Washer method and shell method (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?
 A: 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.
