More Volume Problems — Question 3

Question 3 – Solution

1. Identify the diameter and radius.

Use $dx$ slices for $0\le x\le\pi$. The base segment is a diameter, so$$d(x)=\sin x-0=\sin x,\qquad r(x)=\frac{d(x)}2=\frac{\sin x}{2}.$$

2. Find the area of a semicircle.

$$A(x)=\frac12\pi r(x)^2=\frac12\pi\left(\frac{\sin x}{2}\right)^2=\frac\pi8\sin^2x.$$

3. Set up the volume and use a trigonometric identity.

$$V=\frac\pi8\int_0^\pi\sin^2x\,dx,\qquad\sin^2x=\frac{1-\cos(2x)}2.$$$$V=\frac\pi{16}\int_0^\pi(1-\cos(2x))\,dx.$$

4. Integrate with the chain-rule factor.

Since $\int\cos(2x)\,dx=\sin(2x)/2$,$$V=\frac\pi{16}\left[x-\frac{\sin(2x)}2\right]_0^\pi.$$

5. Apply the bounds.

$$V=\frac\pi{16}\left[\left(\pi-\frac{\sin(2\pi)}2\right)-\left(0-\frac{\sin0}2\right)\right]=\boxed{\frac{\pi^2}{16}}.$$

All volumes are in cubic units.