# Double integral

If function is ##f(-x,-y)=f(x,y)##, is then

##\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=0##?

CompuChip
Homework Helper
No, consider for example
$$\int_{-a}^a \int_{-a}^a x^2 y^2 \, dx dy$$

Could you tell me some explanation why this is valid only for one integral?

jgens
Gold Member
Could you tell me some explanation why this is valid only for one integral?
The result does not hold for 1-dimensional integrals either. If f(x) = x2, then f(x) = f(-x) and integrating over any interval of the form [-a,a] (where a ≠ 0) gives you a non-zero number.

If you have a function that satisfies f(-x,y) = -f(x,y) then the integral over [-a,a] × [-a,a] should be zero. So you should probably look for a condition like this.

If function is ##f(-x,-y)=f(x,y)##, is then

##\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=0##?

Is it always valid?
Also is it valid in case ##f(-x,-y)=f(x,y)## that
##\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=2\int^{a}_{0}f(x,y)dxdy##

lurflurf
Homework Helper
For double integrals there are four cases
f(x,y)
f(-x,y)
f(x,-y)
f(-x,-y)

f(-x,-y)=-f(x,y)
implies that
f(x,-y)=-f(x,-y)
thus
the integral would be zero

f(-x,-y)=f(x,y)
implies
f(-x,y)=f(x,-y)
##\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=2\int^{a}_{-a}\int^{a}_{0}f(x,y)dxdy##

Tnx.