consider the region $r$ bounded above by $y = \\frac{\\pi}{2}$ and below by the curve $\\sin y=\\sqrt{x}$…

Explanation:

Step1: Recall the washer - method formula

The formula for the volume $V$ of a solid of revolution about the $x$-axis using the washer method is $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer - radius and $r(x)$ is the inner - radius. We need to express $y$ in terms of $x$ for the curve $\sin y=\sqrt{x}$, so $y = \sin^{- 1}(\sqrt{x})$. The upper - bound of $x$ is $x = 1$ (when $y=\frac{\pi}{2}$, $\sin y = 1=\sqrt{x}$ implies $x = 1$) and the lower - bound is $x = 0$. The outer - radius $R(x)=\frac{\pi}{2}$ and the inner - radius $r(x)=\sin^{-1}(\sqrt{x})$.

Step2: Set up the integral for washer method

$$V=\pi\int_{0}^{1}\left(\left(\frac{\pi}{2}\right)^{2}-(\sin^{-1}(\sqrt{x}))^{2}\right)dx$$

Step3: Recall the shell - method formula

The formula for the volume $V$ of a solid of revolution about the $x$-axis using the shell method is $V = 2\pi\int_{c}^{d}y\left(x_{right}-x_{left}\right)dy$. We need to express $x$ in terms of $y$. From $\sin y=\sqrt{x}$, we have $x=\sin^{2}y$. The lower - bound of $y$ is $y = 0$ and the upper - bound is $y=\frac{\pi}{2}$. Here, $x_{right}=\sin^{2}y$ and $x_{left}=0$.

Step4: Set up the integral for shell method

$$V = 2\pi\int_{0}^{\frac{\pi}{2}}y\sin^{2}y;dy$$

Answer:

a) $\pi\int_{0}^{1}\left(\left(\frac{\pi}{2}\right)^{2}-(\sin^{-1}(\sqrt{x}))^{2}\right)dx$ b) $2\pi\int_{0}^{\frac{\pi}{2}}y\sin^{2}y;dy$