Volumes of Solids of Revolution Method of Rings — Question 6

Question 6 – Solution

1. Choose slices and identify the bounds.

Use vertical slices over $0\le x\le\pi$. Since $\sin x\ge0$ on this interval and the region touches the $x$-axis, rotation gives disks.

2. Identify radii and set up the volume.

$$R(x)=\sin x,\qquad r(x)=0.$$$$V=\pi\int_0^\pi\bigl((\sin x)^2-0^2\bigr)dx=\pi\int_0^\pi\sin^2x\,dx.$$

3. Use the power-reduction identity.

$$\sin^2x=\frac{1-\cos(2x)}2\quad\Longrightarrow\quad V=\frac\pi2\int_0^\pi(1-\cos(2x))\,dx.$$

4. Integrate, including the chain-rule factor.

Because $\dfrac{d}{dx}\sin(2x)=2\cos(2x)$,$$\int\cos(2x)\,dx=\frac{\sin(2x)}2,\qquad V=\frac\pi2\left[x-\frac{\sin(2x)}2\right]_0^\pi.$$

5. Evaluate at $\pi$ and $0$.

$$\begin{align*} V&=\frac\pi2\left[\left(\pi-\frac{\sin(2\pi)}2\right)-\left(0-\frac{\sin0}2\right)\right]\\&=\frac\pi2(\pi-0)=\boxed{\frac{\pi^2}{2}}. \end{align*}$$

All volumes are in cubic units.