Answer
$\dfrac{4 \sqrt 2-2}{3}$
Work Step by Step
Here we have $\iint_S y dS =\int_{0}^{\pi} \int_{0}^{1} (u \sin v) \sqrt {1+u^2} dA$
or, $=\int_{0}^{\pi} \sin v dv \int_{0}^{1} u [ \sqrt {1+u^2}] du$
or, $=\int_{0}^{1} 2u \sqrt {1+u^2} du$
$a=1+u^2$ and $2u du=da$
Now, $\iint_S y dS =\int_1^2 \sqrt {a} da$
or, $=\dfrac{2}{3}a^{3/2}$
Thus, we have $\iint_S y dS =\dfrac{4 \sqrt 2-2}{3}$
