Answer
$0$
Work Step by Step
The surface integral over $S_1$
$\iint_S xz dS= \iint_{D} (x 3 \sin \theta) (3) dx d\theta$
$=\int_{0}^{ 2\pi} \int_0^{5-3 \cos \theta} 9 x \sin \theta dx d\theta$
$=(3/2) \times \int_{0}^{ 2\pi} (5-3 \cos \theta)^2 \times 3 \sin \theta dx d\theta$
$=0$
The surface integral over $S_2$ is $0$ because $\iint_{S_2} xz dS=0$ when $x=0$
The surface integral over $S_3$ is:
$\iint_{S_3} xz dS=\iint_D (5-y) z \sqrt {1+0+1} dA= \int_{0}^{ 2\pi} \int_0^{3} (5-r \cos \theta) (r \sin \theta) \sqrt 2r dr d\theta$
$=\int_{0}^{ 2\pi} \int_0^{3} (5 \sqrt 2r^2 \sin \theta dr d\theta-\int_{0}^{ 2\pi} \int_0^{3} \sqrt 2 r^3 \cos \theta \sin \theta dr d\theta$
$=5 \sqrt 2 \int_{0}^{ 2\pi} \sin \theta d \theta \int_0^{3} r^2 dr -\sqrt 2 \int_{0}^{ 2\pi} \cos \theta \sin \theta d\theta \cdot \int_0^3 r^3 dr$
$=0$
Thus, we have $\iint_{S} xz dS=\iint_{S_1} xz dS+\iint_{S_2} xz dS+\iint_{S_3} xz dS=0$
