# Need help finding Volume bound by curves.

## Homework Statement

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

$$x=1-y^{4}, x=0$$

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

## 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.

\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*}

Hmm can't edit my LaTex code. That integral should have Pi out front.

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Dick
Homework Helper
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.

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Should the inner radius be 1+y^4 ?

Dick
Homework Helper
No, no, no. It should be the distance between (1-y^4) and 10. Shouldn't it?

Okay, Thank you! I think I'm on the right path now. No?

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

Dick
Homework Helper
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?

Yeah, I must have caught that while you were posting.

I think I messed up my algebra somewhere. Could someone check over this for me? Thanks.

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

$$\pi \; \int_{-1}^{1}{\left( -21-22y^{4}-y^{8} \right)}dy$$

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

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

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

I'm pretty sure answer should be:

$$\frac{1376\pi }{45}$$

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Dick