_{0}^{1}$$

$$= \frac{1}{2}e^{3} - \frac{1}{2}e^{3} + \frac{1}{6}e^{3} - (\frac{1}{2}e^{2} - \frac{1}{2}e + \frac{1}{6})$$

$$= \frac{1}{6}e^3 - \frac{1}{2}e^2 + \frac{1}{2}e - \frac{1}{6} = \frac{1}{6}(e - 1)^3$$

Theorem 5..6   The transformation $\Phi : \left\{\begin{array}{c} x = \phi(u,v,w)\\ y = \psi(u,v,w)\\ z = \xi(u,v,w) \end{array}\right.$ is one-to-one mapping of points in the close bounded region $T$ in $xyz$-space into points in the closed bounded region $T'$ in $uvw$-space. Suppose that $\phi,\psi,\xi$ are the class $C^1$ and

$$J = J(u,v,w) = \left\vert\begin{array}{ccc} x_{u}&x_{v}&x_{w}\\ y_{u}&y_{v}&y_{w}\\ z_{u}&z_{v}&z_{w} \end{array}\right \vert \neq 0.$$

If $f(x,y,z)$ is continuous on $T$, then

$$\iiint_{T}f(x,y,z)dxdydz$$

$$= \iiint_{T^{\prime}}f(\phi(u,v,w),\psi(u,v,w),\xi(u,v,w))\vert J\vert dudvdw.$$

Corollary 5..1   Let $(x,y,z)$ be a point of the cylinder in the rectangular coordinates. Now express this by the cylindrical coordinates, $\left\{\begin{array}{c} x = r\cos{\theta}\\ y = r\sin{\theta}\\ z = z \end{array}\right.$

$$J(r,\theta,z) = \left\vert\begin{array}{ccc} \cos{\theta}& -r\sin... ...eta}&0\\ \sin{\theta}& r\cos{\theta} & 0\\ 0&0&1 \end{array}\right \vert = r$$

Thus

$$\iiint_{T}f(x,y,z)dxdydz = \iiint_{T^{\prime}}f(r\cos{\theta},r\sin{\theta},z)dz\vert r\vert drd\theta$$

Figure 5.12: Cylindrical Coordinates

Example 5..10   For $${T = \{(x,y,z) : x^2 + y^2 \leq 1, -\sqrt{1 - x^2 - y^2} \leq z \leq \sqrt{1 - x^2 - y^2}\}}$$, evaluate the triple integral

$$I = \iiint_{T} \frac{1}{\sqrt{1 - x^2 - y^2}} dxdydz$$

SOLUTION In the cylindrical coordinates, a point $(r,\theta,z)$ is expressed by the distance $r$ from the origin , the angle $\theta$ from the polar axis , and $z$. Thus, by letting $x = r\cos{\theta}, y = r\sin{\theta}$, $x^2 + y^2 \leq 1$ is written as $r^2 \leq 1$. a circle with the radius 1. Thus, $\{(r,\theta): 0 \leq \theta \leq 2\pi, 0 \leq r \leq 1\}$ and $-\sqrt{1-x^2-y^2} = -\sqrt{1-r^2}$. From this, the region $T$ is mapped into the region

$$T' = \{(r,\theta, z) : 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi, - \sqrt{1 - r^2} \leq z \leq \sqrt{1 - r^2} \}$$

$$I = \iiint_{T'}\frac{1}{\sqrt{1 - r^2}} rdzdrd\theta = \int_{0}^{2\pi... ... \sqrt{1 - r^2}}^{\sqrt{1 - r^2}} \frac{r}{\sqrt{1 - r^2}}dz\big)dr\big)d\theta$$

$$= \int_{0}^{2\pi}\big( \int_{0}^{1} \big[\frac{rz}{\sqrt{1 - r^2}}\... ... = 2\pi \cdot \left[r^{2} \right ]_{0}^{1} = 2\pi \ensuremath{ \blacksquare}$$

Exercise 5..10   For $${T = \{(x,y,z) : x^2 + y^2 \leq a^2, 0 \leq z \leq x \}}$$, evaluate the following triple integral.

$$\iiint_{T} dx dydz$$

SOLUTION The region $T$ is enclosed by the cylinder $x^2 + y^2 \leq a^2$ and the surface $z = 0, z = x$. Use the cylindrical coordinate, $x = r\cos{\theta}, y = r\sin{\theta}, z = z$. Then

$$T= \{(r,\theta,z) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq a, 0 \leq z \leq r\cos{\theta}\}$$

Thus

$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{a}\int_0^{r\cos{\t... ...i}{2}}^{\frac{\pi}{2}}\int_{0}^{a}\left[z\right]_0^{r\cos{\theta}} r dt d\theta$$

$$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{a} r^2\cos{\theta}... ...\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos{\theta}d\theta )(\int_{0}^{a} r^2 dr)$$

$$= 2\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\left[\frac{r^3}{3}\right]_{0}^{a} = \frac{2a^3}{3}\ensuremath{ \blacksquare}$$

Corollary 5..2   The first coordinate, $\rho$, is the distance from the origin to the point $(x,y,z)$. The second coordinate, $\phi$, is the angle measured from the positive $z$-axis. The third coordinate, $\theta$, is the angle measured from the $x$-axis. Then $\left\{\begin{array}{c} x = \sin{\phi}\cos{\theta}\\ y = \rho\sin{\phi}\sin{\theta}\\ z = \rho\cos{\phi} \end{array}\right .$ Jacobian is

$$J(\rho,\phi,\theta) = \left\vert\begin{array}{ccc} \sin{\phi}\cos... ...\\ \cos{\phi}&-\rho\sin{\phi}&0 \end{array}\right \vert = \rho^{2} \sin{\phi}$$

$$\iiint_{T}f(x,y,z)dxdydz$$

$$= \iiint_{T^{\prime}}f(\rho\sin{\phi}\cos{\theta},\rho\sin{\phi}\sin{\theta},\rho\cos{\phi})\rho^{2} \sin{\phi} d\rho d\phi d\theta$$

NOTE

Figure 5.13: Spherical coordinates

Example 5..11   For $${T = \{(x,y,z) : x^2 + y^2 + z^2 \leq 1\}}$$, evaluate the following triple integral.

$$I = \iiint_{T}\frac{dxdydz}{\sqrt{1 - x^2 - y^2 - z^2}}$$

SOLUTION Note that the region $T$ is the sphere of the radius 1. Now use spherical coordinates to express $(\rho,\phi,\theta)$. The angle $\theta$ is measured from the $x$-axis, so to cover the sphere, $0 \leq \theta \leq 2\pi$. The angle $\phi$ is measured from the $z$-axis, so to cover the sphere, $0 \leq \phi \leq \pi$. The distance $\rho$ is measured from the origin, so to cover the sphere, $0 \leq \rho \leq 1$. Thus, the region $T$ is mapped into the region $T'$

$$T' = \{(\rho,\phi,\theta) : 0 \leq \rho \leq 1, 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi \}$$

Thus

$$I = \iiint_{T'}\frac{\rho^{2} \sin{\phi}}{\sqrt{1 - \rho^2}} d\rho d\... ...{0}^{\pi} \sin{\phi} d\phi \int_{0}^{1} \frac{\rho^2}{\sqrt{1 - \rho^2}} d \rho$$

Now note that the integral with respect to $\rho$ is improper integral. So, we let $\rho = \sin{t}$. Then $d\rho = \cos{t}dt$ and $$\begin{array}{l\vert lll} \rho & 0 & \to & 1\ \hline t & 0 & \to & \frac{\pi}{2} \end{array}$$. Thus

$$\int_0^1 \frac{\rho^2}{\sqrt{1 - \rho^2}}d\rho = \int_0^{\pi/2} \frac{\sin^2{t}\cos{t}}{\sqrt{ 1 - \sin^2{t}}}dt = \int_0^{\pi/2} \frac{\sin^2{t}\cos{t}}{\cos{t}}dt$$

$$= \int_0^{\pi/2} \frac{1 - \cos{2t}}{2}dt = \frac{1}{2}\big[t - \frac{\sin{2t}}{2}\big]_0^{\pi/2} = \frac{\pi}{4}$$

Therefore,

$$I = \frac{\pi}{4} \int_{0}^{2\pi}d\theta \int_{0}^{\pi} \sin{\phi} d\phi = \frac{\pi}{4} \cdot 2\pi \cdot 2 = \pi^2 \ensuremath{ \blacksquare}$$

Exercise 5..11   For $${T = \{(x,y,z) : \frac{x^2}{2^2} + \frac{y^2}{3^2} + \frac{z^2}{4^2} \leq 1\}}$$, evaluate the following triple integral.

$$I = \iiint_{T}(x^2 + y^2 + z^2)\:dxdydz$$

SOLUTION

$$T = \{(x,y,z) : \frac{x^2}{2^2} + \frac{y^2}{3^2} + \frac{z^2}{4^2} \leq 1\}$$

is ellipsoid. To use spherical coordinates, we use the following change of variables. Let

$$x = 2u, y = 3v, z = 4w$$

Then the region $T$ is mapped into the region $T' = \{(u,v,w) : u^2 + v^2 + w^2 \leq 1\}$. Then the Jacobian is

$$J(u,v,w) = \vert\frac{\partial(x,y,z)}{\partial(u,v,w)}\vert = \left\vert\be... ...left\vert\begin{array}{ccc} 2&0&0\\ 0&3&0\\ 0&0&4 \end{array}\right\vert = 24$$

Thus

$$I = \iiint_{T}(x^2+y^2 + z^2)dxdydz = \iiint_{T'} (4 u^2 + 9 v^2 + 16 w^2)(24) dudvdw$$

Now we use spherical coordinates. By the transformation

$$u = \rho \sin{\phi}\cos{\theta}, v = \rho \sin{\phi}\sin{\theta}, w = \rho \cos{\theta}$$

$T'$ is mapped into $T''$.

$$T'' = \{(\rho,\phi,\theta) : 0 \leq \rho \leq 1, 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi\}.$$

Thus,

$$I = 24\iiint_{T''}(4 u^2 + 9 v^2 + 16 w^2) dudvdw$$

$$= 96\iiint_{T''}u^2 dudvdw + 216 \iiint_{T''}v^2 dudvdw + 384\iiint_{T''}w^2 dudvdw$$

Since $\vert J\vert = \rho^2 \sin{\phi}$,

$$96\iiint_{T''}u^2 dudvdw = 96\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^2 \sin^{2}{\phi}\cos^{2}{\theta}\vert J\vert d\rho d\phi d\theta$$

$$= 96\int_{0}^{2\pi}\cos^{2}{\theta}d\theta\int_{0}^{\pi}\sin^{3}{\phi}d\phi\int_{0}^{1}\rho^4\:d\rho$$

$$= 96\cdot\pi\cdot\frac{4}{3}\cdot\frac{1}{5} = \frac{4\cdot 96}{15}\pi = \frac{128}{5}\pi$$

$$216\iiint_{T''}v^2 dudvdw = 216\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^2 \sin^{2}{\phi}\sin^{2}{\theta}\vert J\vert d\rho d\phi d\theta$$

$$= 216\int_{0}^{2\pi}\sin^{2}{\theta}d\theta\int_{0}^{\pi}\sin^{3}{\phi}d\phi\int_{0}^{1}\rho^4\:d\rho$$

$$= 216\cdot\pi\cdot\frac{4}{3}\cdot\frac{1}{5} = \frac{4\cdot 216}{15}\pi = \frac{288}{5}\pi$$

$$384\iiint_{T''}w^2 dudvdw = 384\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^2 \cos^{2}{\theta}\vert J\vert d\rho d\phi d\theta$$

$$= 384\int_{0}^{2\pi}\sin{\theta}\cos^{2}{\theta}d\theta\int_0^\pi \sin{\phi}d\phi \int_{0}^{1}\rho^4\:d\rho$$

$$= 384\cdot \pi\cdot 2 \cdot \frac{1}{5} = \frac{768}{5}\pi.$$

Therefore,

$$I = (\frac{128}{5} + \frac{288}{5} + \frac{768}{5})\pi = \frac{1184}{5}\pi \ensuremath{ \blacksquare}$$

Suppose the density $\sigma = f(x,y,z)$. Then the mass $m$ of the solid $T$ is given by

$$m = \iiint_{T}f(x,y,z)dxdydz$$

Especially when the density $\sigma = f(x,y,z) = 1$,

$$\iiint_{T} dx dydz$$

We can think of this as the volume of $T$.

Example 5..12   Find the volume of the tetorahedoron in the figure.

SOLUTION To evaluate the triple integral, first find the projection of $T$ onto $xy$-plane. The projection $\Omega_{xy}$ is the triangle region bounded by $x= 0, y= 0, y\leq 1-x$. Now expressing by vertically simple region.

$$\Omega_{xy} = \{(x,y) : 0 \leq x \leq 1, 0 \leq y \leq 1-x \}$$

Thus

$$T = \{(x,y,z) : 0 \leq x \leq 1, 0 \leq y \leq 1-x, 0 \leq z \leq 1-x-y \}$$

$$V = \iint_{\Omega_{xy}}[\int_{0}^{1-x-y}dz]dxdy = \int_{0}^{1}\int_{0}^{1-x}(1-x-y)dydx$$

$$= \int_{0}^{1} \left[(1-x)y - \frac{y^2}{2}\right ]_{0}^{1-x} dx = \int_{0}^{1}\frac{(1-x)^{2}}{2}dx$$

$$= - \left[\frac{(1-x)^{3}}{6}\right ]_{0}^{1} = \frac{1}{6}$$

$\blacksquare$

In this question, the volume is simply given by the base area $\times$ the height $\div$ 3. Thus $${\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}}$$ .

Exercise 5..12   In Example5.11, when the density is given by $\sigma(x,y,z) = xy$, find the mass.

SOLUTION

$$m = \iiint_{T}xydxdydz = \int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}xy dzdydx$$

Now

$$\int_{0}^{1-x-y}xydz = xy(1-x-y) = x(1-x)y - xy^2$$

Then

$$\int_{0}^{1-x}\int_{0}^{1-x-y}xydzdy = \int_{0}^{1-x}(x(1-x)y - xy^2)dy$$

$$= \left[\frac{1}{2}x(1-x)y^2 - \frac{1}{3}xy^3\right ]_{0}^{1-x} = \frac{1}{6}x(1-x)^{3}$$

Thus

$$m = \int_{0}^{1}\frac{1}{6}x(1-x)^{3}dx = \frac{1}{6}\int_{0}^{1}(x-3x^2 + 3x^3 - x^4)dx$$

$$= \frac{1}{6}\left[\frac{1}{2}x^2 - x^3 + \frac{3}{4}x^4 -\frac{1}{5}x^5\right ]_{0}^{1} = \frac{1}{120} \ensuremath{ \blacksquare}$$

consider the system of $n$ particles, ${\rm P}_{1}, {\rm P}_{2}, \ldots, {\rm P}_{n}$. Now the cartesian coordinates of ${\rm P}_{i}$ is ${\rm P}_{i}(x_{i},y_{i})$ and the mass of the particle ${\rm P}_{i}$ is $m_{i}$. Now draw a line perpendicular to the $x$-axis so that the rotation moment is equal. Thus

$$\sum_{i=1}^{n}m_{i}(x_{i} - \bar x) = 0$$

or

$$\bar x = \frac{\sum_{i=1}^{n}m_{i}x_{i}}{\sum_{i=1}^{n}m_{i}}$$

Similarly,

$$\bar y = \frac{\sum_{i=1}^{n}m_{i}y_{i}}{\sum_{i=1}^{n}m_{i}}$$

The point $(\bar x, \bar y)$ is called centroid of a system of particles.

Figure 5.14: Centroid of a system of particles

If every point ${\rm P}(x,y)$ in the closed bounded region $\Omega$ is given a density $\sigma(x,y)$, we partition $\Omega$ into $\Delta : \Omega_{1},\Omega_{2},\ldots,\Omega_{n}$ and select an arbitrary point ${\rm P}_{i}(x_{i},y_{i})$ in $\Omega_{i}$. Consider the centroid $(x_{\Delta},y_{\Delta})$ of particles ${\rm P}_{1}, {\rm P}_{2}, \ldots, {\rm P}_{n}$.

Let the area of $\Omega_{i}$ be $\Delta S_{i}$. Then

$$x_{\Delta} = \frac{\sum_{i=1}^{n}\sigma({\rm P}_{i})x_{i}\Delta S... ...({\rm P}_{i})y_{i}\Delta S_{i}}{\sum_{i=1}^{n}\sigma({\rm P}_{i})\Delta S_{i}}$$

Now by letting $\vert\Delta\vert \rightarrow 0$, $x_{\Delta},y_{\Delta}$ converge to $\bar x, \bar y$. Thus,

$$\bar x = \frac{1}{m}\iint_{\Omega}\sigma(x,y)xdxdy, \bar y = \frac{1}{M}\iint_{\Omega}\sigma(x,y)ydxdy$$

where $m = \iint_{\Omega}\sigma(x,y)dxdy$ represents mass.

If the density $\sigma = \rho(x,y,z)$ is given to each point in the closed region $T$, we can find $(\bar x, \bar y, \bar z)$ as follows.

$$\bar x = \frac{1}{m}\iiint_{T}\rho(x,y,z)xdxdydz, \bar y = \frac{1}{m}\iiint_{T}\rho(x,y,z)ydxdydz$$

$$\bar z = \frac{1}{m}\iiint_{T}\rho(x,y,z)zdxdydz, m = \iiint_{T}\rho(x,y,z)dxdydz$$

Example 5..13   Find the centroid of the figure below if the density at each point is $\sigma(x,y,z) = xy$.

SOLUTION

$$\bar x = \frac{1}{m}\iiint_{T}xyxdxdydz$$

$$= \frac{1}{m}\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}x^2 ydzdydx$$

$$= \frac{1}{m}\int_{0}^{1}\int_{0}^{1-x}x^{2}y(1-x-y)dydx$$

$$= \frac{1}{m}\int_{0}^{1}\left[(\frac{x^{2}(1-x)y^{2}}{2} - \frac{x^{2}y^{3}}{3})\right ]_{0}^{1-x}dx$$

$$= \frac{1}{m}\int_{0}^{1}\frac{x^{2}(1-x)^{3}}{6} dx = \frac{1}{6m}\int_{0}^{1}(x^{2} - 3x^{3} + 3x^{4} - x^{5}) dx$$

$$= \frac{1}{6m}\left[\frac{x^{3}}{3} - \frac{3x^{4}}{4} + \frac{3x^{5}}{5} - \frac{x^{6}}{6} \right ]_{0}^{1} = \frac{1}{360m}$$

Since by Exercise5.12, $${m = \frac{1}{120}}$$. Thus, $${\bar x = \frac{1}{3}}$$.

We next find $\bar{y}$. Interchange $x$ and $y$, we get the same figure. Thus $\bar y = \frac{1}{3}$. Similarly, $\bar z = \frac{1}{6}$ $\blacksquare$

Exercise 5..13   Find the centroid of the right cone with the radius $a$ and the height $h$ if the density at each point is constant.

SOLUTION By symmetry, $\bar{x} = \bar{y} = 0$. So we only need to find $\bar{z}$.

$$\bar{z} = \frac{1}{V}\iiint_{T}z dx dy dz$$

$V = \frac{1}{3}\pi a^2 h$. Note that the triangle with the base $a$ and the height $h$ and the triangle with the base $r$ and the height $k$ is similar. Thus,

$$\frac{h}{a} = \frac{k}{r}$$

and $k = \frac{hr}{a}$. Therefore,

$$T' = \{(r.\theta,z) : 0 \leq r \leq a, 0 \leq \theta \leq 2\pi, \frac{hr}{a} \leq z \leq h\}$$

Thus

$$\iiint_{T'}z\vert J\vert dz dr d\theta = \iiint_{T'}zrdz dr d\theta$$

$$= \int_{0}^{2\pi} \int_{0}^{a} \int_{\frac{hr}{a}}^{h}z dz rdr d\theta$$

$$= \int_{0}^{2\pi}\int_{0}^{a}\left[\frac{z^2}{2}\right]_{\frac{hr}{a}}^{h} r dr d\theta$$

$$= \int_{0}^{2\pi}\int_{0}^{a}\frac{1}{2}(h^2 - \frac{h^2 r^2}{a^2})r dr d\theta$$

$$= \frac{1}{2}\int_{0}^{2\pi}d\theta\int_{0}^{a}(h^2 r - \frac{h^2 r^3}{a^2})dr$$

$$= \pi\left[\frac{h^2 r^2}{2} - \frac{h^2 r^4}{4a^2}\right]_{0}^{a} = \pi (\frac{h^2 a^2}{2} - \frac{h^2 a^4}{4a^2})$$

$$= \frac{\pi h^2 a^2}{4}.$$

Therefore,

$$\bar{z} = \frac{\pi h^2 a^2}{4} \cdot \frac{3}{\pi a^2 h} = \frac{3h}{4}\ensuremath{ \blacksquare}$$

Let $\sigma = \sigma(x,y,z)$ be the density at each point $(x,y,z)$ of region. Then the moment of inertia of $R$ about $x$-axis, $y$-axis, $z$-axis is give by the following.

$$I_x = \iiint_R \sigma (y^2 + z^2)dxdydz\hskip 0.5cm I_y = \iiint_R \sigma(x^2 + z^2)dxdydz$$

$$I_z = \iiint_R \sigma (x^2 + y^2)dxdydz$$

Exercise A

1.

Evaluate the following triple integrals．

(a) $${\int_{0}^{a}\int_{0}^{b}\int_{0}^{c} dxdydz}$$ (b) $${\int_{0}^{1}\int_{0}^{x}\int_{0}^{y} ydzdydx}$$

2.

Express the followings using triple integrals.

(a) The mass of the ball $x^{2} + y^{2} + z^{2} \leq r^{2}$ provided the density is proportional to the distance from the origin．

(b) The mass of the cone $z = 1$ and $z = \sqrt{x^{2} + y^{2}}$ provided that the density is proportional to the distance from the origin．

(c) The volume common to $z = 4 - x^{2} - y^{2}$ and $z = 2 + y^{2}$．

3.

Find the center of mass of the following closed region bounded by the following curves．

(a) $y = x$ and $${y = x^2}$$ provided the density is constant. (b) $x^{2} = 4y$ and $${x - 2y + 4 = 0}$$.

(c) $y = x^{2} - 2x$ and $${y = 6x - x^{2}}$$.

Exercise B

1.

Evaluate the following triple integrals for $${T = \{(x,y,z) : 0 \leq x \leq y \leq z \leq 1 \}}$$．

(a) $${\iiint_{T} dx dydz}$$ (b) $${\iiint_{T}e^{x+y+z} dxdydz }$$

2.

Evaluate the following triple integrals.．

(a) $${\iiint_{T} dx dydz, T = \{(x,y,z):\sqrt{x^2 + y^2} \leq z \leq 3 \}}$$

(b) $${\iiint_{T}(x^2 + y^2 + z^2) dxdydz, T = \{(x,y,z):\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1\} }$$

3.

Find the center of mass of the following region．

(a) $y = x$ and $${y = 6x - x^2}$$

(b) Semishpher $${x^2 + y^2 + z^2 \leq a^2, z \geq 0}$$ provided the density is proportional to the distanace from the origin.

(c) The right circular cone with the bottom radius $a$ and the hight $h$．

(d) $${ax \leq x^2 + y^2 \leq a^2}$$

(e) Find the center $\bar y, \bar z$ of the trapezoid given in Example 5.13

(f) $${ax \leq x^2 + y^2 \leq a^2}$$ provided the density is proportional to the distance from the origin.