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Double integral boundaries.

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    D = {x>0, x^2 < y < 10-x^2)


    integral (integral D of y^2 sqrt x)

    2. Relevant equations

    3. The attempt at a solution

    I'm having trouble figuring out the bounds of the integral. y goes from x^2 to 10-x^2 but I think I have to split this integral up into two parts. I am not sure how to bound x. The parabolas intersect at the point sqrt 5
  2. jcsd
  3. Nov 26, 2011 #2
    What do you mean probably at sqrt(5)? They intersect at two points
  4. Nov 26, 2011 #3
    They intersect at (sqrt 5, 5) and (-sqrt 5, 5). x > 0 so we don't need the negative point.
  5. Nov 26, 2011 #4
    Yes you got, it
  6. Nov 26, 2011 #5
    So x goes from 0 to sqrt 5? thats the bound for x?
  7. Nov 26, 2011 #6
    What do you think it should be? Let's put it this way, if it isn't x = 0, where would you start from? You mentioned splitting the integral, how do you plan to do that?
  8. Nov 26, 2011 #7
    I guess I'm overthinking. It looks like x goes from 0 to x^2 and then stops when x reaches sqrt 5, then goes from 10-x^2 back to 0. This is just from the drawing I mean.
  9. Nov 27, 2011 #8


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    Yes, provided that you integrate over y first -- which, as you said, goes from x^2 to 10-x^2.
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