Volumes of Solids of Revolution Method of Rings — Question 5

Question 5 – Solution

1. Identify the bounds and slice direction.

The given left boundary is $x=0$. The lines meet where $2-x=x$, giving $x=1$. Use vertical slices, perpendicular to the $x$-axis.

2. Measure the radii from the $x$-axis.

On $[0,1]$, $2-x\ge x\ge0$, so$$R(x)=2-x,\qquad r(x)=x.$$The rotated slice is a washer with area $\pi(R^2-r^2)$.

3. Set up and simplify the volume integral.

$$V=\pi\int_0^1\left[(2-x)^2-x^2\right]dx.$$$$(2-x)^2-x^2=(4-4x+x^2)-x^2=4-4x.$$

4. Integrate term by term.

$$\int4\,dx=4x,\qquad\int4x\,dx=4\frac{x^2}{2}=2x^2.$$$$V=\pi\left[4x-2x^2\right]_0^1.$$

5. Apply the bounds.

$$V=\pi\bigl[(4(1)-2(1)^2)-(4(0)-2(0)^2)\bigr]=\pi(4-2)=\boxed{2\pi}.$$

All volumes are in cubic units.