# How do you find the volume of the solid obtained by rotating the region bounded by the curves y=x^2 and y=2-x^2 and x=0 about the line x=1?

Oct 24, 2015

I would use shells (to avoid doing two integrals).

#### Explanation:

Here is a picture of the region. I have included the axis of rotation ($x = 1$) as a dashed black line, a representative slice taken parallel to the axis of rotation (solid black segments inside the region) and the radius of the rotation (red line segment taken at about $y = 1.7$).

When we rotate the slice, we'll get a cylindrical shell.

The representative cylindrical shell will have volume:

$2 \pi \times \text{radius"xx"height"xx"thickness}$.

Here we have:
$\text{radius} = r = \left(1 - x\right)$.

The height will be the greater $y$ value minus the lesser (the top y minus the bottom y).
$\text{height" = y_"top"-y_"bottom} = \left(2 - {x}^{2}\right) - \left({x}^{2}\right) = \left(2 - 2 {x}^{2}\right)$

$\text{thickness} = \mathrm{dx}$

We also not that in the region, $x$ varies from $0$ to $1$

Using shell, the volume of the solid is found by evaluating:

$V - {\int}_{a}^{b} 2 \pi \times \text{radius"xx"height"xx"thickness}$

So we need

$V - {\int}_{0}^{1} 2 \pi \left(1 - x\right) \left(2 - 2 {x}^{2}\right) \mathrm{dx}$

$= 2 \pi {\int}_{0}^{1} \left(2 - 2 x - 2 {x}^{2} + 2 {x}^{3}\right) \mathrm{dx}$

$= \frac{5}{3} \pi$