Volumes of Solids of Revolution Method of Rings — Question 9

Question 9

Find the volume of the solid obtained by rotating the region bounded by $$y=x^2 \quad\text{and}\quad y=2x,$$ about the horizontal line $$y=-1.$$

Question 9 – Solution

1. Find the bounds and choose vertical slices.

The curves meet where $x^2=2x$, or $x(x-2)=0$. Thus $0\le x\le2$. The horizontal rotation axis $y=-1$ requires vertical slices of thickness $dx$.

2. Measure both radii from $y=-1$.

On $[0,2]$, $2x\ge x^2$. The outer radius reaches the line, and the inner radius reaches the parabola:$$R(x)=2x-(-1)=2x+1,\qquad r(x)=x^2-(-1)=x^2+1.$$

3. Set up the washer integral and expand.

$$V=\pi\int_0^2\left[(2x+1)^2-(x^2+1)^2\right]dx.$$$$\begin{aligned}(2x+1)^2-(x^2+1)^2&=(4x^2+4x+1)-(x^4+2x^2+1)\\&=-x^4+2x^2+4x.\end{aligned}$$

4. Integrate each power.

$$V=\pi\left[-\frac{x^5}{5}+\frac{2x^3}{3}+2x^2\right]_0^2.$$The coefficients follow from $\int x^n\,dx=x^{n+1}/(n+1)$.

5. Evaluate and use a common denominator.

$$\begin{align*} V&=\pi\left(-\frac{32}{5}+\frac{16}{3}+8-0\right)\\&=\pi\left(\frac{-96+80+120}{15}\right)=\boxed{\frac{104\pi}{15}}. \end{align*}$$

All volumes are in cubic units.