Question

Evaluate the double integral. $$ \int_{-1}^{1} \int_{-2}^{2}\left(x^{2}-y^{2}\right) d y d x $$

Short answer

The value of the double integral is -4.

Step-by-step solution

Set Integral Boundaries

First, consider the region bounded by \(x\) from -1 to 1 and \(y\) from -2 to 2. The function to be integrated is \(x^{2} - y^{2}\). Arrange the problem such that you solve for \(dy\) first and then \(dx\). That is, \[\int_{-1}^{1} \int_{-2}^{2}\left(x^{2}-y^{2}\right) d y d x\]

Integrate with respect to y

Integrate the inner integral first with respect to \(y\). That is, \[\int_{-2}^{2} (x^{2}-y^{2})d y\]This integral yields \[x^{2}y - \frac{1}{3}y^{3}\]Calculated from \(y=-2\) to \(y=2\), this becomes \[2x^{2} - \frac{8}{3}\]

Integrate with respect to x

Finally, integrate the result with respect to x:\[\int_{-1}^{1} (2x^{2} - \frac{8}{3}) dx\]This integral yields \[\frac{2}{3}x^{3} - \frac{8}{3}x\]Calculated from \(x=-1\) to \(x=1\), this becomes \[\frac{2}{3} - \frac{8}{3} + \frac{2}{3} - \frac{8}{3} = -4\]