Disc method and shell method giving different solutions

I have the function f(x)=x^{2}

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*r^{2}*h

r=y=f(x)=x^{2}

h=dx

v=pi(x^{4}) dx

Then I integrate and get that v=pi[(1/5)x^{5}]^{3}_{1 }which gives me (242pi)/5.

Shell method:

v=2pi*r*h dy

r=y

h=3-f(y)=3-sqrt(y)

v=2pi(3y-y^{3/2}) dy

I then integrate, and get that v=2pi[(3/2)y^{2}-(2/5)y^{5/2}]^{9}_{1} which then gives me (232pi)/5.

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

Re: Disc method and shell method giving different solutions

Check your limits for the shells ;)

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-y^{3/2}) dy

I then integrate, and get that v=2pi[(3/2)y^{2}-(2/5)y^{5/2}]^{9}_{1} 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.

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.