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Tried U-sub and anti-differentiation

  1. Mar 11, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy[/tex]

    3. The attempt at a solution

    Tried U-sub and anti-differentiation and none of them got the answer it should be. Help?
  2. jcsd
  3. Mar 11, 2007 #2


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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?
  4. Mar 11, 2007 #3
    Yeah I got 6/64y^6 for the first part. Is that right?
  5. Mar 11, 2007 #4
    No, it should not have a y in it.
  6. Mar 11, 2007 #5


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    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]
  7. Mar 11, 2007 #6
    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: Mar 11, 2007
  8. Mar 11, 2007 #7


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    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.
  9. Mar 11, 2007 #8
    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.
  10. Mar 12, 2007 #9

    Gib Z

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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.
  11. Mar 12, 2007 #10


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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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