Volumes of Solids of Revolution Method of Rings — Question 3

Question 3

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

Question 3 – Solution

1. Find the bounds and choose slices.

The intersections satisfy $x^2=1$, so $x=-1$ and $x=1$. Use vertical slices of thickness $dx$ because the rotation axis is horizontal.

2. Identify outer and inner radii.

On $[-1,1]$, $0\le x^2\le1$. Measured from the $x$-axis,$$R(x)=1,\qquad r(x)=x^2.$$The missing inner disk must be subtracted from the outer disk.

3. Set up the washer integral.

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

4. Integrate term by term.

$$\int1\,dx=x,\qquad\int x^4\,dx=\frac{x^5}{5}.$$$$V=\pi\left[x-\frac{x^5}{5}\right]_{-1}^1.$$

5. Evaluate both endpoints.

$$\begin{align*} V&=\pi\left[\left(1-\frac15\right)-\left(-1+\frac15\right)\right]\\&=\pi\left(\frac45+\frac45\right)=\boxed{\frac{8\pi}{5}}. \end{align*}$$

All volumes are in cubic units.