Volumes of Solids of Revolution Method of Cylinders — Question 3

Question 3 – Solution

1. Locate the region and rotation axis.

The parabola meets $y=0$ at $x=\pm2$. The axis $x=-1$ passes through the region. Vertical strips on opposite sides can rotate into the same shell.

2. Count each radius once.

Let $r$ be distance from $x=-1$. The right strip is at $x=-1+r$, with $0\le r\le3$. Its height is $4-(r-1)^2$. For $0\le r\le1$, a left strip also exists at $x=-1-r$, with height $4-(r+1)^2$.$$\bigl[4-(r-1)^2\bigr]-\bigl[4-(r+1)^2\bigr]=4r\ge0.$$Both strips start at $y=0$, so the taller right strip includes the entire left strip after rotation. Use only the right strip.

3. Set up the shell integral.

$$h(r)=4-(r-1)^2=3+2r-r^2.$$$$V=2\pi\int_0^3 r(3+2r-r^2)\,dr=2\pi\int_0^3(3r+2r^2-r^3)\,dr.$$

4. Integrate each power.

$$V=2\pi\left[\frac32r^2+\frac23r^3-\frac14r^4\right]_0^3.$$

5. Apply the bounds.

$$V=2\pi\left(\frac{27}{2}+18-\frac{81}{4}\right)=2\pi\left(\frac{54+72-81}{4}\right)=\boxed{\frac{45\pi}{2}}.$$

All volumes are in cubic units.