7.3 Answer

7.3

1. The region of integration is a circle, ellipse, or rhombus. Then use the change of variables.

(a)

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

Then we use the polarcoordinates, $x = r\cos{\theta}, y = r\sin{\theta}$. Then $x^2 + y^2 = r^2 \leq 4$．Thus, $\Omega$ is maps to

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

$$\iint_{\Omega} x^2 dxdy = \iint_{\Gamma}r^2 \cos^{2}{\theta} \vert J(r,\theta)\vert dr d\theta \ ({\rm where}, \vert J(r,theta)\vert = r)$$

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

$$= \int_{0}^{2\pi}\frac{1 + \cos{2\theta}}{2} d\theta ([\frac{r^4}{4}]_{0}^{2} = \frac{1}{2}([\theta + \frac{\sin{2\theta}}{2}]_{0}^{2\pi})(4)$$

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

(b)

$$\Omega = \{(x,y) : 1 \leq x^2 + y^2 \leq 4\}$$

Then we use the polarcoordinates. $x = r\cos{\theta}, y = r\sin{\theta}$ implies $1 \leq r^2 \leq 4$．Then $\Omega$ is maps to

$$\Gamma = \{(r,\theta) : 0 \leq \theta \leq 2\pi, \ 1 \leq r \leq 2\}$$

，

$$I = \iint_{\Omega} \log(x^2 + y^2) dxdy$$

$$= \iint_{\Gamma}\log(r^2) \vert J(r,\theta)\vert dr d\theta \ ({\rm where},\vert J(r,theta)\vert = r)$$

$$= \int_{0}^{2\pi}\int_{1}^{2}r\log{r^2} dr d\theta \ {\rm where}\ \... ...r^2 \log{r} - \int{r}dr\\ = r^2 \log{r} - \frac{r^2}{2} + c \end{array}\right.$$

$$= 2\pi[r^2\log{r} - \frac{r^2}{2}]_{1}^{2} = 2\pi(4\log{2} - 2 + \frac{1}{2})$$

$$= 2\pi(4\log{2} - \frac{3}{2}) = 8\pi\log{2} - 3\pi$$

(c)

$$\Omega = \{(x,y) : x+y \leq 1, x \geq 0, y \geq 0 \}$$

Then we use the change of variables $u = x + y, v = x - y$． Since $u = x, v =x$, a line $y=0$ maps to $u = v$，Also since $u = y, v = -y$, $x= 0$ maps to $v = -u$. Aline $x+y=1$ maps to $u = x+y = 1$. Then $\Omega$ maps to

$$\Gamma = \{(u,v) : 0 \leq u \leq 1, - u \leq v \leq u\}$$

Thus,，

$$I = \iint_{\Omega} e^{(y-x)/(y+x)} dxdy$$

$$= \iint_{\Gamma} e^{-v/u} \vert J(u,v)\vert du dv \ , \left\{\begin{array}{ll} \vert J(u,v)\vert &= \vert\frac{\parti... ...\right\vert \vert\\ &= \vert-\frac{1}{2}\vert = \frac{1}{2} \end{array}\right.$$

$$= \frac{1}{2}\int_{0}^{1}\int_{-u}^{u}e^{-v/u}dvdu$$

$$= \frac{1}{2}\int_{0}^{1}[-ue^{-v/u}]_{-u}^{u} du$$

$$= \frac{1}{2}\int_{0}^{1}(-ue^{-1} + ue)du$$

$$= \frac{1}{2}[-\frac{u^2}{2e} + \frac{u^2 e}{2}]_{0}^{1}$$

$$= \frac{1}{4}(-\frac{1}{e} + e)$$

(d)

$$\Omega = \{(x,y) : 1 < x^2 + y^2 < 4\}$$

Then we use the polarcoordinates. $x = r\cos{\theta}, y = r\sin{\theta}$ implies $1 < r^2 < 4$．Thus $\Omega$ maps to

$$\Gamma = \{(r,\theta) : 0 \leq \theta \leq 2\pi, \ 1 < r < 2\}$$

，

$$I = \iint_{\Omega} e^{x^2 + y^2} dxdy$$

$$= \iint_{\Gamma} e^{r^2} \vert J(r,\theta)\vert dr d\theta \ ({\rm where}, \vert J(r,theta)\vert = r)$$

$$= \int_{0}^{2\pi}\int_{1}^{2}re^{r^2} dr d\theta \ {\rm where} \lef... ...e^{t}}dt\\ = \frac{1}{e^{t}} + c\\ = \frac{1}{e^{r^2}} + c \end{array}\right.$$

$$= (\int_{0}^{2\pi}\ d\theta)[\frac{1}{2}{e^{r^2}}]_{1}^{2}$$

$$= 2\pi(\frac{1}{2}(e^4 - e)) = \pi(e^4 - e)$$

(e)

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

Then we use the polarcoordinates. $x = r\cos{\theta}, y = r\sin{\theta}$ implies $r^2 \leq 1$．Then $\Omega$ maps to

$$\Gamma = \{(r,\theta) : 0 \leq \theta \leq 2\pi, \ 0 \leq r \leq 1\}$$

，

$$I = \iint_{\Omega} \sqrt{1 - x^2 - y^2} dxdy$$

$$= \iint_{\Gamma} \sqrt{1 - r^2} \vert J(r,\theta)\vert dr d\theta \ ({\rm where}, \vert J(r,theta)\vert = r)$$

$$= \int_{0}^{2\pi}d\theta \int_{0}^{1}\sqrt{1 - r^2}r dr \ {\rm wher... ...t^{\frac{3}{2}}\\ = -\frac{1}{3}(1 - r^2)^{\frac{3}{2}} + c \end{array}\right.$$

$$= 2\pi[-\frac{1}{3}(1 -r^2)^{\frac{3}{2}}]_{0}^{1}$$

$$= 2\pi(\frac{1}{3}) = \frac{2\pi}{3}$$

(f)

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

Then we use the polarcoordinates. $x = r\cos{\theta}, y = r\sin{\theta}$ implies $r^2 \leq 1$．Also， $x = r\cos{\theta} \geq 0, y = r\sin{\theta} \geq 0$implies $0 \leq \theta \leq \frac{\pi}{2}$．Thus $\Omega$ maps to

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

$$I = \iint_{\Omega} (1 - x - 2y) dxdy$$

$$= \iint_{\Gamma} (1 - r\cos{\theta} - 2r\sin{\theta}) \vert J(r,\theta)\vert dr d\theta \ ({\rm ただし}, \vert J(r,\theta)\vert = r)$$

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

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

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

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

$$= [\frac{\theta}{2} - \frac{\sin{\theta}}{3} + \frac{2\cos{\theta}}{3}]_{0}^{\frac{\pi}{2}}$$

$$= \frac{\pi}{4} - \frac{1}{3}$$

2.

$$\Omega = \{(x,y) : -1 \leq x+y \leq 1, -1 \leq x-y \leq 1 \}$$

Then we use the change of variables $u = x + y, v = x - y$． Aline $x+y = -1$ maps to $u = -1$ by $u= x+y$，A line $x+y=1$ mpas to $u = x+y = 1$，A line $x-y = -1$ maps to $v = -1$ by $v = x-y$， A line $x-y = 1$ maps to $v = x-y = 1$. Then $\Omega$ maps to

$$\Gamma = \{(u,v) : -1 \leq u \leq 1, -1 \leq v \leq 1\}$$

Also, $x = \frac{u+v}{2}, y = \frac{u-v}{2}$ implies

$$x^2 + y^2 = (\frac{u+v}{2})^2 + (\frac{u-v}{2})^2 = \frac{1}{2}(u^2 + v^2)$$

Thus,

$$I = \iint_{\Omega} (x^2 + y^2)e^{-x+y} dxdy$$

$$= \frac{1}{2}\iint_{\Gamma} (u^2 + v^2)e^{-v} \vert J(u,v)\vert du ... ...{array}\right\vert \vert\\ &= \vert-\frac{1}{2}\vert = \frac{1}{2} \end{array}$$

$$= \frac{1}{4}\int_{v=-1}^{1}\int_{u=-1}^{1}(u^2 + v^2)e^{-v}dudv$$

$$= \frac{1}{4}\int_{-1}^{1}[\frac{u^3}{3}e^{-v} + uv^2 e^{-v}]_{-1}^{1} dv$$

$$= \frac{2}{4}\int_{-1}^{1}(\frac{1}{3}e^{-v} + v^2e^{-v})dv$$

ここで，

$$\int{v^2 e^{-v}}dv = \int{v^2 (-e^{-v})'}dv$$

$$= -v^2 e^{-v} + \int{e^{-v}(2v)}dv$$

$$= -v^2e^{-v} + 2\int{(-e^{-v})' v} dv$$

$$= -v^2 e^{-v} + 2[-e^{-v}v + \int{e^{-v}}dv]$$

$$= -v^2 e^{-v} - 2v e^{-v} -2e^{-v} + c$$

したがって，

$$I = \frac{1}{2}[-\frac{1}{3} e^{-v} - v^2 e^{-v} - 2v e^{-v} -2e^{-v}]_{-1}^{1}$$

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

$$= \frac{1}{2}(\frac{4e}{3} - \frac{16}{3e}) = \frac{2e}{3} - \frac{8}{16e}$$