Volumes of Solids of Revolution Method of Rings — Question 1

Question 1 – Solution

1. Choose slices and bounds.

The rotation axis is horizontal, so use vertical slices of thickness $dx$. The stated interval is $0\le x\le4$. Each slice reaches the axis, producing a disk.

2. Measure the radii from the axis.

The distance from $y=0$ to $y=\sqrt{x}$ is the outer radius. There is no central hole.$$R(x)=\sqrt{x}-0=\sqrt{x},\qquad r(x)=0.$$

3. Write the cross-sectional area and volume.

$$A(x)=\pi R(x)^2-\pi r(x)^2=\pi\bigl((\sqrt{x})^2-0^2\bigr)=\pi x.$$$$V=\int_0^4 A(x)\,dx=\pi\int_0^4 x\,dx.$$

4. Integrate using the power rule.

$$\int x\,dx=\frac{x^{1+1}}{1+1}=\frac{x^2}{2},\qquad V=\pi\left[\frac{x^2}{2}\right]_0^4.$$

5. Apply the bounds and simplify.

$$V=\pi\left(\frac{4^2}{2}-\frac{0^2}{2}\right)=\pi\left(\frac{16}{2}\right)=\boxed{8\pi}.$$

All volumes are in cubic units.