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Homework Help: Need help finding Volume bound by curves.

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data

    The region bounded by the given curves is rotated about x = 10

    [tex]
    x=1-y^{4}, x=0
    [/tex]

    Find the Volume V of the resulting solid by any method.

    2. Relevant equations



    3. The attempt at a solution

    I'm using the washer method. Not sure if it is being setup properly as I'm getting the wrong answer. I will show my integration steps if the problem is setup correctly. Thanks.

    [tex]
    \begin{alignl*}
    1-y^{4}=0 \\ \\

    y^{4}=-1 \\ \\

    \Rightarrow y=\pm1 \\ \\

    10^{2}-((1-y^{4}))^2 \\ \\

    \pi \int_{-1}^{1} (10^{2}-((1-y^{4}))^2 dy \\ \\

    = \frac{8936\pi}{45} \\
    \end{alignl*}
    [/tex]

    Hmm can't edit my LaTex code. That integral should have Pi out front.
     
    Last edited: Feb 28, 2010
  2. jcsd
  3. Feb 28, 2010 #2

    Dick

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    10 is the outer radius alright. I don't think (1-y^4) is the inner radius, though. You are rotating the inner curve around x=10 as well. Shouldn't it be larger. Check you picture again.
     
    Last edited: Feb 28, 2010
  4. Feb 28, 2010 #3
    Should the inner radius be 1+y^4 ?
     
  5. Feb 28, 2010 #4

    Dick

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    No, no, no. It should be the distance between (1-y^4) and 10. Shouldn't it?
     
  6. Feb 28, 2010 #5
    Okay, Thank you! I think I'm on the right path now. No?

    [tex]

    \pi \int ((10^{2}) - (10-(1-y^{4}))^{2})dy

    [/tex]
     
  7. Feb 28, 2010 #6

    Dick

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    Almost. I think you've got the square in the wrong place. Isn't the inner radius (10-(1-y^4))?? Shouldn't the square be on the outside?
     
  8. Feb 28, 2010 #7
    Yeah, I must have caught that while you were posting.
     
  9. Feb 28, 2010 #8
    I think I messed up my algebra somewhere. Could someone check over this for me? Thanks.

    [tex]
    \pi \; \int_{-1}^{1}{\left( 100\; -\; \left( 100+10+10y^{4}+10+1+y^{4}+10y^{4}+y^{4}+y^{8} \right) \right)}dy
    [/tex]

    [tex]
    \pi \; \int_{-1}^{1}{\left( -21-22y^{4}-y^{8} \right)}dy
    [/tex]

    [tex]
    \pi \; \left[ -21y\; -\frac{22y^{5}}{5}-\frac{y^{9}}{9} \right]\; from\; -1\; to\; 1
    [/tex]

    [tex]
    \pi \; \left[ \left( -21\; -\frac{22}{5}-\frac{1}{9}\; \right)\; -\; \left( 21+\frac{22}{5}+\frac{1}{9} \right) \right]
    [/tex]

    [tex]
    \pi \; \left( -42-\frac{44}{5}-\frac{2}{9} \right)\; =\; \pi \left( -\frac{1890}{45}-\frac{396}{45}-\frac{10}{45} \right)=\; \frac{1484\pi }{45}
    [/tex]

    I'm pretty sure answer should be:

    [tex]
    \frac{1376\pi }{45}
    [/tex]
     
    Last edited: Mar 1, 2010
  10. Mar 1, 2010 #9

    Dick

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    The integrand is just 100-(9+y^4)^2. It doesn't look to me like you expanded that correctly.
     
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