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Homework Help: Volume of a Solid: Washers/Disks/Shells

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data
    The volume of a solid obtained by rotating the region enclosed by
    x=0, y=1, x=y^5 about the line y=1
    can be computed using the method of disks or washers via an integral.

    2. Relevant equations
    V= ∏[itex]\int[/itex](R^2-r^2)dx

    3. The attempt at a solution
    I have attempted this problem many times, and have come up with a lot of different approaches. When I drew this out, it seemed like I needed to calculate the entire cylinder, and then subtract the 'bowl' from the inside.

    This gave me V= ∏[itex]\int[/itex](1^2-(1-(x^(1/5)))^2)

    I know that it needs to be integrated with respect to x (dx) and that the limits of integration are from 0 to 1 because I got those correct.

    I have also thought of the problem as a simple disk problem using V=∏[itex]\int[/itex](x^(1/5))^2 to no avail.

    I would really appreciate any help!
    Last edited: Sep 30, 2012
  2. jcsd
  3. Sep 30, 2012 #2
    Couple of quick questions:

    * is ∏ supposed to be pi? Because the ∏ symbol is actually the product operator (similar to Ʃ, but with multiplication instead of addition).

    *My understanding is that the approach that you talked about in your attempt at a solution is correct, but... it looks like you may have gotten off into the weeds from that point, because I don't see where you're actually DOING that.
    I think that there should be two integrals, with the inner integral being subtracted from the outer integral.
  4. Sep 30, 2012 #3
    Yes, I did mean pi, sorry about that.

    Thanks for the reply! On "WebWork" (the software my school uses) we are required (for this homework) to report it all as one integral.

    I have drawn out the two "interpretations" I had of this problem and took a photo of it. I hope it is clear enough to be able to see.


    The first, one shown, seems like it actually makes more sense to me now. It seems like y=1 would be the upper function and y=x^(1/5) would be the lower, making it a simple disk problem.

    It still should have the same radius as if it were rotated around the x-axis, but that apparently is not the case as I am not getting the correct solution when I enter pi*(x^(1/5))^2*dx as my answer.
  5. Sep 30, 2012 #4
    The pic helps, I'm looking at it now...
    I know what you mean by "seems like it actually makes more sense to me now", though. That's the way everything has been going for me as well.

    Looking over my notes and textbook though, it should be one integral, just... humm...
  6. Sep 30, 2012 #5
    Thanks so much for helping me. I have spent far too much time working on this problem!
  7. Sep 30, 2012 #6
    I think that you need to integrate this horizontally... fairly certain of that, actually.
  8. Sep 30, 2012 #7
    It is saying correct for dx and incorrect for dy. I am sure you could do it both ways, but they clearly want me to use dx.
  9. Sep 30, 2012 #8
    well... I'm getting [itex]5\pi/6[/itex] as an answer... I think.
    (to [itex]\int(\pi x^1/5)dx[/itex] that is.)
  10. Sep 30, 2012 #9
    Could you tell me how you did it? I have to enter the integral for the answer.
  11. Sep 30, 2012 #10
    well, the pi just goes out in front of the integral. Then you use the power rule for the x to the 1/5th dx, which gives you x^(6/5) divided by 6/5. pi times all of that between 1 and 0 (the 0 side just goes to zero though, so just drop it) = pi((1^(6/5))/6/5). 1 to any exponent is 1, so you're left with pi divided by 6/5, which equals 5pi / 6.
  12. Sep 30, 2012 #11
    The integral is not pi(x^(1/5)^2) or pi(x^(1/5))
  13. Sep 30, 2012 #12
    humm... wait a second, are you trying to find the shaded area or the white area, in your diagram? I thought that it was the white area, from the problem description you gave in the OP. If it's the shaded area, then... my textbook says that you have to use two integrals (Method of washers). so... I'm confused.
  14. Sep 30, 2012 #13
    oh, wait... you can use the Cylindrical Shells method for this...

    v=[itex]\int(\pi(outer radius - inner radius))dx[/itex]

    Give me a minute to figure it out, though.
  15. Sep 30, 2012 #14
  16. Sep 30, 2012 #15
    The limits of integration are definitely 0 to 1. (x=0 is given, and x=y^5 intersects y=1 at (1,1)).

    Using the cylinders method:
    Since we're revolving around the y=1 line, that will be the "inner". so, "inner" f(x)=1.
    "Outer" f(x)=x^1/5.

    [itex]\int(\pi (x^(1/5))^2 - (1)^2) dx = \int(\pi x^(2/5) - 1) dx = \pi-1 \int( x^(2/5)) dx = \pi-1((x^(7/5))/(7/5))[/itex]
    going from 1 to 0 (zero just goes away, which makes things easier),
    1 to an exponent is 1, so:
    [itex](\pi-1)/(7/5) = (5\pi-1)/7[/itex]

  17. Sep 30, 2012 #16
    ...but, I see that you can't use cylinders there.
  18. Sep 30, 2012 #17
    did you try pi times x to the 2/5th? sometimes those things don't like it when you use two expoenets like: (x^1/5)^2
    Maybe it wants you to actually square the exponent?

    Looks like all they want you to do is set the integral up, which should be pi r^2. And, since the region enclosed by those three lines is y=x^1/5 (the white area in your drawing), we're definitely on the right track here.
  19. Sep 30, 2012 #18
    Hmm. I have tried many combinations (I am on attempt 37), to no avail. I have tried (x^(2/5)) as well.
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