Answer
$$\approx 70.9963$$
Work Step by Step
We are given that$$z=x \sin y$$
$Surface \space Area = \int_{-\pi}^{\pi}\int_{-3}^3 \sqrt {1+[\sin y]^2 +[x \cos y]^2} dx dy$
Now, using a calculator, we get:
$$Surface \space area= \int_{-\pi}^{\pi}\int_{-3}^3 \sqrt {1+\sin^2 y +x^2 \cos^2 y} dx dy \approx 70.9963$$
