Volumes of Solids of Revolution Method of Rings — Question 4

Question 4 – Solution

1. Find the bounds and choose slices.

The curves meet where $\sqrt{x}=x$. With $x\ge0$, squaring gives $x=x^2$, so $x=0$ or $x=1$. Use vertical slices for rotation about the $x$-axis.

2. Measure the radii.

For $0\le x\le1$, $\sqrt{x}\ge x\ge0$. Thus the square-root curve is farther from the axis:$$R(x)=\sqrt{x},\qquad r(x)=x.$$

3. Set up the washer integral.

Subtract the squared radii, then multiply by $\pi$:$$V=\pi\int_0^1\bigl(R^2-r^2\bigr)\,dx=\pi\int_0^1\bigl((\sqrt{x})^2-x^2\bigr)\,dx=\pi\int_0^1(x-x^2)\,dx.$$

4. Find the antiderivative.

$$\int x\,dx=\frac{x^2}{2},\qquad\int x^2\,dx=\frac{x^3}{3}.$$$$V=\pi\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1.$$

5. Apply the bounds and simplify.

$$V=\pi\left[\left(\frac12-\frac13\right)-0\right]=\pi\left(\frac{3-2}{6}\right)=\boxed{\frac\pi6}.$$

All volumes are in cubic units.