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Homework Help: Evaluate this surface Integral

  1. Aug 15, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is attached in the picture.

    3. The attempt at a solution

    The suggested solution went straight into the hardcore integration. I was trying a different approach by changing the variables (x,y) into (u,v) which appear to make the integration much easier...

    The Jacobian was found to be J = p

    Not sure where I went wrong!

    Attached Files:

  2. jcsd
  3. Aug 15, 2012 #2
    Where did you get the upper limit for ρ integral? Now you're integrating over a circle, not a rectangle.
  4. Aug 15, 2012 #3
    Hmm if you draw a rectangle with lengths ab, the p ranges from 0 to maximum of √(a2+b2) which is at the corner.

    The angle ranges from 0 to ∏/2..
  5. Aug 15, 2012 #4


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    You cannot set the [itex]\rho[/itex] and [itex]\theta[/itex] limits independently. The upper limit on [itex]\rho[/itex] will be a rather complicated function of [itex]\theta[/itex].
  6. Aug 15, 2012 #5
    This is true, but if you integrate ρ from 0 to √(a2+b2) for each value of the angle, you are integrating over a circle. Actually the calculation is probably a bit simpler anyway in polar coordinates, so you should definitely work it out. The easiest way is to split the angular integral into two parts: consider first [itex] \tan(\theta) < b/a [/itex]. You should be able to find a simple enough expression for the upper bound of ρ in this region.
  7. Aug 15, 2012 #6
    Not sure how to do it... I know the limit of ∅ is from 0 to ∏/2 but I don't even know the limits of p..
  8. Aug 15, 2012 #7
    You need to calculate the distance from the rectangle's corner to an arbitrary point on an opposing edge. Can you do that?
  9. Aug 15, 2012 #8
    If tan ∅ < (b/a) then p = a/cos ∅.

    If tan ∅ > (b/a) then p = b/sin ∅.

    If tan ∅ = (b/a) then p = √a2+b2
  10. Aug 15, 2012 #9
    Perfect. Those are your integration limits for the radius: now you just need to split the angle integral into
    [tex]\int_0^{\pi/2} d\phi = \int_0^{\arctan(b/a)} d\phi + \int_{\arctan(b/a)}^{\pi/2} d\phi [/tex] plug in the radius limits and integrate.
  11. Aug 16, 2012 #10
    Nice. But this really looks more complex than doing it the 'hardcore' way..
  12. Aug 16, 2012 #11
    I'm sure it does, if you're not used to changing variables in multidimensional integrals. Did you actually do the calculation? I think you would have found that the integrals you need to compute in spherical coordinates are a lot easier than those in cartesian coordinates. For the first term, you'd get
    [tex] \int_0^{\arctan(b/a)} d\phi \int_0^{a/\cos(\phi)} dr \frac{r^2 \cos(\phi)}{r^2} = a \int_0^{\arctan(b/a)} d\phi = a \arctan(b/a) [/tex]
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