Volumes of Solids of Revolution Method of Rings — Question 8

Question 8 – Solution

1. Find the intersections and $y$-bounds.

$$x^2=2x\quad\Longrightarrow\quad x(x-2)=0\quad\Longrightarrow\quad x=0,2.$$The intersection points are $(0,0)$ and $(2,4)$, so integrate from $y=0$ to $y=4$.

2. Use horizontal slices and identify radii.

Rewrite the curves as $x=\sqrt{y}$ and $x=y/2$. For $0\le y\le4$, $\sqrt{y}\ge y/2$. Distances from the $y$-axis give$$R(y)=\sqrt{y},\qquad r(y)=\frac y2.$$

3. Set up the washer integral.

$$V=\pi\int_0^4\left[(\sqrt{y})^2-\left(\frac y2\right)^2\right]dy=\pi\int_0^4\left(y-\frac{y^2}{4}\right)dy.$$

4. Integrate term by term.

$$\int y\,dy=\frac{y^2}{2},\qquad\int\frac{y^2}{4}\,dy=\frac14\frac{y^3}{3}=\frac{y^3}{12}.$$$$V=\pi\left[\frac{y^2}{2}-\frac{y^3}{12}\right]_0^4.$$

5. Apply the bounds and simplify.

$$V=\pi\left(\frac{16}{2}-\frac{64}{12}-0\right)=\pi\left(8-\frac{16}{3}\right)=\boxed{\frac{8\pi}{3}}.$$

All volumes are in cubic units.