7.6 Answer

7.6

1.

(a) Projection of

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

onto $xy$-plane. Then

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

Using V-simple, we have

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

Thus,

$$V = \iint_{\Omega}\int_{z=y}^{1}dz dx dy$$

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

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

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

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

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

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

(b) Projection of

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

onto $xy$-plane.

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

Using V-simple,

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

Then，

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

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

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

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

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

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

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

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

2.

(a) Since

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

, $T$ is bounded by the upper semisphere $z = \sqrt{x^2 + y^2}$ and the plane $z = 3$．Also，the line of intersection of $z = \sqrt{x^2 + y^2}$ and $z = 3$ is $\sqrt{x^2 + y^2} = 3$．Thus,

$$\Omega_{xy} = \{(x,y) : \sqrt{x^2 + y^2} \leq 3\}$$

Using the polarcoordinates,

$$\Gamma = \{(r,\theta) : 0 \leq \theta \leq 2\pi, \ r \leq 3\}$$

Thus

$$V = \iiint_{T}dxdydz = \iint_{\Omega_{xy}}\int_{z = \sqrt{x^2 + y^2}}^{3} dz dxdy$$

$$= \iint_{\Omega_{xy}}(3 - \sqrt{x^2 + y^2})dx dy$$

$$= \iint_{\Gamma}(3 - r)r dr d\theta$$

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

$$= \int_{0}^{2\pi}\left[\frac{3r^2}{2} - \frac{r^3}{3}\right]_{0}^{3} d\theta$$

$$= \int_{0}^{2\pi}(\frac{27}{2} - \frac{27}{3})d\theta$$

$$= \frac{27}{6}[\theta]_{0}^{2\pi} = 9\pi$$

(b)

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

Then $T$ is an ellipsoid．To be able to use the shperical coordinatess, we use the following change of variable.

$$x = au, \ y = bv, \ z = cw$$

Then $T$ is transfered to $T' = \{(u,v,w) : u^2 + v^2 + w^2 \leq 1\}$．Then

$$J(u,v,w) = \vert\frac{\partial(x,y,z)}{\partial(u,v,w)}\vert$$

$$= \left\vert\begin{array}{ccc} x_{u}&x_{v}&x_{w}\\ y_{u}&y_{v}&y_{w}\\ z_{u}&z_{v}&z_{w} \end{array}\right\vert$$

$$= \left\vert\begin{array}{ccc} a&0&0\\ 0&b&0\\ 0&0&c \end{array}\right\vert = abc$$

Thus

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

$$= \iiint_{T'} (a^2 u^2 + b^2 v^2 + c^2 w^2)abc dudvdw$$

Now using the spherical coordinates,

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

Then $T'$ maps to

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

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

$$V = abc\iiint_{T'}(a^2 u^2 + b^2 v^2 + c^2 w^2) dudvdw$$

Now let

$$V_{1} = abc\iiint_{T'}(a^2 u^2) dudvdw$$

$$V_{2} = abc\iiint_{T'}(b^2 v^2) dudvdw$$

$$V_{3} = abc\iiint_{T'}(c^2 w^2) dudvdw$$

Then we find $V_{1}$. Then $V_{2} = \frac{b^2}{a^2}V_{1}$ , $V_{3} = \frac{c^2}{a^2}V_{1}$．

$$V_{1} = abc\iiint_{T'}(a^2 u^2) dudvdw$$

$$= abc\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}(a^2 \rho^2 \sin^{2}{\phi}\cos^{2}{\phi})\vert J\vert d\rho d\phi d\theta$$

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

$$= a^{3}bc\cdot\pi\cdot\frac{4}{3}\cdot\frac{1}{5} = \frac{4}{15}\pi a^{3}bc$$

Note that

$$\int_{0}^{\pi}\sin^{3}{\phi}d\phi = \int_{0}^{\pi}\sin^{2}{\phi}\sin{\phi}d\phi$$

$$= \int_{0}^{\pi}(1 - \cos^{2}{\phi})\sin{\phi}d\phi \ \left(\begin{array}{l} t = \cos{t}\\ dt = -\sin{t}dt \end{array}\right)$$

$$= 2\int_{0}^{1}(1 - t^2)dt$$

$$= 2\left[t - \frac{t^3}{3} \right]_{0}^{1} = \frac{4}{3}$$

3.

(a) Let the $\sigma$ be density. Then $\sigma =$ constant．

$$\Omega = \{(x,y) : 0 \leq x \leq 1, x^2 \leq y \leq x\}$$

Then

$$\bar{x} = \frac{\iint_{\Omega}\sigma x dx dy}{\iint_{\Omega}\sigma dx dy}$$

$$\frac{\int_{0}^{1}\int_{x^2}^{x}xdy dx}{\int_{0}^{1}\int_{x^2}^{x}dy dx}$$

$$= \frac{\int_{0}^{1}[xy]_{x^2}^{x}dx}{\int_{0}^{1}[y]_{x^2}^{x}dx}$$

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

$$= \frac{\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{0}^{1}}{\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{1}}$$

$$= \frac{\frac{1}{3} - \frac{1}{4}}{\frac{1}{2} - \frac{1}{3}} = \frac{\frac{1}{12}}{\frac{1}{6}} = \frac{1}{2}$$

$$\bar{y} = \frac{\iint_{\Omega}\sigma y dx dy}{\iint_{\Omega}\sigma dx dy}$$

Thus

$$\iint_{\Omega}y dx dy = \int_{0}^{1}\int_{x^2}^{x}y dy dx$$

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

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

$$= \frac{1}{2}\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{0}^{1}$$

$$= \frac{1}{2}\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{1}{15}$$

Therefore,

$$\bar{y} = \frac{\frac{1}{15}}{\frac{1}{6}} = \frac{2}{5}$$

(b) By symmetry, $\bar{x} = \bar{y} = 0$．Let $\sigma$ be the density.

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

Note that $\sigma = k\sqrt{x^2 + y^2+ z^2}$．Express

$$T = \{(x,y,z) : x^2 + y^2 + z^2 \leq a^2, z \geq 0\}$$

using the spherical coordinates.

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

Then

$$\iiint_{T}\sigma z dx dy dz = \iiint_{T'}k\rho (\rho\cos{\phi})\vert J\vert d\rho d\phi d\theta$$

$$= k\iiint_{T'}\rho^2 \cos{\phi}(\rho^2 \sin{\phi})d\rho d\phi d\theta$$

$$= k\int_{0}^{2\pi}d\theta \int_{0}^{\frac{\pi}{2}}\sin{\phi}\cos{\phi}d\phi \int_{0}^{a} \rho^4 d\rho$$

$$= 2k\pi \left[\frac{\sin^{2}{\phi}}{2}\right]_{0}^{\frac{\pi}{2}} \left[\frac{\rho^5}{5}\right]_{0}^{a}$$

$$= 2k\pi (\frac{1}{2})\frac{a^5}{5} = \frac{k\pi a^5}{5}$$

Also，

$$\iiint_{T}\sigma dx dy dz = \iiint_{T'}k\rho \vert J\vert d\rho d\phi d\theta$$

$$= k\iiint_{T'}\rho(\rho^2 \sin{\phi})d\rho d\phi d\theta$$

$$= k\int_{0}^{2\pi}d\theta \int_{0}^{\frac{\pi}{2}}\sin{\phi}d\phi \int_{0}^{a} \rho^3 d\rho$$

$$= 2k\pi \left[-\cos{\phi}\right]_{0}^{\frac{\pi}{2}} \left[\frac{\rho^4}{4}\right]_{0}^{a}$$

$$= 2k\pi\cdot 1 \cdot \frac{a^4}{4} = \frac{k\pi a^4}{2}$$

Thus，

$$\bar{z} = \frac{k\pi a^5}{5} \cdot \frac{2}{k\pi a^4} = \frac{2a}{5}$$

(c) The right circular cone $T$ with the radius $a$ and the height $h$ is expressed by

$$T = \{(x,y,z) : x^2 + y^2 \leq \frac{a^2}{k^2}z^2 \ (0 \leq z \leq h)\}$$

By symmetry, $\bar{x} = \bar{y} = 0$．

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

Note that $V = \frac{1}{3}\pi a^2 h$． Express $T$ using the cylindrical coordinates. Then

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

Thus

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

$$= \iiint_{T'}zrdz dr d\theta$$

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

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

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

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

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

$$= \pi (\frac{h^2 a^2}{2} - \frac{h^2 a^4}{4a^2}) = \frac{\pi h^2 a^2}{4}$$

Therefore,

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

(d) The area $A$ of the region

$$\Omega = \{(x,y) : ax \leq x^2 + y^2 \leq a^2 \}$$

is

$$A = \pi a^2 - \pi(\frac{a}{2})^2 = \frac{3\pi a^2}{4}$$

By symmetry, $\bar{y} = 0$．Now using the porlarcoordinates $\Omega$を表わすと

$$\Gamma = \{(r,\theta) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi... ...\{(r,\theta) : \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}, 0 \leq r \leq a\}$$

Thus

$$\bar{x} = \frac{1}{A}\iint_{\Omega}x dx dy$$

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

$$= \frac{2}{A} \left(\int_{0}^{\frac{\pi}{2}}\left[\frac{r^3 \cos{\t... ...rac{\pi}{2}}^{\pi}\left[\frac{r^3 \cos{\theta}}{3}\right]_{0}^{a}d\theta\right)$$

$$= \frac{2}{A} \left(int_{0}^{\frac{\pi}{2}}\frac{a^3}{3}(\cos{\thet... ...})d\theta + \int_{\frac{\pi}{2}}^{\pi}\frac{a^3 \cos{\theta}}{3} d\theta\right)$$

$$= \frac{2}{A}\left(\frac{a^3}{3}(1 - \frac{3\cdot 1}{4\cdot2}\frac{\pi}{2}) + \left[\frac{a^3\sin{\theta}}{3}\right]_{\frac{\pi}{2}}^{\pi}\right)$$

$$= \frac{2}{A}\left(\frac{a^3}{3}(1 - \frac{3\pi}{16}) - \frac{a^3}{3}\right)$$

$$= \frac{8}{3\pi a^2}(-\frac{\pi a^3}{16}) = -\frac{a}{6}$$

(e) To evaluate the triple interal, project $T$ onto the appropriate axes plane．Here project $T$ to $xy$ plane．Using V-simple,

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

Then

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

Thus

$$\bar y = \frac{1}{M}\iiint_{T}xyydxdydz$$

$$= \frac{1}{M}\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}x y^2 dzdydx$$

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

$$= \frac{1}{M}\int_{0}^{1}\int_{0}^{1-x}(x(1-x)y^2 - xy^3)dy dx$$

$$= \frac{1}{M}\int_{0}^{1}\left[\frac{x(1-x)y^{3}}{3} - \frac{xy^{4}}{4})\right ]_{0}^{1-x}dx$$

$$= \frac{1}{M}\int_{0}^{1}\left(\frac{x(1-x)^{4}}{3} - \frac{x(1-x)^4}{4}\right) dx$$

$$= \frac{1}{12M}\int_{0}^{1}(x(1-x)^4) dx \ \left(\begin{array}{l} t = 1 - x\\ dt = - dx \end{array}\right)$$

$$= \frac{1}{12M}\int_{0}^{1}(t^4(1 - t)dt = \frac{1}{12M}\int_{0}^{1}(t^4 - t^5)dt$$

$$= \frac{1}{12M}\left[\frac{t^5}{5} - \frac{t^6}{6}\right]_{0}^{1} = \frac{1}{360M}$$

Note that $${M = \frac{1}{120}}$$ implies $${\bar y = \frac{1}{3}}$$． Also，

$$\bar z = \frac{1}{M}\iiint_{T}xyzdxdydz$$

$$= \frac{1}{M}\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}xyz dzdydx$$

$$= \frac{1}{M}\int_{0}^{1}\int_{0}^{1-x}\left[\frac{xyz^2}{2}\right]_{0}^{1-x-y}dydx$$

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

$$= \frac{1}{2M}\int_{0}^{1}\int_{0}^{1-x}\left(xy(1-x)^2 - 2xy^2(1-x) + xy^3\right)dy dx$$

$$= \frac{1}{2M}\int_{0}^{1}\left[\frac{x(1-x)^2 y^{2}}{2} - \frac{2x(1-x)y^{3}}{3}) + \frac{xy^4}{4}\right ]_{0}^{1-x}dx$$

$$= \frac{1}{2M}\int_{0}^{1}\left(\frac{x(1-x)^{4}}{2} - \frac{2x(1-x)^4}{3} + \frac{x(1-x)^4}{4}\right) dx$$

$$= \frac{1}{24M}\int_{0}^{1}(x(1-x)^4) dx \ \left(\begin{array}{l} t = 1 - x\\ dt = - dx \end{array}\right)$$

$$= \frac{1}{24M}\int_{0}^{1}(t^4(1 - t)dt = \frac{1}{24M}\int_{0}^{1}(t^4 - t^5)dt$$

$$= \frac{1}{24M}\left[\frac{t^5}{5} - \frac{t^6}{6}\right]_{0}^{1} = \frac{1}{720M}$$

Note that $${M = \frac{1}{120}}$$ implies $${\bar y = \frac{1}{6}}$$．

(f) The density $\sigma$ is proportional to the distance from the origin. Then， $\sigma = k\sqrt{x^2 + y^2}$．Also,the mass $M$ of

$$\Omega = \{(x,y) : ax \leq x^2 + y^2 \leq a^2 \}$$

is

$$M = \iint_{\Omega}\sigma dx dy$$

Using the polar coordinates, we express $\Omega$. Then

$$\Gamma = \{(r,\theta) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi... ...\{(r,\theta) : \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}, 0 \leq r \leq a\}$$

Thus

$$M = k\iint_{\Gamma}r \vert J\vert dr d\theta$$

$$= k\int_{\Gamma}r^2 dr d\theta$$

$$= 2k\int_{0}^{\frac{\pi}{2}}\int_{a\cos{\theta}}^{a}r^2 dr d\theta + 2k \int_{\frac{\pi}{2}}^{\pi}\int_{0}^{a}r^2 dr d\theta$$

$$= 2k\int_{0}^{\frac{\pi}{2}}\left[\frac{r^3}{3}\right]_{a\cos{\thet... ...d\theta + 2k\int_{\frac{\pi}{2}}^{\pi}\left[\frac{r^3}{3}\right]_{0}^{a}d\theta$$

$$= 2k\int_{0}^{\frac{\pi}{2}}\frac{a^3(1 - \cos^{3}{\theta})}{3} d\theta + 2k\int_{\frac{\pi}{2}}^{\pi}\frac{a^3}{3}d\theta$$

$$= \frac{2ka^3}{3}(\frac{\pi}{2} - \frac{2}{3}) + \frac{2ka^3}{3}\frac{\pi}{2}$$

$$= \frac{2ka^3}{3} - \frac{4ka^3}{9} = \frac{2ka^3(3\pi - 2)}{9}$$

By the symmetry, $\bar{y} = 0$．Thus

$$\bar{x} = \frac{1}{M}\iint_{\Omega}\sigma x dx dy$$

$$= \frac{1}{M}\iint_{\Gamma}kr\cdot r\cos{\theta}\vert J\vert dr d\theta$$

$$= \frac{2k}{M}\left(\int_{0}^{\frac{\pi}{2}}\int_{a\cos{\theta}}^{a... ...theta + \int_{\frac{\pi}{2}}^{\pi}\int_{0}^{a}r^3 \cos{\theta}dr d\theta\right)$$

$$= \frac{2k}{M} \left(\int_{0}^{\frac{\pi}{2}}\left[\frac{r^4 \cos{\... ...rac{\pi}{2}}^{\pi}\left[\frac{r^4 \cos{\theta}}{4}\right]_{0}^{a}d\theta\right)$$

$$= \frac{ka^4}{2M} \left(int_{0}^{\frac{\pi}{2}}(\cos{\theta} - \cos^{5}{\theta})d\theta + \int_{\frac{\pi}{2}}^{\pi}\cos{\theta}d\theta\right)$$

$$= \frac{ka^4}{2M}\left(1 - \frac{4\cdot 2}{5\cdot3\cdot 1} + \left[\sin{\theta}\right]_{\frac{\pi}{2}}^{\pi}\right)$$

$$= \frac{ka^4}{2M}\left(-\frac{8}{15}\right) = -\frac{4ka^4}{15M}$$

$$= -\frac{4ka^4}{15M}\cdot\frac{9}{2ka^3(3\pi - 2)} = -\frac{6a}{5(3\pi -2)}$$