Volumes of Solids of Revolution Method of Rings — Question 2

Question 2 – Solution

1. Choose slices and confirm the bounds.

Use vertical slices, perpendicular to the $x$-axis. The curve meets $y=0$ where$$4-x^2=0\quad\Longrightarrow\quad x=\pm2.$$

2. Identify the radii.

On $[-2,2]$, $4-x^2\ge0$. The region touches the axis, so each cross-section is a disk:$$R(x)=4-x^2,\qquad r(x)=0.$$

3. Set up the volume and expand the square.

$$V=\pi\int_{-2}^2\bigl(R^2-r^2\bigr)\,dx=\pi\int_{-2}^2(4-x^2)^2\,dx.$$$$(4-x^2)^2=16-8x^2+x^4.$$

4. Use symmetry and integrate.

The integrand is even, so double the integral over $[0,2]$:$$V=2\pi\int_0^2(16-8x^2+x^4)\,dx=2\pi\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_0^2.$$

5. Substitute and combine the fractions.

$$\begin{align*} V&=2\pi\left(32-\frac{64}{3}+\frac{32}{5}\right)\\&=2\pi\left(\frac{480-320+96}{15}\right)=\boxed{\frac{512\pi}{15}}. \end{align*}$$

All volumes are in cubic units.