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

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SUMMARY

The average value of the function f(x,y) = x^2 + y^2 was calculated over two regions: a square defined by -a ≤ x ≤ a and -a ≤ y ≤ a, and a disk defined by x^2 + y^2 ≤ a^2. For the square, the integral resulted in (8/3)a^4, leading to an average value of (2/3)a^2 after dividing by the area, 4a^2. For the disk, the conversion to polar coordinates yielded an integral of 2π(sqrt(a)^3)/3, confirming the average value calculation aligns with the symmetry of the region.

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  • Understanding of double integrals in calculus
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  • Knowledge of average value calculations for functions
  • Basic concepts of symmetry in mathematical functions
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Homework Statement


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

Homework Equations





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]
 
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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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