## Calculus 8th Edition

$241 \pi$
Surface integral over $S_1$ is given as: $\iint_{S_1} x^2+y^2+z^2 dS= \iint_{D} (9 \sin^2 u+9 \cos^2 u+v^2) (3) dA=(3) \times \iint_{D}9+v^2 dA= (3) \times \int_{0}^{2} \int_0^{2 \pi} 9+v^2 du dv=124 \pi$ Surface integral over $S_2$ is given as: $\iint_{S_2} x^2+y^2+z^2 dS= \iint_{D} (v^2 \sin^2 u+v^2 \cos^2 u+2^2) v dA$ or, $= \int_{0}^{3} \int_0^{2 \pi} [v^3+4v] du dv$ or, $=\dfrac{153 \pi}{2}$ Surface integral over $S_3$ is given as: $\iint_S x^2+y^2+z^2 dS= \iint_{D} (v^2 \sin^2 u+v^2 \cos^2 u+(0)^2) v dA$ or, $=\iint_{D} [v^3] du dv$ or, $=2 \pi \int_{0}^{3} v^3 du dv$ or, $=\dfrac{81 \pi}{2}$ Our result is: $\iint_{S} xz dS=\iint_{S_1} xz dS+\iint_{S_2} xz dS+\iint_{S_3} xz dS=124 \pi+\dfrac{153 \pi}{2}+\dfrac{81 \pi}{2}=241 \pi$