Answer
(a) Using Theorem 4, we show that
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y \le \frac{1}{4}$
(b) Using a computer algebra system, we evaluate the double integral and the result is
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y \simeq 0.240$
Work Step by Step
(a) From the double integral $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y$, we see that the domain is a horizontally simple region whose description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 1,0 \le x \le 1} \right\}$
Using the inequality $0 \le \sin x \le x$ for $x \ge 0$, we have
$0 \le \sin \left( {xy} \right) \le xy$, ${\ \ \ \ }$ for $x,y \ge 0$
By part (a) of Theorem 4, we obtain
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sin \left( {xy} \right){\rm{d}}A \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A$
Next, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}A = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 0}^1 xy{\rm{d}}x} \right){\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 \left( {{x^2}y|_0^1} \right){\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^1 y{\rm{d}}y$
$ = \frac{1}{4}\left( {{y^2}|_0^1} \right)$
$ = \frac{1}{4}$
Hence, $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y \le \frac{1}{4}$.
(b) Using a computer algebra system, we evaluate the double integral and the result is
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \sin \left( {xy} \right){\rm{d}}x{\rm{d}}y \simeq 0.240$
