(1) Sketch the region enclosed between the curve $y=sin^2x$ and the straight line $y =2x/π$
(2) Find the volume of the solid $S$ obtained by revolving the region in part (1) about the $y$-axis by using
(a)the washer method
(b) the shell method
For (1),
For (2), I have no idea what I should do. Would anyone mind telling me how to solve it by using washer method and shell method?
For the shell method, the height $h(x)$ of the shell is $$ h(x) = \left\{\begin{array}{ll} 2x/\pi-\sin^{2}(x), & 0 \le x \le \pi/4 \\ \sin^{2}(x)-2x/\pi, & \pi/4 \le x \le \pi/2. \end{array}\right. $$ So the volume is $$ \begin{align} V = \int_{0}^{\pi/2}2\pi h(x)x^{2}\,dx & = \int_{0}^{\pi/4}2\pi(2x/\pi-\sin^{2}(x))x^{2}\,dx \\ & +\int_{\pi/4}^{\pi/2}2\pi(\sin^{2}(x)-2x/\pi)x^{2}\,dx. \end{align} $$ This can be solved using $\cos(2x)=\cos^{2}(x)-\sin^{2}(x)=1-2\sin^{2}(x)$ or $\sin^{2}(x)=\frac{1}{2}(1-\cos(2x))$. So, $$ \int \sin^{2}(x)x^{2}\,dx = \frac{1}{2}\int(1-\cos(2x))x^{2}\,dx = \frac{x^{3}}{6}-\frac{1}{2}\int\cos(2x)x^{2}\,dx, $$ which can be evaluated using integration by parts.
For the washer method, the area of the washer is $$ A(y) = \left\{\begin{array}{cc} \pi(\sin^{-1}(\sqrt{y}))^{2}-\pi(\pi y/2)^{2}, & 0 \le y \le 1/2 \\ \pi(\pi y/2)^{2}-\pi(\sin^{-1}(\sqrt{y}))^{2}, & 1/2 \le y \le 1. \end{array}\right. $$ So the volume is $$ \begin{align} V = \int_{0}^{1}A(y)\,dy & = \pi\int_{0}^{1/2}(\sin^{-1}(\sqrt{y}))^{2}-(\pi y/2)^{2})\,dy \\ & + \pi\int_{1/2}^{1}((\pi y/2)^{2})^{2}-(\sin^{-1}(\sqrt{y}))^{2})\,dy. \end{align} $$ The substitution $x = \sin^{-1}(\sqrt{y})$, or $y=\sin^{2}(x)$ seems to be suggested.