Answer
(a) The sine function is an odd function, therefore the inner integral is zero. Hence, the double integral is zero.
(b) The cosine function is an even function. The inner integral and the outer integral is always positive, therefore
$\mathop \smallint \limits_{ - 1}^1 \mathop \smallint \limits_{ - 1}^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y > 0$
Work Step by Step
(a) Evaluate:
$\mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = - 1}^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y = - \mathop \smallint \limits_{y = - 1}^1 \frac{1}{y}\left( {\cos \left( {xy} \right)|_{ - 1}^1} \right){\rm{d}}y$
$ = - \mathop \smallint \limits_{y = - 1}^1 \frac{1}{y}\left( {\cos y - \cos \left( { - y} \right)} \right){\rm{d}}y$
Since $\cos \left( { - y} \right) = \cos y$, so $\mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = - 1}^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y = 0$.
The sine function is an odd function, therefore the inner integral is zero. Hence, the double integral is zero.
(b) Evaluate
$\mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = - 1}^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{y = - 1}^1 \frac{1}{y}\left( {\sin \left( {xy} \right)|_{ - 1}^1} \right){\rm{d}}y$
$\mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = - 1}^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y = 2\mathop \smallint \limits_{y = - 1}^1 \frac{1}{y}\sin y{\rm{d}}y$
Notice that $\sin y$ and $\frac{1}{y}$ are both odd functions. However, the product $\frac{1}{y}\sin y$ is an even function. Therefore, we can write
$\mathop \smallint \limits_{y = - 1}^1 \mathop \smallint \limits_{x = - 1}^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y = 4\mathop \smallint \limits_{y = 0}^1 \frac{1}{y}\sin y{\rm{d}}y$
Since $\frac{1}{y}$ and $\sin y$ are both positive in the domain $(0,1]$, the integral is positive. Hence,
$\mathop \smallint \limits_{ - 1}^1 \mathop \smallint \limits_{ - 1}^1 \cos \left( {xy} \right){\rm{d}}x{\rm{d}}y > 0$
