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Homework Help: Finding the volume using double integrals

  1. Jul 9, 2010 #1
    1. The problem statement, all variables and given/known data
    R be the region enclosed by x=y^2,x=y=1 and y=2 hence evaluate
    [tex]]

    \iint e^{x/y^{2}} da
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    i have setup the integral as
    [tex]
    \int_{1}^{2} \int_{1}^{y^2} e^{x/y^{2}} dxdy
    [/tex]
    so far i got here
    [tex]
    \int_{1}^{2} ey^{2}dy - \int_{1}^{2} y^{2} e^{1/y^2} dy
    [/tex]


    I want to know that if i m right upto this point.
    Thanks
     
    Last edited: Jul 9, 2010
  2. jcsd
  3. Jul 9, 2010 #2
    anyone please :|
     
  4. Jul 9, 2010 #3
    Where did you get the function
    [tex]
    e^{x/y^2}?
    [/tex]

    I'm confused a little by your latex formatting - use { and not ( for enclosing the things in the super/subscript.

    Are you sure that's the right function? If so, what you've shown in the post looks correct. Unfortunately, I don't think you're going to get a simple answer (just in terms of elementary functions) from the integral
    [tex]
    \int_1^2 y^2e^{1/y^2}\textrm{d}\,y
    [/tex]
     
  5. Jul 9, 2010 #4

    LCKurtz

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    It isn't clear what region you are trying to describe. What are the boundaries of your region?
     
  6. Jul 9, 2010 #5
    The term [tex]e^{x/y^2}[/tex] was given in the question i was asked to integrate between the region limits.Then could you please tell me how to get the solution of [tex]
    \int_1^2 y^2e^{1/y^2}\textrm{d}\,y
    [/tex]
    Thanks for the latex info i m still learning.

    The region is actually in the first quadrant [tex] y=\sqrt{x}[/tex] ,x=1, y=1 and y=2 and the intersection of those lines and curve region looks like a triangle.
     
  7. Jul 9, 2010 #6

    LCKurtz

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    y = sqrt(x) is a piece of a parabola, x = 1 is a vertical straight line, y = 1 and y = 2 are two horizontal straight lines. You have given 4 "sides" and are talking about a "triangular" area?? Is y = 1 one of the sides?
     
  8. Jul 10, 2010 #7
    i have uploaded the picture of the region the shaded region i m talking about.

    Code (Text):
    http://i30.tinypic.com/2920l6a.png
     
  9. Jul 10, 2010 #8
    i know and i guess my integration limits are right :smile:
     
  10. Jul 10, 2010 #9

    LCKurtz

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    So y = 1 is not a boundary of the region and shouldn't be used in the description. You might have described it as the region above y = sqrt(x), below y = 2 and to the right of x = 1.
     
  11. Jul 10, 2010 #10
    Sorry OP, I misread your question, ignore what I said.
     
  12. Jul 10, 2010 #11
    i m using y=1 as a lower limit of my y limit. Well i wanna know if my double integral setup is correct or not and solution too please.
     
  13. Jul 10, 2010 #12

    LCKurtz

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    Now that I understand the region, yes you have it set up correctly. And as tmccullough has already noted, you aren't going to be able to find a simple antiderivative for that second integral.
     
  14. Jul 10, 2010 #13
    can you tell me how to solve it or refer me to any topics from which i can learn how to deal with these type of problems.

    Thanks
     
  15. Jul 10, 2010 #14

    LCKurtz

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    Well, you apparently know how to set them up. When you wind up with a definite integral you can't do in a nice closed form, you must use numerical methods to get a decimal answer as accurate as you need. You could use Simpson's rule or if you have access to a high level mathematics program like Maple or Mathematica it will give you an immediate decimal answer. There are also resources on the internet such as Wolfram Alpha you can use.
     
  16. Jul 10, 2010 #15
    thank you :smile:
     
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