# Finding the volume of a region rotated about the y-axis.

I'm having trouble trying to find the volume of the region formed by $y = x^2-7x+10$ and $y = x+3$ rotated about the y-axis. I was able to graph it, but I'm having difficulty when trying to come up with the integral for it.

Any help showing me how to get the integral I should use would be a huge help.

## 2 Answers

This one may be easier to handle using the Method of Cylindrical Shells. Setting $x+3=x^2-7x+10$ we find that the curves meet at $x=1$ and $x=7$. Over the interval $[1,7]$, the line $y=x+3$ is the upper curve. When we take a vertical strip of width "$dx$" from $x$ to $x+dx$, and rotate it, we get a cylindrical shell of radius $x$ and height $(x+3)-(x^2-7x+10)=-x^2+8x-7$. Thus the volume is $$\int_1^7 2\pi x(-x^2+8x-7)\,dx.$$ For the integration, multiply out and integrate term by term.

• that is much tidier! – David Holden Aug 12 '14 at 11:41
• Thanks so much, this cleared a lot up! Seems that I was mistaking the y axis with a boundary which skewed a lot. – Jeffrey Aug 14 '14 at 11:50
• You are welcome. With a clear understanding of the geometry, the rest is often straightforward. – André Nicolas Aug 14 '14 at 14:01

hint: you must split the integration. for $y \in [-\frac94,4]$ the radii of the inner and outer annuli are the two roots of $x^2-7x+(10-y)=0$, whereas for $y \in [4,10]$ the outer radius is the larger root of that equation, but the inner radius is given by $y-3$