Is the function $z=x^{3}y$ even or odd?

Question

I just wanted to know because I am trying to calculate $\int_{-1}^{1}\int_{-1}^{1}x^{3}ydxdy$.

Answer 1

As for parity, see generalization to multivariate calculus, which states that for $f: \mathbb{R}^n \rightarrow \mathbb{R}$, a function is even if for $X \in \mathbb{R}^n, f(X)=f(-X)$, so in this case, the function is "even".

However, if your question is about integrals, you can't use this generalization of the definition. You need to look at the parity of the function you are integrating at the moment. I assume you are using the method of considering $x^3$ as constant while you integrate with respects to $y$ and viceversa. Doing that means that first you'll be integrating a function of the sort $g(y)=k.y$ (with $k=x^3$ constant), this function is obviously odd so the answer would be $0$.

Why do we get this result?

As you can see, the definition of parity in $\mathbb{R}^2$ will compare the value of the function valued at $X=(x,y)$ with the function valued at $-X=(-x,-y)$. So it compares the value of the function valued in a vector in the first quadrant, with one in the third quadrant, or one in the second quadrant with one in the fourth quadrant.

So for integration, it's actually more useful to look at the function's parity with respect to individual variables. For this case $f(x,y)=-f(-x,y)$ so the function is odd in the $x$ variable (same can be shown for $y$). Consequently, integrating a symmetrical interval with respects to a variable for which the function is odd, (in this case $x$) will lead to the integral being equal to $0$.

Answer 2

It is even because $(-x)^3(-y)=x^3y$.

Despite that, this is unnecessary for calculating your integral, where you can use that $x^3$ and $y$ are odd, which results in the value $0$.