math challenge (intermediate)?

Question

For a 1D integral over a symmetric interval we have
Int(f(x),x = -a..a) = Int(f(x),x = -a..0) + Int(f(x),x = 0..a)
Now break up the double integral similarly:
Int(Int(f(x,y),x = -a..a),y = -a..a)
The first one to get the correct expansion get 10 points.

Answer 1

∫ f(x,y) {-a, a} {-a, a} dx dy =

∫ f(x,y) {-a, 0} {-a, 0} dx dy
+ ∫ f(x,y) {-a, 0} {0, a} dx dy
+ ∫ f(x,y) {0, a} {-a, 0} dx dy
+ ∫ f(x,y) {0, a} {0, a} dx dy

Answer 2

It's just the sum of the four surface integrals (one in each quadrant) of the square [(-a,a), (-a,a)] centered at the origin.
int(int(f(x,y)x=-a,a)y=-a,a) =
int(int(f(x,y)x=-a,0)y=-a,0) +
int(int(f(x,y)x=0,a)y=-a,0) +
int(int(f(x,y)x=-a,0)y=0,a) +
int(int(f(x,y)x=0,a)y=0,a)

If it was a volume integral of the cube [(-a,a),(-a,a),(-a,a)], it could be broken into 8 sub-integrals. And, in general, any N-dimensional integral of a symmetric function over a symmetric range (if the function posesses continuous derivatives -and- meets the Cauchy-Riemann requirements for continuity) then it can always be reduced to the sum of 2^N sub-integrals.

Doug

Answer 3

There is no solution to this