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

Aug 10, 2017

#### Explanation:

Here is a graph of the region.

I've taken a slice perpendicular to the axis of rotation. The slice is taken at a variable value of $y$.

The thickness of the slice is $\mathrm{dy}$, so we need the equations in the form $x =$ a function of $y$.

The curve on the left ($y = \sqrt{x}$) is $x = {y}^{2}$

on the right is the line $x = y$

Rotating the slice will generate a washer of thickness $\mathrm{dy}$ and
volume $\pi \left({R}^{2} - {r}^{2}\right) \mathrm{dy}$
where $r$ is the outer radius and $r$ the inner.

The outer radius of the washer is the distance between the curve on the left and the line $x = 2$.
So $R = 2 - {y}^{2}$

The inner radius is the distance from the line on the right and the line $x = 2$.
So $R = 2 - y$

$y$ varies from $0$ to $1$.

The volume of the representative washer is

${\int}_{0}^{1} \pi \left({\left(2 - {y}^{2}\right)}^{2} - {\left(2 - y\right)}^{2}\right) \mathrm{dy} = \pi {\int}_{0}^{1} \left({y}^{4} - 5 {y}^{2} + 4 y\right) \mathrm{dy}$

evaluate to get

$= \pi \left(\frac{8}{15}\right) = \frac{8 \pi}{15}$