Answer
$4 \sqrt {22}$
Work Step by Step
We have $Surface \ area = \iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|$
and $\iint_{D} dA$ is the area of the region $D$
Now, $\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=\lt 3,2,3\gt $
and $|\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|=\sqrt {(3)^2+(2)^2+(3)^2}=\sqrt {22}$
Therefore, $Surface \ area=\int_{-1}^1 \int_0^2 \sqrt {3^2+2^2+3^2} \ du \ dv =\int_{-1}^1 \int_0^2 \sqrt {22} du dv$
$=\int_{-1}^1 [\sqrt {22} u]_0^2 dv$
$=[2 \sqrt {22}]_{-1}^1$
So, $Surface \ area=4 \sqrt {22}$
