Answer
$0$
Work Step by Step
The surface integral over $S_1$ can be calculated as:
$\iint_S xz dS= \iint_{D} x \times (3 \sin \theta) (3) dx d\theta=\int_{0}^{ 2\pi} \int_0^{5-3 \cos \theta} 9 x \sin \theta dx d\theta=(3/2) \times \int_{0}^{ 2\pi} (5-3 \cos \theta)^2 \times 3 \sin \theta dx d\theta=0$
The surface integral over $S_2$ when $x=0$ is $\iint_{S_2} xz dS=0$
Now, the surface integral over $S_3$ is:
$\iint_{S_3} xz dS= \int_{0}^{ 2\pi} \int_0^{3} (5-r \cos \theta) \times (r \sin \theta) \times \sqrt 2r dr d\theta$
or, $=\int_{0}^{ 2\pi} \int_0^{3} 5 \sqrt 2r^2 \times [\sin \theta dr d\theta] -\int_{0}^{ 2\pi} \int_0^{3} [\sqrt 2 r^3] \times [\cos \theta \times \sin \theta \times dr d\theta]=0$
Total surface integral is given by: $\iint_{S} xz dS=\iint_{S_1} xz dS+\iint_{S_2} xz dS+\iint_{S_3} xz dS=0$
