Some kind of system of equations with double integrals

[benji55545]

Hey there, this is my first post, hopefully I don't screw anything up.

Homework Statement

Suppose that ∫ ∫D f(x, y) dA = 4 where D is the disk x2 +y2 ≤ 16. Now suppose E is the disk x2 + y2 ≤ 144 and g(x,y) = 3 f( [x/3], [y/3] ). What is the value of ∫ ∫E g(x, y) dA?

Homework Equations

The Attempt at a Solution

Well, I figured switching the surface of integration into polar coordinates might be a good idea, but that didn't really lead anywhere. I figured that ∫ (0,4) f(x/3,y/3) would probably be 4/2pi since the limits of integration of the outside integral are usually 0 to 2pi and often have no variables in the function. I also noticed that the fuction in the second double integral was just multiplied by three but didn't know if I could just say that 3*f(x/3,y/3) was equal to f(x,y)... I'm thinking no. That's as far as my thinking went, I couldn't fathom where to go.
Thanks for any help.

 

[Dick]

Just take the g integral and do the change of variables u=x/3, v=y/3. What does the domain look like in u,v variables?

 

[benji55545]

I'm afraid I don't follow what you mean by take the g integral and do the change of variables. Do I somehow take the g integral first? I think my biggest stumbling block is the lack of concrete numbers...

and for the limits in x and y, would they be
0<y<(+-sqrt(144-x^2))
-6<x<6

so u and v might be...
-2<u<2
0<v<+-sqrt(144-9u^2)

right?

 

[Dick]

Nooo. -12<x<12 so -4<u<4. But more concretely the u,v domain is your original disk D. Convert the g integral into something that looks like the f integral using the change of variables.

 

[HallsofIvy]

Don't worry about the limits of integration! If u= x/3 and v= y/3, then x= 3u and y= 3v. The circle x²+ y²= 144, in the "xy-plane" becomes (3u)²+ 3v²= 9u²+ 9v²= 144 or, dividing by 9, u²+ v²= 16, in the "uv-plane". Now do you see the point?