# Disc method and shell method giving different solutions

• Oct 4th 2013, 01:05 PM
Latsabb
Disc method and shell method giving different solutions
I have the function f(x)=x2

I am to give the volume of the shape when this function is spun 360 degrees around the x axis by using the disc method, and the shell method, from x=1 and x=3.

When I work out the disc method, I get (242pi)/5. But when I use the shell method, I get (232pi)/5. I have been looking over my work, and it all seems to work out, so I am curious why they are different. I will show my setups below:

Disc method:
v=pi*r2*h
r=y=f(x)=x2
h=dx
v=pi(x4) dx

Then I integrate and get that v=pi[(1/5)x5]31 which gives me (242pi)/5.

Shell method:
v=2pi*r*h dy
r=y
h=3-f(y)=3-sqrt(y)
v=2pi(3y-y3/2) dy

I then integrate, and get that v=2pi[(3/2)y2-(2/5)y5/2]91 which then gives me (232pi)/5.

Have I done something wrong here? Or is there a reason for the discrepancy?
• Oct 4th 2013, 03:23 PM
tom@ballooncalculus
Re: Disc method and shell method giving different solutions
Check your limits for the shells ;)
• Oct 5th 2013, 07:34 AM
SlipEternal
Re: Disc method and shell method giving different solutions
Quote:

Originally Posted by Latsabb
Shell method:
v=2pi*r*h dy
r=y
h=3-f(y)=3-sqrt(y)
v=2pi(3y-y3/2) dy

I then integrate, and get that v=2pi[(3/2)y2-(2/5)y5/2]91 which then gives me (232pi)/5.

Have I done something wrong here? Or is there a reason for the discrepancy?

As tom said, your limits of integration are wrong for the shell method. Additionally, the h you use is only correct on the limits you currently have for the integration.

Hint: Currently, you are finding the volume of a shape that has a cylindrical hole through the center.
• Oct 5th 2013, 08:38 AM
Latsabb
Re: Disc method and shell method giving different solutions
Yes, I worked it over, and saw my mistake, that I needed to take it from 0 to 9, and then subtract the small "spike" from the tip of the cone. Thank you for the hints.