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Need help finding Volume bound by curves.

  • Thread starter revolve
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  • #1
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Homework Statement



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.

Homework Equations





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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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:
  • #3
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Should the inner radius be 1+y^4 ?
 
  • #4
Dick
Science Advisor
Homework Helper
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No, no, no. It should be the distance between (1-y^4) and 10. Shouldn't it?
 
  • #5
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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]
 
  • #6
Dick
Science Advisor
Homework Helper
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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?
 
  • #7
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Yeah, I must have caught that while you were posting.
 
  • #8
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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:
  • #9
Dick
Science Advisor
Homework Helper
26,258
618
The integrand is just 100-(9+y^4)^2. It doesn't look to me like you expanded that correctly.
 

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