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Weird shell method problem to find volume

  1. Jun 4, 2014 #1
    Y=x^(2/3) and y=x^2 rotated about the x=4.

    First I equate the equations giving me x=0,-1,1. The problem is I have exactly 2 graphs that are symmetric in respect to the y axis. N I have not encountered a problem in stewart dealing with this.

    I know that the volume on the right side of the X axis is going to be. 2pie∫(4-x)(x^2-x^(2/3)) dx. And the limits of integration are from 0 to 1.

    What about the side left from origin?

    After I find the integral for both of the areas. Do I right of origin- left of origin or right of origin +LEFT of origin?

    Thanks
     
  2. jcsd
  3. Jun 4, 2014 #2
    Here is a pic of my work so far.
     

    Attached Files:

  4. Jun 4, 2014 #3

    adjacent

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    I have to lie on my side to read that. It will be easier for you and others if you make that picture upright.
     
  5. Jun 4, 2014 #4
    Ok here it is
     

    Attached Files:

  6. Jun 4, 2014 #5

    CAF123

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    I think you have the left hand portion of your graph incorrect. y = x2/3 is not defined in the reals when x is negative.

    I am guessing the complete problem statement is find the volume of revolution of the solid formed from rotating the area enclosed by the two curves around the line x=4? I agree with the integral you wrote down, up to a sign.

    You can check your answer using disks.
     
  7. Jun 4, 2014 #6
    So x^(2/3)=x^(2)

    Becomes. X^(2/3)(1-x^(4/3))

    Which gives we the interceptions of x=0,-1,1. So I do not take into account the left hand side of the graph even tho the algebra is correct? We would have the sum of two volumes?
     
  8. Jun 4, 2014 #7
    X^(2/3) is defined because of cubic root holds no restrictions provided it is not in the denominator.
     
  9. Jun 4, 2014 #8

    CAF123

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    How did you get x=-1? It is not a solution. So the only intersections of the graphs are at 0 and 1. No need to consider two volumes. The volume of revolution is generated by this area only:
    http://m.wolframalpha.com/input/?i=sketch+x^2+=+x^(2/3)&x=3&y=1

    Suppose there exists a negative solution. Then ##-p = x^{2/3}##, p > 0. Take the root of both sides gives ##(-p)^{3/2} = x = (-1)^{3/2} p^{3/2} = -i p^{3/2} \in \mathbb{C}##.
     
  10. Jun 4, 2014 #9

    verty

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    I would just assume we are talking about the first quadrant only. Otherwise you should use an absolute value and have different limits of integration. Anyway, it seems like you know what you are doing here.
     
  11. Jun 4, 2014 #10

    LCKurtz

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    And once you decide whether you are going to use [-1,1] or [0,1], note that you have the upper and lower curves reversed in your setup integral. You will get a negative answer as it stands.
     
  12. Jun 5, 2014 #11

    verty

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    Whereas with an absolute value sign in the integrand, the statement is at least conceptually correct, which is always what you want to aim for, getting a conceptually correct formula that one can proceed to treat mathematically.
     
    Last edited: Jun 5, 2014
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