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Triple Integral Cartesian Coordinates

  1. Mar 6, 2012 #1
    Ok I have a quick question. I have this problem that is doable with polar coordinates and triple integrals but I was wondering if it would be possible to do this problem in the cartesian coordinate system (odd question I know...).
    1. The problem statement, all variables and given/known data

    A sprinkler distributes water in a circular pattern, supplying water to a depth of e^(-r) feet per hour at a distance of r feet from the sprinkler.

    A. What is the total amount of water supplied per hour inside of a circle of radius 10?
    2pi-2pie^(-10)
    B. What is the total amount of water that goes throught the sprinkler per hour?
    2pi



    2. Relevant equations
    Just integration techniques I guess. pi*R^2 is the equation for a circle area. x^2+y^2=100 is the equation in standard form for this circle.


    3. The attempt at a solution

    Here's where I get lost. In cartesian coordinates the bounds for the resulting double integral should be 0<=y<=sqrt(100-x^2) and 0<=x<=10, right? Then from there I take the double integral of the equation of the circle and...?

    Any help is appreciated, thanks!

    EDIT: Should I take the double integral of pi(x^2+y^2) with the bounds I have above? or is that wrong?
     
  2. jcsd
  3. Mar 6, 2012 #2

    LCKurtz

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    Those bounds would be for a quarter of the circle. You could use those limits and multiply the answer by 4. But your integrand would be ##e^{-r}=e^{-\sqrt{x^2+y^2}}##. I don't think anyone would want to work it in rectangular coordinates. If you really must try it, you might have success with a trig substitution on the inside integral. Not sure, I haven't tried it.
     
    Last edited: Mar 6, 2012
  4. Mar 6, 2012 #3
    Wait, why would those limits only be for a quarter of the circle?
     
  5. Mar 6, 2012 #4

    LCKurtz

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    Because your limits have both x and y positive.
     
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