Volume of revolution where am i going wrong?

[helpmeplz!]

Hey guys I'm stuck on this problem. Its an easy one but I need some help..

It's asking for the volume of the solid obtained by rotating the region bounded by y=x^2, y=4, x=0 about the line x= -2 using the shell method.

I got the answer correct using the disk method (answer is 136 pi/3)..

With shell method I get integral 2pi (2+x) (4-x^2) dx from x=0 to x=2 which is 88pi/3. Can someone please help me out of this mess?

 

[mathman]

You need to supply your work details.

 

[helpmeplz!]

What do you mean? My question is what part of the integral that I set up is wrong and why? Not the evaluating of the integral which I'm using a calculator for anyway.

 

[helpmeplz!]

Hmmm... are you sure that is the correct answer? I got $\frac{88 \pi}{3}$ for both

Using slices, I get integral pi * (2+sqrty)^2 dy from y=0 to 4. which gives 136 pi/3?

 

[jbrussell93]

There should be a hole in the center due to the gap from x=-2 to x=0. I'll leave the rest up to you

 

[helpmeplz!]

There should be a hole in the center due to the gap from x=-2 to x=0. I'll leave the rest up to you

Right, but the shell method takes that into account doesn't it? My integral only foes from 0 to 2..

 

[jbrussell93]

Yes the shell method does, so I think you got the correct answer there. Think about the disk method and what your volume would look like. (Yours is missing the hole)

 

[helpmeplz!]

Yes the shell method does, so I think you got the correct answer there. Think about the disk method and what your volume would look like. (Yours is missing the hole)

Oh yes, that's true I guess khan academy is wrong here?
http://www.khanacademy.org/math/cal...d/v/disc-method-rotating-around-vertical-line

 

[jbrussell93]

Yeah, it looks like it... Unless I'm wrong, which is entirely possible :P

 

[jbrussell93]

Actually, I don't think he ever specifies x=0 as a bound. So assuming x=-2 is the bound instead, it would be correct.

 

[Redbelly98]

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