Find the average squared distance between the points of R={(x,y): 0<=x<=2, 0<=y<=2} and the point (2,2).

Question

The average squared distance is?

I need help setting up this double integral and also possibly some hints on simplifying this out. But mostly setting it up is my struggle here. Thanks in advance!

Answer 1

Distance² = f(x,y) = (x-2)² + ( y-2)²

Average Distance² = ∫∫_Rf(x,y)dxdy/Area

Area in this case is 2 x 2 = 4

The limits for both integrals are 0 to 2

D² = ∫∫(x-2)² + (y-2)² dx dy

The inner integral evaluates to (x-2)³/3 + (y-2)²x from 0 to 2 = 0 + 2(y-2)² - (-8/3 + 0) = 2(y-2)² + 8/3

∫2(y-2)² + 8/3 dy = (2/3)(y-2)³ + 8y/3 from 0 to 2 = 0 +16/3 - (-16/3) = 32/3

Ave D² = D²/Area = (32/3)/4 = 8/3