How can I prove a function is odd? I have a problem of calculus and I can compute it.
The problem is 
$\iiint_V x^5 y^7 z^9 dxdydz$, where $V = \{ (x, y, z) : x^2 + y^2 + z^2 \leq  1 \}$.
And the solution is 0. this solution is exactly right. If you want I can show whole my solution.
I solved it using by polar coordinates.
But I heard if prove $\iiint_V x^5 y^7 z^9 dxdydz$ = -$\iiint_V x^5 y^7 z^9 dxdydz$
(because $x^5y^7z^9$ is odd ft),
I don't need to compute it.
If someone know How can prove it please help me.
I have no idea, how can start it..
 A: A function $f(t)$ is odd when $f(-t)=-f(t)$. In this case $t=(x,y,z)$. So you just need to show that 
$$-f(t)=-\int\int\int_{V}x^{5}y^{7}z^{9}=\int\int\int_{V}(-x)^{5}(-y)^{7}(-z)^{-9}=f(-t)$$
Which should immediately follow when you begin. 
A: Try thinking in just one dimension-it works the same as in three.  If you are interested in $\int_{-1}^1x^5dx$, you can compute it by the usual technique of $\int_{-1}^1x^5dx=\frac 16x^6|_{-1}^1=\frac 16(1^6-(-1)^6)=0$  Alternately, you can observe that $x^5$ is odd and say $\int_{-1}^1x^5dx=\int_{-1}^0x^5dx+\int_{0}^1x^5dx=\int_{0}^1(-x)^5dx+\int_{0}^1x^5dx=0$ where the last step is essentially the $u$ substitution $u=-x$  The intuitive way to see this is that the bit from $x$ to $x+dx$ is counteracted by the bit from $-x-dx$ to $-x$
In your 3D problem, you could reflect across any one of the coordinate planes or through the origin as your function is odd in each variable separately.
A: Just note this, you can write the integral as

$$ I=\int_{-1}^{1}x^5dx\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}y^7\,dy\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}z^9\,dz. $$

Now, you can see that each integral is an integral of an odd function over a symmetrical interval.
