# Thread: Stuck on setting up a cylindrical shells problem!

1. ## Stuck on setting up a cylindrical shells problem!

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curve about the x-axis.
y = 4x^2, 2x+y=6

I have set these two equations equal to each other and found their points of intersection is (-3/2, 9) and (1,4). Since this is rotating about the x-axis dy should be used.

So I have set up the problem as

However when I integrate that the answer is not like the back of the book. The answer according to the back of the book is (250pi)/3.

What am I doing wrong? I could get that answer easily with the disk/washers method but I don't know where I went wrong on doing this with the cylindrical shells. Please help me. Thank you.

2. Originally Posted by florx
Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curve about the x-axis.
y = 4x^2, 2x+y=6

I have set these two equations equal to each other and found their points of intersection is (-3/2, 9) and (1,4). Since this is rotating about the x-axis dy should be used.

So I have set up the problem as

However when I integrate that the answer is not like the back of the book. The answer according to the back of the book is (250pi)/3.

What am I doing wrong? I could get that answer easily with the disk/washers method but I don't know where I went wrong on doing this with the cylindrical shells. Please help me. Thank you.
Shells is really not a good way to do this problem. You need to break this into two different integrals!

$\int_{0}^{4}2\pi y \left( \frac{\sqrt{y}}{2}-\frac{-\sqrt{y}}{2}\right)dy+\int_{4}^{9}2\pi y \left( \frac{6-y}{2}-\frac{-\sqrt{y}}{2}\right)dy=\frac{250}{3}\pi$

3. I've had similar problems to this and I could solve the problem with just one integral but what indication told you it would need two integrals? Please explain. Thank you thus far for helping me.

4. Originally Posted by florx
I've had similar problems to this and I could solve the problem with just one integral but what indication told you it would need two integrals? Please explain. Thank you thus far for helping me.
Since we are rotating around the x axis the height of the shells are parallel to the x axis and bounded by the curves.

Notice that when y=4 the curve that is on top changes. Remember that we are integrating with respect to y, so the curve "on top" is the one on the right. Notice the blue and red lines in the attachment.

Shells.pdf