Answer
$$\int_{-1}^1\int_{-2}^2(x^2-y^2)\hspace{0.5mm}dydx=-8$$
Work Step by Step
We will start with the integral with respect to y:
$\int_{-1}^1\int_{-2}^2(x^2-y^2)\hspace{0.5mm}dydx=\int_{-1}^1\left(x^2y-\frac{y^3}{3}\right)\bigg\vert_{-2}^2dx=\int_{-1}^1\left((2x^2-\frac{2^3}{3})-\left(-2x^2-\frac{(-2)^3}{3}\right)\right)dx=\int_{-1}^1\left(2x^2-\frac{8}{3}+2x^2-\frac{8}{3}\right)dx=\int_{-1}^1\left(4x^2-\frac{16}{3}\right)dx$
(We could have also gotten to this point by realizing that $x^2-y^2$ is an even function and so integrating over (-2,2) is the same thing as integrating over (0,2) and doubling the result.) Continuing we get,
$\int_{-1}^1\left(4x^2-\frac{16}{3}\right)dx=\left(\frac{4}{3}x^3-\frac{16}{3}x\right)\bigg\vert_{-1}^1$
$=\left(\frac{4}{3}-\frac{16}{3}\right)-\left(-\frac{4}{3}+\frac{16}{3}\right)=-\frac{12}{3}-\frac{12}{3}=-8$
