7.5 Answer

7.5

1.

(a) Express $\Omega$ using V-simple.，

$$\Omega = \{(x,y) : 0 \leq x \leq 1, 0 \leq y \leq (1 - x^{2/3})^{3/2}\}$$

Then

$$I = \iint_{\Omega}dx dy = \int_{x=0}^{1}\int_{y=0}^{(1 - x^{2/3})^{3/2}}dy dx$$

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

Now we let $x = \cos^{3}{t}$. Then $dx = -3\cos^{2}{t}\sin{t}dt$

$$\begin{array}{l\vert lll} x&0&\to&1\\ \hline t&\frac{\pi}{2}&\to&0 \end{array}$$

Thus

$$I = \int_{\frac{\pi}{2}}^{0}{(1 - \cos^{2}{t})^{3/2}}(-3\cos^{2}{t}\sin{t}) dt$$

$$= -3\int_{\frac{\pi}{2}}^{0}\sin^{4}{t}\cos^{2}{t}dt$$

$$= 3\int_{0}^{\frac{\pi}{2}}\sin^{4}{t}(1 - \sin^{2}{t})dt$$

$$= 3\left(\frac{3\cdot1}{4\cdot2}\frac{\pi}{2} - \frac{5\cdot3\cdot1}{6\cdot4\cdot2}\frac{\pi}{2}\right) = \frac{3\pi}{32}$$

(b) We find the range of enclosure of $r = a\cos{3\theta}$．To do so，we find the angle satisfying $r = 0$. $\cos{3\theta} = 0$ implies $\theta = \pm\frac{\pi}{6},\pm\frac{\pi}{2},\cdots,$．Also, $r = \cos{3\theta}$ is symmetric about $x$ axis．Thus，the area we need to find is 12 times the area of region between $\theta = 0$ and $\theta = \frac{\pi}{6}$．Let $\Omega$ be the region between $\theta = 0$ and $\theta = \frac{\pi}{6}$. Then using the polarcoordinates, $\Omega$ maps to

$$\Gamma = \{(r,\theta) : 0 \leq \theta \leq \frac{\pi}{6}, 0 \leq r \leq a\cos{3\theta}\}$$

Thus

$$I = \iint_{\Omega}dx dy = \iint_{\Gamma}r dr d\theta$$

$$= \int_{0}^{\frac{\pi}{6}}\int_{0}^{a\cos{3\theta}} rdrd\theta$$

$$= \int_{0}^{\frac{\pi}{6}}\left[\frac{r^2}{2}\right]_{0}^{a\cos{\theta}} d\theta$$

$$= \frac{a^2}{2}\int_{0}^{\frac{\pi}{6}}\cos^{2}{3\theta}d\theta = \frac{a^2}{2}\int_{0}^{\frac{\pi}{6}}\frac{1 + \cos{6\theta}}{2}d\theta$$

$$= \frac{a^2}{4}\left[\theta + \frac{\sin{6\theta}}{6}\right]_{0}^{\frac{\pi}{6}} = \frac{a^2}{4}(\frac{\pi}{6}) = \frac{a^2 \pi}{24}$$

Therefore, the area we need to find is $${\frac{a^2 \pi}{2}}$$.

(c)

First, we find the points of intersection.，

$$\frac{8}{x^2 + 4} = \frac{x^2}{4}$$

implies $x^4 + 4x^2 - 32 = (x^2 + 8)(x^2 - 4) = 0$．Then $x = \pm 2$．Thus we express $\Omega$ using V-simple.

$$\Omega = \{(x,y) : -2 \leq x \leq 2, \frac{x^2}{4} \leq y \leq \frac{8}{x^2 + 4}\}$$

Therefore

$$I = \iint_{\Omega}dx dy = \int_{-2}^{2}\int_{\frac{x^2}{4}}^{\frac{8}{x^2 + 4}} dy dx$$

$$= \int_{-2}^{2}\left[y\right]_{\frac{x^2}{4}}^{\frac{8}{x^2 + 4}} dx$$

$$= \int_{-2}^{2}\left(\frac{8}{x^2 + 4} - \frac{x^2}{4}\right) dx$$

$$= \left[8\cdot\frac{1}{2}\tan^{-1}{\frac{x}{2}} - \frac{x^3}{12}\right]_{-2}^{2}$$

$$= 4\tan^{1} - \frac{8}{12} - (4\tan^{-1}{(-1)} - \frac{-8}{12})$$

$$= 2\pi - \frac{3}{4} - (-2\pi + \frac{3}{4}) = 4\pi - \frac{3}{2}$$

2.

(a) To find the area of a surface，we need a function $z = f(x,y)$ and the $\Omega$ which is created by the orthogonal projection of $z = f(x,y)$. Note that $z = \pm \sqrt{a^2 - (x^2 + y^2)}$．Now take an orthogona projection of the curve. Then $z = 0$ and

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

Thus, we let

$$S = 2\iint_{\Omega}\sqrt{z_{x}^2 + z_{y}^2 + 1} dx dy$$

Note that，

$$z_{x} = \frac{-2x}{2\sqrt{a^2 - (x^2 + y^2)}}, \ z_{y} = \frac{-y}{a^2 -(x^2 +y^2)}$$

Then

$$z_{x}^2 + z_{y}^2 + 1 = \frac{x^2 + y^2 + a^2 - (x^2 + y^2)}{a^2 - (x^2 + y^2)} = \frac{a^2}{a^2 - (x^2 + y^2)}$$

$$S = 2\iint_{\Omega}\frac{a}{\sqrt{a^2 - (x^2 + y^2)}} dx dy$$

The region $\Omega$ is a circle. Using the polarcoordinates, we have

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

Therefore, the area we need to find is

$$S = 2a\iint_{\Gamma}\frac{1}{\sqrt{a^2 - r^2}}rdrd\theta$$

$$= 2a \int_{0}^{2\pi}\int_{0}^{r}\frac{r}{\sqrt{a^2 - r^2}}dr d\thet... ...ac{1}{2}}dt\\ = -t^{\frac{1}{2}} + c\\ = -\sqrt{a^2 - r^2} \end{array}\right)$$

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

$$= 2a \int_{0}^{2\pi}a d\theta = 2a^2 \left[\theta\right]_{0}^{2\pi}$$

$$= 2a^2 (2\pi) = 4a^2 \pi$$

(b) To find the area of a surface，we need a function of the surface $z = f(x,y)$ and the $\Omega$ which is created by the orthogonal projection of $z = f(x,y)$. We use the orthogonal projection of $z = \pm \sqrt{a^2 - (x^2 + y^2)}$．$xy$. In other words, we let $z = 0$. Then

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

Thus,

$$S = 2\iint_{\Omega}\sqrt{z_{x}^2 + z_{y}^2 + 1} dx dy$$

Note that，

$$z_{x} = \frac{-2x}{2\sqrt{a^2 - (x^2 + y^2)}}, \ z_{y} = \frac{-y}{a^2 -(x^2 +y^2)}$$

Thus,

$$z_{x}^2 + z_{y}^2 + 1 = \frac{x^2 + y^2 + a^2 - (x^2 + y^2)}{a^2 - (x^2 + y^2)} = \frac{a^2}{a^2 - (x^2 + y^2)}$$

$$S = 2\iint_{\Omega}\frac{a}{\sqrt{a^2 - (x^2 + y^2)}} dx dy$$

Using the polarcoordinates, we have

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

Therefore,

$$S = 2a\iint_{\Gamma}\frac{1}{\sqrt{a^2 - r^2}}rdrd\theta$$

$$= 2a \int_{0}^{2\pi}\int_{0}^{r}\frac{r}{\sqrt{a^2 - r^2}}dr d\thet... ...ac{1}{2}}dt\\ = -t^{\frac{1}{2}} + c\\ = -\sqrt{a^2 - r^2} \end{array}\right)$$

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

$$= 2a \int_{0}^{2\pi}a d\theta = 2a^2 \left[\theta\right]_{0}^{2\pi}$$

$$= 2a^2 (2\pi) = 4a^2 \pi$$

This is exactly the surface area of the sphere.

(c) To find the area of a surface，we need a function of the surface $z = f(x,y)$ and the $\Omega$ which is created by the orthogonal projection of $z = f(x,y)$.． In this problem, we cut the surface by $x^2 + y^2 = a^2$. Then we take orthogonal projection. Then,

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

Note that $z = \pm \sqrt{a^2 -x^2}$ implies

$$S = 2\iint_{\Omega}\sqrt{z_{x}^2 + z_{y}^2 + 1} dx dy$$

$$z_{x} = \frac{-2x}{2\sqrt{a^2 - x^2}}, \ z_{y} = 0$$

implies

$$z_{x}^2 + z_{y}^2 + 1 = \frac{x^2}{a^2 - x^2} + 1 = \frac{a^2}{a^2 - x^2}$$

$$S = 2\iint_{\Omega}\frac{a}{\sqrt{a^2 - x^2}} dx dy$$

$$= 2\int_{-a}^{a}\int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} \frac{a}{\sqrt{a^2 - x^2}}dy dx$$

$$= 2\int_{-a}^{a}\left[\frac{ay}{\sqrt{a^2 - x^2}}\right]_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} dx$$

$$= 4\int_{-a}^{a}a dx = 8a\int_{0}^{a} dx = 8a[x]_{0}^{a} = 8a^2$$

(d) The surface area $S$ of revolving solid rotating $y = f(x), \ a \leq x \leq b$ around $x$ axis is given by

$$S = \int_{a}^{b} 2\pi y \sqrt{1 + (y')^2} dx$$

よって

$$S = \int_{0}^{k}2\pi mx\sqrt{1 + m^2}dx = 2\pi m \sqrt{1 + m^2}\int_{0}^{k}x dx$$

$$= 2\pi m \sqrt{1 + m^2}[x]_{0}^{k} = \pi m \sqrt{1 + m^2} k^2$$

3. Given the continuos function $z = f(x,y) > z = f(x,y)$ on $\Omega$, Then the volume of the solid bounded by the sufaces $f,g$ and the suface parallel to the $z$ axis is given by

$$V = \iint_{\Omega}(f(x,y) - g(x,y))dx dy$$

(a) $\Omega = \{(x,y) : x^2 + y^2 \leq a^2, \ x \geq 0\}$. Using the polarcoordinates, $x = r\cos{\theta} > 0$ implies that $\Omega$ maps to

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

Thus,

$$V = \iint_{\Omega}x dx dy$$

$$= \iint_{\Gamma}r\cos{\theta} r dr d\theta$$

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

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

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

$$= \frac{2a^3}{3}$$

(b) $\Omega = \{(x,y) : x \leq 1 - y^2, \ x \geq 0, \ y \geq 0 \}$. Using -simple,

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

Thus,

$$V = \iint_{\Omega}(f(x,y) - g(x,y)) dx dy$$

$$= \iint_{\Omega}(1 - x^2)dx dy$$

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

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

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

$$= \int_{0}^{1}(1 - y^2 - \frac{1}{3}(1 - 3y^2 + 3y^4 - y^6))dy$$

$$= \int_{0}^{1}(\frac{2}{3} - y^4 + \frac{1}{3}y^6)dy$$

$$= \left[\frac{2}{3}y - \frac{y^5}{5} + \frac{1}{21}y^7\right]_{0}^{1}$$

$$= \frac{2}{3} - \frac{1}{5} + \frac{1}{21} = \frac{54}{105} = \frac{18}{35}$$

(c) $\Omega = \{(x,y) : x^2 + y^2 \leq ax \}$．Using the polarcoordinte, we express the boundary of $\Omega$. Then $x^2 + y^2 = ax$ implies $r^2 = ar\cos{\theta}$．Thus， $r = a\cos{\theta}$ becomes 0 at， $\theta = -\frac{\pi}{2}, \frac{\pi}{2}$．Thus,

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

Therefore,，

$$V = \iint_{\Omega}(f(x,y) - g(x,y)) dx dy$$

$$= 2\iint_{\Omega}\sqrt{a^2 - (x^2 + y^2)}dx dy$$

$$= 2\iint_{\Gamma}\sqrt{a^2 - r^2}r dr d\theta$$

$$= 4\int_{0}^{\frac{\pi}{2}}\int_{0}^{a\cos{\theta}}\sqrt{a^2 - r^2}... ...ac{3}{2}} + c\\ = -\frac{1}{3}(a^2 - r^2)^{\frac{3}{2}} + c \end{array}\right)$$

$$= 4\int_{0}^{\frac{\pi}{2}}\left[-\frac{1}{3}(a^2 - r^2)^{\frac{3}{2}}\right]_{0}^{a\cos{\theta}} d\theta$$

$$= -\frac{4}{3}\int_{0}^{\frac{\pi}{2}}((a^2 - a^2 \cos^{2}{\theta})^{\frac{3}{2}} - a^3)d\theta$$

$$= -\frac{4}{3}\int_{0}^{\frac{\pi}{2}}((a^2 \sin^{2}{\theta})^{\frac{3}{2}} - a^3)d\theta$$

$$= -\frac{4}{3}\int_{0}^{\frac{\pi}{2}}(a^3 \sin^{3}{\theta} - a^3)d\theta$$

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

(d) The line of intersection of the cone $z = 1 - \sqrt{x^2 + y^2}$ and the plane $z = x$ is $x = 1 - \sqrt{x^2 + y^2}$．Then the volume of the solid is given by the group of straight lines that pass through the boundary of $\Omega = \{(x,y) : 0 \leq x \leq 1 - \sqrt{x^2 + y^2} \}$ and are parallel to the $z$ axis．Using the polarcoordintate, we express the boundary of $\Omega$. Then $r\cos{\theta} = 1 - r$ implies $r = \frac{1}{1 + \cos{\theta}}$．$x \geq 0$ implies $r\cos{\theta} \geq 0$．Thus, $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$．Thus,，

$$\Gamma = \{(r,\theta) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, \ 0 \leq r \leq \frac{1}{1 + \cos{\theta}}\}.$$

Therefore，

$$V = \iint_{\Omega}(f(x,y) - g(x,y)) dx dy$$

$$= \iint_{\Omega}(1 - \sqrt{x^2 + y^2} - x)dx dy$$

$$= \iint_{\Gamma}(1 - r - r\cos{\theta})r dr d\theta$$

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

$$= 2\int_{0}^{\frac{\pi}{2}}\left[\frac{r^2}{2} - \frac{r^3}{3}(1 + \cos{\theta})\right]_{0}^{\frac{1}{1 + \cos{\theta}}} d\theta$$

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

$$= 2\int_{0}^{\frac{\pi}{2}} \frac{1}{6(1+\cos{\theta})^2} d\theta \... ...ray}{l} Note that\\ 1 + \cos{\theta} = 2\cos^{2}{\frac{\theta}{2}} \end{array}$$

$$= \frac{1}{3}\int_{0}^{\frac{\pi}{2}}\frac{1}{4\cos^{4}{\theta}} d\theta$$

$$= \frac{1}{12}\int_{0}^{\frac{\pi}{2}} \sec^{4}{\frac{\theta}{2}} d\theta$$

$$= \frac{1}{12}\int_{0}^{\frac{\pi}{2}} \sec^{2}{\frac{\theta}{2}}\sec^{2}{\frac{\theta}{2}} d\theta$$

$$= \frac{1}{12}\int_{0}^{\frac{\pi}{2}} (1 + \tan^{2}{\frac{\the... ...{\pi}{2}\\ \hline t & 0 & \to & 1 \end{array}\end{array}\right)$$

$$= \frac{1}{12}\int_{0}^{1}(1 + t^2)(2dt)$$

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

$$= \frac{1}{6}(1 + \frac{1}{3}) = \frac{2}{9}$$