Answer
$\approx 70.9963$
Work Step by Step
We are given that $z=x \sin y$
Now, $Surface \ Area = \int_{-\pi}^{\pi}\int_{-3}^3 \sqrt {1+[\sin y]^2 +[x \cos y]^2} \ dx \ dy$
We will use calculator, so we have
$$Surface \ Area = \int_{-\pi}^{\pi}\int_{-3}^3 \sqrt {1+\sin^2 y +x^2 \cos^2 y} \ dx \ dy \approx 70.9963$$
