TITLE: Basic integration question: integration bounds for $\iint x^2y^2 \ dx\,dy$
QUESTION [6 upvotes]: I have the integral
$$\iint x^2y^2 \ dx\,dy$$
but I am meant to evaluate it at the limits $0<y<1$ and $-2y<x<2y$. I am wondering what terminals of integration I should put in for $x$. Do I evaluate between $x=-2y$ and $x=2y$ or between $x=-2$ and $x=2$?

REPLY [5 votes]: $$
\iint x^2y^2\,dx\,dy=\int_0^1y^2\left[\int_{-2y}^{2y}x^2\,dx\right]\,dy=\int_0^1y^2\left[\frac{x^3}{3}\right]_{-2y}^{2y}\,dy=\int_0^1\frac{16}{3}y^5\,dy=\frac{16}{3\cdot6}y^6\Big|_0^1=\frac89.
$$

REPLY [4 votes]: First evaluate
$$\int_{-2y}^{2y}x^2y^2dx$$
treating $y$ as some contant with a value between $0$ and $1$.
You will get an expresion on $y$, which now you should integrate between $0$ and $1$, say
$$\int_0^1 f(y)dy$$