Volume of a bounded region

Question

How do you find the volume of a region bounded by $y=\sqrt{x}, y=x-2$ and $y=0$ when rotated around the $x$-axis?

Answer 1

Here is the region you need to revolve about the $x$-axis:

Your volume integral will therefore be $$V=\int_0^2 \pi(\sqrt{x})^2 \, dx+\int_2^4 \pi[(\sqrt{x})^2-(x-2)^2] \, dx.$$