3.10 Anser

3.10

1.

(a) Find the intersection of $x = y^2$ and $x = 3 - 2y^2$. Then $y^2 = 3 - 2y^2$ implies $y = \pm 1$. Thus these two curves intersects at $(1,-1)$ and $(1,1)$．Now think of this figure's are as a sum of $\Delta A$．Cut this figure by the line perpendicular to $y$-axis. Then the width is given by the right-hand curve $-$ the left-hand curve. The hight is given by the $\Delta y$. Thus,

$$\Delta A = (3 - 2y^2 - y^2) \Delta y$$

Therefore,

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

$$= 3y - y^3 \mid_{-1}^{1} = (3-1) - (-3+1) = 4.$$

(b)

This figure is not differentiable at $t = \frac{\pi}{2}$. So, we integrate from $t = 0$ to $t = \frac{\pi}{2}$ and double the value．Cut the figure by the rectangle with small width. Then the area of the rectangle is given by $\Delta A$ and

$$\Delta A = y \Delta x$$

Thus the area $A$ is

$$A = 2\int_{0}^{1}y dx$$

Note that $x,y$ are parameterized by $t$.

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

Thus,

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

$$= 6\int_{0}^{\pi/2}\sin^{4}{\cos^{2}{t}}dt$$

$$= 6\int_{0}^{\pi/2}\sin^{4}{t}(1 - \sin^{2}{t})dt = 6\int_{0}^{\pi/2}(\sin^{4}{t} - \sin^{6}{t})dt$$

Here we use the following formula.

$$\int_{0}^{\pi/2}\sin^{n}{x}dx = \left\{\begin{array}{ll} \frac{(... ...rac{\pi}{2}, & n \ even\\ \frac{(n-1)!!}{n!!}, & n \ odd， \end{array}\right.$$

Note that $n!! = n(n-2)(n-4)\cdots$. Therefore，

$$A = 6(\frac{3\cdot1}{4 \cdot 2}\frac{\pi}{2} - \frac{5\cdot3\cdot1}{6\cdot4\cdot2}\frac{\pi}{2}) = \frac{3\pi}{16}.$$

2.

(a) Cut this figure by the plane perpendicular to the axis of rotation. Then its cross section becomes a shape called a washer． $(y-2)^2 = 1 - x^2$ implies $y = 2 \pm \sqrt{1 - x^2}$．Then

$$y_{out} = 2 + \sqrt{1 - x^2}, \ y_{in} = 2 - \sqrt{1 - x^2}.$$

Thus the cross setion area $A$ is

$$A = \pi (y_{out}^{2} - y_{in}^{2}) = 8\pi\sqrt{1 - x^2}.$$

Thus, putting small thickness $\Delta x$. Then the volume $\Delta V$ is

$$\Delta V = A \Delta x = 8\pi\sqrt{1 - x^2} \Delta x$$

Thus, the volume is

$$V = \int_{-1}^{1}8\pi\sqrt{1-x^2}dx = 16\pi\int_{0}^{1}\sqrt{1 - x^2}dx$$

Note that by letting $x = \sin{\theta}$，we have $dx = \cos{\theta}$, $\sqrt{1 - x^2} = \sqrt{1 - \sin^{2}{\theta}} = \cos{\theta}$. Therefore,

$$V = 16\pi\int_{0}^{\pi/2}\cos{\theta}\cos{\theta}d \theta = 16\pi \frac{\pi}{4} = 4\pi^2.$$

(b) The graph of this function is called a cycloid．The intersection with the $x$ axis is when $t = 0$ and $t = 2\pi$. When this figure is rotated around the $x$ axis and the rotating body is cut by a plane perpendicular to the $x$ axis, the cross-sectional area is $\pi y^2$. If you add a little thickness $\Delta x$ to this, its volume will be

$$\Delta V = \pi y^2 \Delta x.$$

Thus, the volume is given by

$$V = \int_{0}^{2\pi}\pi y^2 dx$$

This graph is symmetric with $x = \pi$

$$V = 2\int_{0}^{\pi}\pi y^2 dx.$$

So, we express $x，y$ by the parameter $t$. Then

$$V = 2 \int_{0}^{\pi}\pi y^2 dx$$

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

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

Using $1 - \cos{t} = 2\sin^{2}{\frac{t}{2}}$, we have

$$V = 16\pi \int_{0}^{\pi}\sin^{6}{\frac{t}{2}}dt$$

$$= 32\pi \int_{0}^{\pi/2}\sin^{6}{\theta} d\theta \ (\frac{t}{2} = \theta)とおく$$

$$= 32\pi(\frac{5!!}{6!!}\frac{\pi}{2}) = 16\pi^2 (\frac{5\cdot3\cdot1}{6\cdot4\cdot2}) = 5\pi^2.$$

3.

(a) The graph of $x^{2/3} + y^{2/3} = 1$ is called asteroid．Now parametrize this function. Then $x = \cos^{3}{t}, \ y = \sin^{3}{t} \ (0 \leq t \leq 2\pi)$. Then we calculate the length of the section $x \geq 0, y \geq 0$. Then multiply by 4. A part of curve $\Delta L$ is given by

$$\Delta L = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\frac{\Delta x}{\Delta t})^2 + (\frac{\Delta y}{\Delta t})^2} \Delta t$$

Thus the length of the curve is

$$L = 4\int_{0}^{\pi/2}\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$$

$$= 4\int_{0}^{\pi/2}\sqrt{(-3\cos^{2}{t}\sin{t})^2 + (3\sin^{2}{t}\cos{t})^2} dt$$

$$= 4\int_{0}^{\pi/2}\sqrt{9(\cos^{4}{t}\sin^{2}{t} + \sin^{4}{t}\cos^{2}{t})} dt$$

$$= 12 \int_{0}^{\pi/2}\sqrt{\cos^{2}{t}\sin^{2}{t}(\cos^{2}{t} + \sin^{2}{t})} dt$$

$$= 12 \int_{0}^{\pi/2} \cos{t}\sin{t} dt = 12 \frac{\sin^{2}{t}}{2}\mid_{0}^{\pi/2} = 6$$

(b) We parametrize．Let $x = \cos^{4}{t}, y = \sin^{4}{t} \ (0 \leq t \leq \frac{\pi}{2})$. Then

$$\sqrt{x} + \sqrt{y} = \cos^{2}{t} + \sin^{2}{t} = 1$$

Thus the length of the curve is

$$L = \int_{0}^{\pi/2}\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$$

$$= \int_{0}^{\pi/2}\sqrt{(-4\cos^{3}{t}\sin{t})^2 + (4\sin^{3}{t}\cos{t})^2} dt$$

$$= 4\int_{0}^{\pi/2}\sqrt{\cos^{6}{t}\sin^{2}{t} + \sin^{6}{t}\cos^{2}{t}} dt$$

$$= 4\int_{0}^{\pi/2}\sqrt{\cos^{4}{t} + \sin^{4}{t}} \ \cos{t}\sin{t}dt$$

$$= 4\int_{0}^{\pi/2}\sqrt{(1 - \sin^{2}{t})^2 + \sin^{4}{t}} \ \sin{t}\cos{t}dt$$

Now let $u = \sin^{2}{t}$. Then $du = 2\sin{t}\cos{t}dt$．Note that

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

Then

$$L = 2\int_{0}^{1}\sqrt{(1 - u)^2 + u^2} du = 2\int_{0}^{1}\sqrt{2u^2 - 2u + 1}du$$

$$= 2\sqrt{2}\int_{0}^{1}\sqrt{(u - \frac{1}{1})^2 + \frac{1}{4}}du$$

$w = u - \frac{1}{2}$とおくと$dw = du$

$$\begin{array}{l\vert lll} u&0&\to&1\\ \hline w&-\frac{1}{2}&\to&\frac{1}{2} \end{array}$$

Thus,

$$L = 2\sqrt{2}\int_{-1/2}^{1/2}\sqrt{w ^2 + \frac{1}{4}} dw$$

Here we use the following formula.

$$\int \sqrt{x^2 + A} dx = \frac{1}{2}(x\sqrt{x^2 + A} + A \log\vert x + \sqrt{x^2 + A}\vert)$$

Then

$$L = 2\sqrt{2}\frac{1}{2}(w\sqrt{w^2 + \frac{1}{4}} + \frac{1}{4}\log\vert w + \sqrt{w^2 + \frac{1}{4}}\vert)\mid_{-1/2}^{1/2}$$

$$= \sqrt{2}(\frac{1}{2\sqrt{2}} + \frac{1}{4}\log\vert\frac{1}{2} + ... ...(-\frac{1}{2\sqrt{2}} + \frac{1}{4}\log\vert-\frac{1}{2} + \frac{1}{\sqrt{2}}))$$

$$= \sqrt{2}(\frac{1}{\sqrt{2}} + \frac{1}{4}\log\vert\frac{\frac{1}{2} + \frac{1}{\sqrt{2}}}{\frac{-1}{2} + \frac{1}{\sqrt{2}}})$$

$$= 1 + \frac{\sqrt{2}}{4}\log\vert\frac{2 + \sqrt{2}}{2 - \sqrt{2}}\vert$$

$$= 1 + \frac{\sqrt{2}}{4}\log\vert\frac{(2+\sqrt{2})^2}{2}\vert$$

$$= 1 + \frac{sqrt{2}}{2}\log\vert 1 +\sqrt{2}\vert$$

(c) $r = 1 + \cos{\theta}$ is symmetric with $x$-axis and，

$$\Delta L = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\frac{\Delta x}{\Delta \theta})^2 + (\frac{\Delta y}{\Delta \theta})^2} \Delta \theta$$

Using the polar coordinate, $x = r\cos{\theta}, y = r\sin{\theta}$．Then

$$\frac{dx}{d\theta} = -\sin{\theta}\cos{\theta} - \sin{\theta}(1 + \cos{\theta})$$

$$= -\sin{\theta} - 2\sin{\theta}\cos{\theta} = -\sin{\theta} -\sin{2\theta}$$

$$\frac{dy}{d\theta} = -\sin^{2}{\theta} + \cos{\theta}(1 + \cos{\theta})$$

$$= \cos{\theta} + \cos^{2}{\theta} - \sin^{2}{\theta} = \cos{\theta} + \cos{2\theta}$$

Thus,

$$L = 2\int_{0}^{\pi}\sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta$$

$$= 2\int_{0}^{\pi}\sqrt{2 + 2(\sin{\theta}\sin{2\theta} + \cos{\theta}\cos{2\theta})} d\theta$$

$$= 2\int_{0}^{\pi}\sqrt{2 + 2\cos{(2\theta - \theta)}} d\theta$$

$$= 2\int_{0}^{\pi}\sqrt{2 + 2\cos{\theta}} d\theta$$

$$= 2\sqrt{2}\int_{0}^{\pi}\sqrt{1 + \cos{\theta}}d\theta$$

$$= 2\sqrt{2}\int_{0}^{\pi}\sqrt{2\cos^{2}{\frac{\theta}{2}}}d\theta$$

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

$$= 4 \cdot2 \cdot \sin{\frac{\theta}{2}}\mid_{0}^{\pi} = 8$$