More Volume Problems — Question 9

Question 9 – Solution

1. Find the bounds.

The intersections satisfy $|x|=2$, giving $x=-2$ and $x=2$. Use $dx$ slices because the cross-sections are perpendicular to the $x$-axis.

2. Identify the rectangle dimensions.

The base length is top minus bottom, and the height is three times the base:$$b(x)=2-|x|,\qquad h(x)=3(2-|x|).$$

3. Set up the area and volume.

$$A(x)=b(x)h(x)=3(2-|x|)^2,\qquad V=3\int_{-2}^2(2-|x|)^2dx.$$

4. Use symmetry and integrate.

The integrand is even; on $[0,2]$, $|x|=x$. Thus$$V=6\int_0^2(2-x)^2dx=6\int_0^2(4-4x+x^2)dx.$$$$V=6\left[4x-2x^2+\frac{x^3}{3}\right]_0^2.$$

5. Apply the bounds.

$$V=6\left[(8-8+\frac83)-0\right]=6\cdot\frac83=\boxed{16}.$$

All volumes are in cubic units.