Let's solve the problem step by step: --- ### **Step 1: Find the Divergence of \( \mathbf{F} \)** Given: \[ \mathbf{F} = (z-y, x-z, x) \] The divergence is: \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(z-y) + \frac{\partial}{\partial y}(x-z) + \frac{\partial}{\partial z}(x) \] Calculate each term: - \( \frac{\partial}{\partial x}(z-y) = 0 \) - \( \frac{\partial}{\partial y}(x-z) = 0 \) - \( \frac{\partial}{\partial z}(x) = 0 \) So, \[ \nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0 \] --- ### **Step 2: Set Up the Volume Integral** By the Divergence Theorem: \[ \iint_{\partial D} \mathbf{F} \cdot d\mathbf{S} = \iiint_D (\nabla \cdot \mathbf{F}) \, dV \] Since the divergence is **0**, \[ \iiint_D 0 \, dV = 0 \] So, the correct answer is: \[ \boxed{0} \] --- ### **Step 3: Surface Integral Parameterization** The region \( D \) is given by: \[ \frac{x^2}{4} + \frac{y^2}{27} + \frac{z^2}{8} = 1 \] This is an ellipsoid. The parameterization is: \[ \mathbf{r} = (a \sin u \cos v,\, b \sin u \sin v,\, c \cos u) \] where \( a = 2, \, b = 3\sqrt{3}, \, c = 2\sqrt{2} \). But since the divergence integral is zero, the surface integral by Divergence Theorem must also be zero. --- ### **Step 4: Match to the Choices** - The volume integral simplifies to \( \iiint_D 0 \, dV \), which is \( 0 \). - For the surface integral, since it must also be zero, the answer is the one where the integrand does **not** simplify to a constant (not \(0\)), so the answer is **A**. --- ### **Final Answers:** #### Volume Integral: \[ \boxed{0} \] #### Surface Integral: Select option **A**. --- **Summary**: - The divergence is zero, so the volume integral is zero. - The surface integral by Divergence Theorem is also zero. - The correct selections are: - The integral simplifies to \( 0 \). - Select option **A** for the surface integral.