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Find the average value of the function f(x,y)=x^2+y^2

  1. Nov 8, 2009 #1

    een

    User Avatar

    1. The problem statement, all variables and given/known data
    Let a>0 be a constant. Find the average value of the function f(x,y)=x^2+y^2
    1) on the square -a[tex]\leq[/tex]x[tex]\leq[/tex]a, -a[tex]\leq[/tex]y[tex]\leq[/tex]a
    2) on the disk x^2+y^2[tex]\leq[/tex]a^2
    2. Relevant equations



    3. The attempt at a solution
    1) I integrated [tex]\int[/tex]a-(-a) [tex]\int[/tex]a-(-a) (x^2+y^2) dxdy and got (8/3)a^4..Is this right?

    2)I converted it to polar coordinates 0[tex]\leq[/tex][tex]\theta[/tex][tex]\leq[/tex]2pi
    and 0[tex]\leq[/tex]r[tex]\leq[/tex]sqrt(a)
    i integrated [tex]\int[/tex]0-2pi[tex]\int[/tex]0-sqrt(a) r^2drd[tex]\theta[/tex]
    and got 2/3pi*(sqrt(a)^3)... is this right???----- 2pi[tex]\frac{\sqrt{a}^3}{3}[/tex]
     
  2. jcsd
  3. Nov 8, 2009 #2

    Mark44

    Staff: Mentor

    For 1, I get (2/3)a^2 for the average value. Because of the symmetry of the region and the integrand, I took a short cut and integrated from 0 to a for both x and y, and multiplied the result by 4. Don't forget that for the average value, you have to divide by the area of the region, which is 4a^2. Your answer divided by 4a^2 equals mine.
     
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