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Volume of revolution where am i going wrong?

  1. Jul 7, 2013 #1
    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?
  2. jcsd
  3. Jul 7, 2013 #2


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    You need to supply your work details.
  4. Jul 7, 2013 #3
    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.
  5. Jul 7, 2013 #4
    Using slices, I get integral pi * (2+sqrty)^2 dy from y=0 to 4. which gives 136 pi/3?
  6. Jul 7, 2013 #5
    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
  7. Jul 7, 2013 #6
    Right, but the shell method takes that into account doesn't it? My integral only foes from 0 to 2..
  8. Jul 7, 2013 #7
    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)
  9. Jul 7, 2013 #8
  10. Jul 7, 2013 #9
    Yeah, it looks like it... Unless I'm wrong, which is entirely possible :P
  11. Jul 7, 2013 #10
    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.
  12. Jul 7, 2013 #11


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    Moderator's note:

    Thread moved from Mathematics > Calculus to
    Homework & Coursework Questions > Calculus & Beyond Homework

    The usual homework forum guidelines are in effect.
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