1. **Problem Statement:** Evaluate the surface integral $$\iint_S \mathbf{A} \cdot \mathbf{n} \, dS$$ where $$\mathbf{A} = z \mathbf{i} + x \mathbf{j} - 3y^2 z \mathbf{k}$$ and $$S$$ is the portion of the cylinder $$x^2 + y^2 = 8$$ lying in the first octant between $$z=0$$ and $$z=4$$. 2. **Formula and Important Rules:** The surface integral of a vector field $$\mathbf{A}$$ over a closed surface $$S$$ can be evaluated using the Divergence Theorem: $$ \iint_S \mathbf{A} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \mathbf{A} \, dV $$ where $$V$$ is the volume enclosed by $$S$$. 3. **Calculate Divergence:** $$ \nabla \cdot \mathbf{A} = \frac{\partial}{\partial x}(z) + \frac{\partial}{\partial y}(x) + \frac{\partial}{\partial z}(-3y^2 z) = 0 + 0 - 3y^2 = -3y^2 $$ 4. **Set up the volume integral:** The volume $$V$$ is bounded by the cylinder $$x^2 + y^2 \leq 8$$ in the first octant (where $$x,y,z \geq 0$$) and between $$z=0$$ and $$z=4$$. Using cylindrical coordinates: $$ x = r \cos \theta, \quad y = r \sin \theta, \quad z = z $$ with $$r \in [0, \sqrt{8}]$$, $$\theta \in [0, \frac{\pi}{2}]$$ (first octant), and $$z \in [0,4]$$. The volume element is $$dV = r \, dr \, d\theta \, dz$$. 5. **Rewrite divergence in cylindrical coordinates:** $$ -3 y^2 = -3 (r \sin \theta)^2 = -3 r^2 \sin^2 \theta $$ 6. **Evaluate the integral:** $$ \iiint_V \nabla \cdot \mathbf{A} \, dV = \int_0^{\pi/2} \int_0^{\sqrt{8}} \int_0^4 -3 r^2 \sin^2 \theta \cdot r \, dz \, dr \, d\theta $$ $$ = \int_0^{\pi/2} \int_0^{\sqrt{8}} \int_0^4 -3 r^3 \sin^2 \theta \, dz \, dr \, d\theta $$ Integrate with respect to $$z$$: $$ \int_0^4 dz = 4 $$ So, $$ = \int_0^{\pi/2} \int_0^{\sqrt{8}} -3 r^3 \sin^2 \theta \cdot 4 \, dr \, d\theta = -12 \int_0^{\pi/2} \sin^2 \theta \, d\theta \int_0^{\sqrt{8}} r^3 \, dr $$ 7. **Calculate the integrals:** $$ \int_0^{\sqrt{8}} r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^{\sqrt{8}} = \frac{(\sqrt{8})^4}{4} = \frac{(8)^2}{4} = \frac{64}{4} = 16 $$ $$ \int_0^{\pi/2} \sin^2 \theta \, d\theta = \frac{\pi}{4} $$ 8. **Final evaluation:** $$ -12 \times \frac{\pi}{4} \times 16 = -12 \times 4 \pi = -48 \pi $$ **Answer:** $$ \boxed{-48 \pi} $$ --- **Slug:** surface integral **Subject:** vector calculus **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 5