Volume of a Solid by Plane Slicing

[Guest]

The problem is:

Let R be the region under the graph of y=x^1.5 from x=1 to x=9. Find the exact volume of the solid generated by rotating R about the x-axis.


Any help would be greatly appreciated! Thank you!

 

[galactus]

Let's see. You can use disks or shells:

I would use the disks.

The slices are parallel to the y-axis. You are adding up all the 'slices' from

1 to 9 along the x-axis. The area of a disk is given by.

\(\displaystyle {\pi}r^{2}\). Therefore, letting f(x) be r and dx the thickness,

we have \(\displaystyle {\pi}(f(x))^{2}dx\).

So, your solid of revolution can be

\(\displaystyle {\pi}\int_{1}^{9}(x^{\frac{3}{2}})^{2}dx\)

Another way is cylindrical shells. It's a little messier than this one.

One way for shells is:

\(\displaystyle 2{\pi}\int_{0}^{27}y(9-y^{\frac{2}{3}})dy-2{\pi}\int_{0}^{1}y(1-y^{\frac{2}{3}})dy\)