More Volume Problems — Question 1

Question 1 – Solution

1. Find the intersections.

$$x^2=2-x\quad\Longrightarrow\quad x^2+x-2=(x+2)(x-1)=0.$$Thus $x=-2$ and $x=1$, giving intersection points $(-2,4)$ and $(1,1)$.

2. Choose washers and identify the radii.

Use vertical slices perpendicular to the $x$-axis. On $[-2,1]$, $0\le x^2\le2-x$, so$$R(x)=2-x,\qquad r(x)=x^2.$$

3. Set up the volume and expand.

$$V=\pi\int_{-2}^1\bigl(R^2-r^2\bigr)dx=\pi\int_{-2}^1\bigl((2-x)^2-x^4\bigr)dx.$$$$(2-x)^2-x^4=4-4x+x^2-x^4.$$

4. Find an antiderivative.

Using the power rule for each term,$$F(x)=4x-2x^2+\frac{x^3}{3}-\frac{x^5}{5},\qquad V=\pi[F(x)]_{-2}^1.$$

5. Evaluate the endpoints and subtract.

$$F(1)=4-2+\frac13-\frac15=\frac{32}{15},\qquad F(-2)=-8-8-\frac83+\frac{32}{5}=-\frac{184}{15}.$$$$V=\pi\left(\frac{32}{15}+\frac{184}{15}\right)=\boxed{\frac{72\pi}{5}}.$$

All volumes are in cubic units.