Volumes of Solids of Revolution Method of Rings — Question 10

Question 10 – Solution

1. Find the bounds and choose horizontal slices.

The intersections satisfy $x=x^2$, so $x=0,1$ and $y=0,1$. Rotation about the vertical line $x=1$ requires horizontal slices of thickness $dy$.

2. Rewrite the curves and measure radii.

The boundaries are $x=y$ on the left and $x=\sqrt{y}$ on the right. The axis is to their right, so the farther, left boundary gives the outer radius:$$R(y)=1-y,\qquad r(y)=1-\sqrt{y}.$$

3. Set up the washer integral and expand.

$$V=\pi\int_0^1\left[(1-y)^2-(1-\sqrt{y})^2\right]dy.$$$$\begin{aligned}(1-y)^2-(1-\sqrt{y})^2&=(1-2y+y^2)-(1-2\sqrt{y}+y)\\&=y^2-3y+2y^{1/2}.\end{aligned}$$

4. Integrate using the power rule.

In particular, $\int2y^{1/2}\,dy=2\cdot\frac23y^{3/2}=\frac43y^{3/2}$. Therefore,$$V=\pi\left[\frac{y^3}{3}-\frac{3y^2}{2}+\frac43y^{3/2}\right]_0^1.$$

5. Apply the bounds and simplify.

$$V=\pi\left(\frac13-\frac32+\frac43-0\right)=\pi\left(\frac{2-9+8}{6}\right)=\boxed{\frac\pi6}.$$

All volumes are in cubic units.