Answer
$241 \pi$
Work Step by Step
The surface integral over $S_1$ is given by:
$\iint_{S_1} x^2+y^2+z^2 dS= \iint_{D} (9 \sin^2 u+9 \cos^2 u+v^2) (3) dA$
or, $=3 \times \iint_{D}9+v^2 dA$
or, $=3 \times \int_{0}^{2} \int_0^{2 \pi} [9+v^2]du dv$
or, $\iint_{S_1} x^2+y^2+z^2 dS=124 \pi$
The surface integral over $S_2$
$\iint_{S_2} x^2+y^2+z^2 dS= \iint_{D} (v^2 \sin^2 u+v^2 \cos^2 u+2^2) v dA$
or, $= \int_{0}^{3} \int_0^{2 \pi}[ v^3+4v] du dv$
or,$ \iint_{S_2} x^2+y^2+z^2 dS=\dfrac{153 \pi}{2}$
The surface integral over $S_3$ is given as:
$\iint_{S_3} x^2+y^2+z^2 dS= \iint_{D} (v^2 \sin^2 u+v^2 \cos^2 u+(0)^2) v dA$
or, $=\iint_{D}v^3 du dv$
or, $=(2 \pi) \times \int_{0}^{3} v^3 dv dv$
$\iint_{S_3} x^2+y^2+z^2 dS=(2 \pi) \times [\dfrac{v^4}{4}]_0^3=\dfrac{81 \pi}{2}$
Thus, we have $\iint_{S} x^2+y^2+z^2 dS=124 \pi+\dfrac{153 \pi}{2}+\dfrac{81 \pi}{2}=241 \pi$
