1. **State the problem:** We want to evaluate the surface integral $$\iint_S \mathbf{F} \cdot d\mathbf{S}$$ where $$\mathbf{F}(x,y,z) = (2x^2 + e^y)\mathbf{i} + 5xy\mathbf{j} + 8z\mathbf{k}$$ and $$S$$ is the surface bounding the region $$E$$ defined by the coordinate planes $$x=0, y=0, z=0$$, the planes $$x=3, y=2$$, and the plane $$x+z=4$$. 2. **Use the Divergence Theorem:** The Divergence Theorem states: $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E \nabla \cdot \mathbf{F} \, dV$$ where $$\nabla \cdot \mathbf{F}$$ is the divergence of $$\mathbf{F}$$. 3. **Calculate the divergence:** $$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(2x^2 + e^y) + \frac{\partial}{\partial y}(5xy) + \frac{\partial}{\partial z}(8z)$$ Calculate each partial derivative: - $$\frac{\partial}{\partial x}(2x^2 + e^y) = 4x$$ (since $$e^y$$ is constant w.r.t. $$x$$) - $$\frac{\partial}{\partial y}(5xy) = 5x$$ - $$\frac{\partial}{\partial z}(8z) = 8$$ So, $$\nabla \cdot \mathbf{F} = 4x + 5x + 8 = 9x + 8$$ 4. **Set up the volume integral:** The region $$E$$ is bounded by: - $$0 \leq x \leq 3$$ - $$0 \leq y \leq 2$$ - $$0 \leq z \leq 4 - x$$ (from $$x + z = 4$$) The volume integral is: $$\iiint_E (9x + 8) \, dV = \int_0^3 \int_0^2 \int_0^{4-x} (9x + 8) \, dz \, dy \, dx$$ 5. **Integrate with respect to $$z$$:** $$\int_0^{4-x} (9x + 8) \, dz = (9x + 8) \times (4 - x) = (9x + 8)(4 - x)$$ 6. **Simplify the integrand:** $$ (9x + 8)(4 - x) = 36x + 32 - 9x^2 - 8x = -9x^2 + 28x + 32 $$ 7. **Integrate with respect to $$y$$:** Since the integrand does not depend on $$y$$, $$\int_0^2 (-9x^2 + 28x + 32) \, dy = 2(-9x^2 + 28x + 32) = -18x^2 + 56x + 64$$ 8. **Integrate with respect to $$x$$:** $$\int_0^3 (-18x^2 + 56x + 64) \, dx = \left[-6x^3 + 28x^2 + 64x\right]_0^3$$ Calculate each term at $$x=3$$: - $$-6(3)^3 = -6 \times 27 = -162$$ - $$28(3)^2 = 28 \times 9 = 252$$ - $$64(3) = 192$$ Sum: $$-162 + 252 + 192 = 282$$ At $$x=0$$, the expression is 0. 9. **Final answer:** $$\iint_S \mathbf{F} \cdot d\mathbf{S} = 282$$