- #1

- 12

- 0

## 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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter clipzfan611
- Start date

- #1

- 12

- 0

[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?

- #2

cristo

Staff Emeritus

Science Advisor

- 8,107

- 73

Can you evaluate these integrals?

- #3

- 12

- 0

Can you evaluate these integrals?

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

- #4

- 370

- 0

No, it should not have a y in it.

- #5

cristo

Staff Emeritus

Science Advisor

- 8,107

- 73

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]

- #6

- 12

- 0

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:

- #7

cristo

Staff Emeritus

Science Advisor

- 8,107

- 73

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.

- #8

- 12

- 0

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

Gib Z

Homework Helper

- 3,346

- 5

They look like they intersect at 4, lets 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. Im 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.

- #10

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

Share: