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Volumes of revolution

  1. Feb 20, 2010 #1
    R(x)=x^3 bounded by x=0, x=2 and y=1.
    a. revolved around x=2
    b. revolved around x=10

    my pathetic attempt:
    a. v=pi[(integral from 0 to 1)(2-y^1/3)^2]dy
    so =pi[4y-3y^(4/3)+(3/5)y^(5/3)]evaluated from 0 to 1
    =(8/5)pi

    b. v=pi[(integral from 0 to 2)(10-x^3)^2]dy
    I must admit that I don't think that equation is right. But I just seem to get how to set this up.
     
  2. jcsd
  3. Feb 21, 2010 #2

    Mark44

    Staff: Mentor

    Are you sure you have copied the description of the region that is being revolved correctly? Could it be the region bounded by the graph of y = x^3, y = 0, x = 1, and y = 1? The region as you wrote it doesn't have a boundary on the bottom.
     
  4. Feb 21, 2010 #3
    o, ummm. No there is no lower bound. I assume that it is y=0. And for b, I meant to say that I just can't seem to get the set up.
     
  5. Feb 21, 2010 #4

    Mark44

    Staff: Mentor

    For a) are you sure it doesn't say the region bounded by the graph of y = x3, y = 0, x = 2, and y = 1? That would make more sense. In problems like these, it's very rare that you have to assume how the region is defined.
     
  6. Feb 21, 2010 #5
    Consider the region R bounded by   3f x  x , y =1, and x = 2 that is what it says.
     
  7. Feb 21, 2010 #6
    oops....copy pasting didnt work, but that is the region
     
  8. Feb 21, 2010 #7

    Mark44

    Staff: Mentor

    What you entered is showing up as squares. What does it say?
     
  9. Feb 21, 2010 #8
    f(x)=x^3, y=1, and x=2. I just got a response back from my teacher and she said that "The limits of integration are determined by the region described. You have to figure them out." So now I am confuzed more than before.
     
  10. Feb 21, 2010 #9

    Mark44

    Staff: Mentor

    OK, that's better. In your first post, you also had x = 0 as one of the bounds. The region being revolve has a somewhat triangular shape. It's above the line y = 1, to the right of the curve, and to the left of the line x = 2.

    For a, your integrand is correct, but your limits of integration aren't. Your values of dy are running from what to what? Here's the integral. All you need to do are to get the right limits of integration, find the antiderivative, and then evaluate the antiderivative at the two limits of integration.
    [tex]\pi \int_?^? (2 - y^{1/3})^2 dy[/tex]
     
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