Proving multivariable result for evaluating odd functions

Question

I'm rather stuck on showing this result (the multivariable analog of the single variable result). What I've tried so far was to evaluate it with respect to a single integral first and show that since that goes to $0$, the entire thing goes to $0$. Got confused trying to split up the integrals though.

Let $r > 0$ and set $R = \{(x,y)\in R^2: -r \leq x, y \leq r\}.$ Let $f$ be an integrable function such that $f(-x, -y) = -f(x,y)$. Show that $$\iint_R f(x,y) \,dx\,dy = 0$$

Also need to show the same result if $f(x, -y) = -f(x,y)$ but I think this will just be a similar process to showing the first one.

Would greatly appreciate any help !

Answer 1

Let $I=\iint_R f(x,y) \,dx\,dy=\int_{x=-r}^0 \int_{y=-r}^0 f(x,y) \,dx\,dy + \int_{x=0}^{r} \int_{y=0}^{r} f(x,y) \,dx\,dy =A+B$

Now $$B=\int_{x=0}^{r} \int_{y=0}^{r} f(x,y) \,dx\,dy$$

Say $u=-x$ and $v=-y$

Then we have $$B=(-1)^2\int_{u=0}^{-r} \int_{u=0}^{-r} f(-u,-v) \,du\,dv$$ $$=(-1)^2\int_{u=0}^{-r} \int_{v=0}^{-r} [-f(u,v)] \,du\,dv$$ $$=-\int_{u=-r}^{0} \int_{v=-r}^{0} f(u,v) \,du\,dv$$ $$=-\int_{x=-r}^0 \int_{y=-r}^0 f(x,y) \,dx\,dy=-A$$

$$A+B=0$$

Hope this helps you.

Answer 2

HINT

If you limit the region of integration to Q1 and Q3, they are opposites of each other. So are Q2 and Q4. Divide the integral into 4 parts and show they pairwise cancel out.

Similar idea with the second problem.

Answer 3

$ - \int_{-r}^r \int_{-r}^r f(x,y) dx dy = \int_{-r}^r \int_{-r}^r (-f(x,y))dxdy = \int_{-r}^r \int_{-r}^r f(-x,-y) dx dy $ $= \int_{-r}^r \int_r^{-r} (-f(s,-y))dsdy = -\int_r^{-r} \int_{-r}^r f(s,-y)dyds = -\int_r^{-r} \int_r^{-r} (-f(s,t))dtds$ $= \int_r^{-r} \int_r^{-r} f(s,t)dsdt = \int_{-r}^r \int_{-r}^r f(x,y) dx dy$

Answer 4

First, understand why this is true when there is only one variable involved. Consider for instance $f(x)=x^3$:

Do you see why $\int_{-r}^r f(x)=0$ ? Since $f(-x)=-f(x)$, $f(x)$ is antisymmetric, and since we are integrating on a symmetric interval, the result follows. Geometrically, the areas between the curve and the $x$-axis cancel one another.

Can you extend this when $2$ variables are involved?