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Tried U-sub and anti-differentiation

  • #1

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?
 

Answers and Replies

  • #2
cristo
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You don't need a substitution; remember that this can be split into two integrals [tex]2\pi\left[\int_0^8\frac{1}{64}y^5 dy-\int_0^8y^2 dy\right][/tex].

Can you evaluate these integrals?
 
  • #3
You don't need a substitution; remember that this can be split into two integrals [tex]2\pi\left[\int_0^8\frac{1}{64}y^5 dy-\int_0^8y^2 dy\right][/tex].

Can you evaluate these integrals?
Yeah I got 6/64y^6 for the first part. Is that right?
 
  • #4
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No, it should not have a y in it.
 
  • #5
cristo
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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
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
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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
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
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The reason your answer is negative is because you got the order wrong. You wanted the larger, higher valued functions area, minus the smaller ones. You will see for values from 0 to 8, your bounds of integration, the y^5/64 may be smaller than the y^2, and may be larger. So you must split up the integral where the functions intersect.

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
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Oop! I recognize that integral. clipzfan66 posted a "volume of revolution" problem in another thread and I suggested that order of integration. I was thinking about x to a power larger than 1 and that y= xn was less than y= x for x< 1! Didn't think about this being a fractional power!
 

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