Finding volume using disks/washers

Question

Find the volume of a solid obtained by rotating the region bounded by $y = \sqrt{x}$ and $y=x$ about $y=2$.

I need help setting up this problem. Thank you!

Answer 1

Here, I am going to use the washer method because that is often easier when there are more than two curves.

First, graph on a graphing calculator or visualize the graphs of $y=\sqrt x$ and $y=x$. It is easy to see that $y=\sqrt x$ is closer to $y=2$ than $y=x$. Also, the region between $y=\sqrt x$ and $y=x$ starts at $x=0$ and ends at $x=1$. Therefore, since $y=\sqrt x$ is closer to $y=2$, the inside function is $2-\sqrt x$ (or $\sqrt x-2$; it does not matter since we are going to square this function, which will lead to the same result both ways). Also, since $y=x$ is farther from $y=2$, the outside function is $2-x$.

Thus, we have to integrate $[R_O(x)]^2-[R_I(x)]^2$ ($R_O$ is the outside function and $R_I$ is the inside function) from the beginning of the region, $0$, to the end of the region, $1$, and then multiply that by $\pi$, which gives us: $$\pi \int_0^1 \{[2-x]^2-[2-\sqrt x]^2\}dx$$

Thus, you need to solve that definite integral. Good luck!