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## Homework Statement

[tex]2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy[/tex]

## The Attempt at a Solution

Tried U-sub and anti-differentiation and none of them got the answer it should be. Help?

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- Thread starter clipzfan611
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In summary, the student attempted to solve a shells problem with respect to y and got the answer wrong because they did not use the correct order of integration. They got the answer -64pi/3.

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[tex]2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy[/tex]

Tried U-sub and anti-differentiation and none of them got the answer it should be. Help?

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- #2

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Can you evaluate these integrals?

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cristo said:

Can you evaluate these integrals?

Yeah I got 6/64y^6 for the first part. Is that right?

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No, it should not have a y in it.

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clipzfan611 said:Yeah I got 6/64y^6 for the first part. Is that right?

As the previous post points out, this is a definite integral and so should not contain y.

But, still, you have computed the indefinite integral incorrectly. Recall: [tex]\int y^ndy=\frac{y^{(n+1)}}{n+1}[/tex]

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cristo said:As the previous post points out, this is a definite integral and so should not contain y.

But, still, you have computed the indefinite integral incorrectly. Recall: [tex]\int y^ndy=\frac{y^{(n+1)}}{n+1}[/tex]

This is a shells problem with respect to y. I know y^5 would come out to 1/6y^6 but the 1/64 thing is confusing me.

Edit: Oh you multiply 1/6 and 1/64?

Last edited:

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clipzfan611 said:This is a shells problem with respect to y. I know y^5 would come out to 1/6y^6 but the 1/64 thing is confusing me.

Well, 1/64 is a constant, and so can be taken out of the integral. I could've done this above to give [tex]\frac{1}{64}\int y^5dy[/tex] Since you know what the value of [tex]\int y^5dy[/tex] is, then simply multiply this by 1/64. Then, you need to use the limits of integration.

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cristo said:Well, 1/64 is a constant, and so can be taken out of the integral. I could've done this above to give [tex]\frac{1}{64}\int y^5dy[/tex] Since you know what the value of [tex]\int y^5dy[/tex] is, then simply multiply this by 1/64. Then, you need to use the limits of integration.

I got the answer--it is -64pi/3. OT: However, can you have a negative volume in the first quadrant? Because I graphed the original equations y=x and y=64x^1/4. Thanks for your help.

- #9

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They look like they intersect at 4, let's say 4 for the sake of simplicity, when you do it, you have to solve [tex]\frac{y^5}{64}=y^2[/tex], you can get that into a simple cubic. I am pretty sure its 4 though.

Before 4, y^2 is larger, so the area here is [tex]\int_0^{4} y^2 - \frac{y^5}{64} dy[/tex], after 4 the area is [tex]\int_4^8 \frac{y^5}{64} - y^2 dy[/tex]...

EDIT: For the washers problem, multiply this result by 2 pi as required.

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