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Double Integral over general region

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the integral using a geometric argument.

    ∫∫D√(16 - x2 - y2)dA
    over the region D where D = {(x,y) : x2 + y2 ≤ 16}

    By the way, the subscript D next to the integral refers to the region over which the function is integrated.


    2. Relevant equations
    ∫∫f(x,y)dxdy = ∫∫f(x,y)dydx

    Don't think it's a polar coordinate question but here is the equation for a double integral in polar coordinates anyway:
    ∫∫Df(x,y)r dA


    3. The attempt at a solution
    Attempted to solve the equation using a double integral in polar coordinates with:
    0 ≤ r ≤ 4
    0 ≤ θ ≤ pi
    got 64π/3 but I'm not certain whether it is correct. I'm not sure whether the graph approaches the x-y plane asymptotically or not. The equation for f(x,y) = √(16 - x2 - y2) describes what seems like the positive z half of a sphere

    Edit: Made a second attempt. Assumed the shape to be the top half of a sphere so the equation became:
    (0.5)∫∫D(16 - x2 - y2)dA
    Where x ranges from 0 to 4 and y ranges from -√(16-x2) to √(16-x2)

    Any help would be much appreciated.
    Thank you
     
    Last edited: Nov 18, 2012
  2. jcsd
  3. Nov 18, 2012 #2

    Zondrina

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    Homework Helper

    Draw your region, its a circle with radius 4 with a center at (0,0).

    There are four curves you need to consider when setting up your integral. You can find these curves by solving x2 + y2 = 16.
     
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