Prove that: double integral over x²+y²≤ a² x^my^n dx dy =
Prove that: $$ \iint_{x^2+y^2\leqslant a^2} x^my^n\,dx\,dy = 0 $$ if $m$ and $n$ are natural numbers and at least one of them is odd.
Explanation of Method
The disk is symmetric under reflection across either axis. If $m$ is odd, reflecting $x\to-x$ flips the sign of the integrand while leaving the disk unchanged as a set, forcing the integral to equal its own negative — hence zero. The same argument applies to $y\to-y$ when $n$ is odd.
- Suppose $m$ is odd (the case $n$ odd is identical by symmetry).
- Split the disk into its left and right halves.
- Show the two halves contribute opposite values via the substitution $x\to-x$.
Step 1 · Split the disk
Let $D^+$ be the right half of the disk ($x\geqslant0$) and $D^-$ the left half ($x\leqslant0$). Since $D$ is symmetric about the $y$-axis:
$$ \iint_D x^my^n\,dx\,dy = \iint_{D^+} x^my^n\,dx\,dy + \iint_{D^-} x^my^n\,dx\,dy. $$
Step 2 · Relate the two halves
In the second integral, substitute $x=-x'$ (with $x'\geqslant0$); this maps $D^-$ bijectively onto $D^+$ with Jacobian $1$ (a reflection has $|{-1}|=1$):
$$ \iint_{D^-} x^my^n\,dx\,dy = \iint_{D^+} (-x')^my^n\,dx'\,dy. $$
Since $m$ is odd, $(-x')^m=-x'^m$, so:
$$ \iint_{D^-} x^my^n\,dx\,dy = -\iint_{D^+} x'^my^n\,dx'\,dy = -\iint_{D^+} x^my^n\,dx\,dy. $$
Step 3 · Combine the results
$$ \iint_D x^my^n\,dx\,dy = \iint_{D^+} x^my^n\,dx\,dy - \iint_{D^+} x^my^n\,dx\,dy = 0. $$
If instead $n$ is odd, the identical argument with the roles of $x$ and $y$ swapped (splitting into upper and lower halves, using the disk's symmetry about the $x$-axis) gives the same conclusion.
Additional Notes
- This can also be seen via polar coordinates: $\iint_D x^my^n\,dx\,dy=\dfrac{a^{m+n+2}}{m+n+2}\displaystyle\int_0^{2\pi}\cos^m\theta\sin^n\theta\,d\theta$, and the angular integral vanishes whenever $m$ or $n$ is odd, since $\cos^m\theta\sin^n\theta$ then changes sign under $\theta\to\theta+\pi$ while $[0,2\pi)$ is symmetric under this shift.
Result
This establishes the claim: