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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 !

4 Answers 4

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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.

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    Are the bounds supposed to change as they do in the last part? Goes from $u = -r$, $v = -r$ to $x = -r$, $y = -r$. – 2017-02-16
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    Actually, I see what's happening now. Subbing it back in we'd get $f(-x,-y)$, or we can change the bounds to be negative. – 2017-02-16
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    The integral does not depend on the variable used. – 2017-02-17
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First, understand why this is true when there is only one variable involved. Consider for instance $f(x)=x^3$:

enter image description here

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?

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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.

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