# Volume of a solid of revolution with multiple bounding functions

I have a homework problem (in a Calc 2 course) that asks me to calculate the volume of the solid of revolution formed by rotating the following three curves around the x axis:

$x=5\sqrt{y}$, $x=-5y$, $y=1$.

I'm comfortable calculating the volumes of these regions when the upper and lower bounds are a single curve. In this case, however, the lower bound of this region is the line $x=-5y$ to the left of the y-axis, and $x=5\sqrt{y}$ to the right of the Y axis. Please help me understand how to set up this integral.

• Just break it up into two integrals, each with single-function bounds, then sum their results. – dfan Mar 3 '14 at 16:23

It is wise to split up the integral into two parts. We can draw the region in a "reversed axis" situation in which you will find that we have two functions, $f: x = 5\sqrt{y}$ and $g: x = -5y$. Then rotate them around the x-axis (which will be vertical in this case) with bounds from 0 to either $f$ or $g$.