## Calculus 8th Edition

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Surface integral over $S_1$ is given as: $\iint_{S_1} xz dS= \iint_{D} x \times 3 \sin \theta \times (3) dx d\theta=\int_{0}^{ 2\pi} \int_0^{5-3 \cos \theta} 9 x \times \sin \theta dx d\theta=\int_{0}^{ 2\pi} (5-3 \cos \theta)^2 \times 3 \sin \theta \times \dfrac{3}{2} dx d\theta=0$ The surface integral over $S_2$ is $0$ because $\iint_{S_2} xz dS=0$ when $x=0$ Surface integral over $S_3$ is given as: $\iint_{S_3} xz dS=\iint_D (5-y) \times z \sqrt {1+0+1} dA= \int_{0}^{ 2\pi} \times \int_0^{3} (5-r \cos \theta) (r \sin \theta) \sqrt 2r dr d\theta=\int_{0}^{ 2\pi} \int_0^{3} 5 \sqrt 2r^2 \times \sin \theta dr d\theta-\int_{0}^{ 2\pi} \int_0^{3} \sqrt 2 r^3 \times \cos \theta \times \sin \theta dr d\theta$ and $\iint_{S_3} xz dS=5 \sqrt 2 \times \int_{0}^{ 2\pi} \sin \theta d \theta \times \int_0^{3} r^2 dr -\sqrt 2 \times \int_{0}^{ 2\pi} \cos \theta \sin \theta d\theta \cdot \int_0^3 r^3 dr=0$ Our result is: $\iint_{S} xz dS=\iint_{S_1} xz dS+\iint_{S_2} xz dS+\iint_{S_3} xz dS=0$