삼각치환

Note that $f(x)$ is an even function, nonnegative over its entire domain.

(a) To find the area of $R$, we evaluate

$$\begin{aligned} \text{Area}&=\int _{-1/2}^{1/2} \dfrac{|x|}{\sqrt[4]{1-x^2}}~\mathrm{d}x = 2\int _{0}^{1/2} \dfrac{x}{\sqrt[4]{1-x^2}}~\mathrm{d}x \\ \text{We use the substitution $u=1-x^2$, $\mathrm{d}u=-2x~\mathrm{d}x$.} \\ &=-\int_{1}^{3/4} \frac{1}{u^{1/4}}~\mathrm{d}u\\ &=-\left[\frac{4}{3}u^{3/4}\right]_1^{3/4} = -\frac{4}{3}\left(\left(\frac{3}{4}\right)^{3/4}-1\right)\\ &=\frac{4}{3} - \sqrt[4]{\frac{4}{3}} \end{aligned}$$

(b) We slice the solid of rotation into circular disks of width $\mathrm{d}x$ and radius $\dfrac{|x|}{\sqrt[4]{1-x^2}}$.

$$\begin{aligned} \text{Volume}&=\int_{-1/2}^{1/2} \pi\left(\dfrac{|x|}{\sqrt[4]{1-x^2}}\right)^2~\mathrm{d}x\\ &=2\pi\int_{0}^{1/2} \dfrac{x^2}{\sqrt{1-x^2}}~\mathrm{d}x \\ \text{We use the substitution $x=\sin \theta$, $\mathrm{d}x = \cos\theta~\mathrm{d}\theta$, so $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta}=\cos \theta$. Note $\sin 0 =0$ and $\sin\frac{\pi}{6}=\frac{1}{2}.$} \\ &=2\pi\int_{0}^{\pi/6} \frac{\sin^2 \theta}{\cos \theta}\cos \theta~\mathrm{d}\theta \\&=2\pi\int_{0}^{\pi/6} \sin^2 \theta~\mathrm{d}\theta \\&=\pi\int_{0}^{\pi/6}\big(1- \cos(2 \theta)\big)~\mathrm{d}\theta\\ &=\pi\left[\theta - \frac{1}{2}\sin(2\theta)\right]_0^{\pi/6}\\ &=\pi\left(\frac{\pi}{6} - \frac{1}{2}\cdot \frac{\sqrt{3}}{2}\right)\\ &=\frac{\pi^2}{6} - \frac{\sqrt{3}\pi}{4} \end{aligned}$$