1. **Problem statement:** Find the surface integral \(\int_{\Sigma} \vec{F} \cdot d\vec{S}\) where \(\Sigma\) is the hemisphere of radius 1 centered at the origin, lying above the \(xy\)-plane, oriented downward. 2. **Recall the vector field:** \[\vec{F}(x,y,z) = (2yz - 3xyz^2) \vec{i} + (x + 2yz - 2xz) \vec{j} + (2 + yz^3 - z^2) \vec{k}\] 3. **Orientation and surface:** The hemisphere \(\Sigma\) is the upper hemisphere \(z \geq 0\) of radius 1, but the orientation is downward, i.e., the normal vector points toward the negative \(z\)-direction. 4. **Use the Divergence Theorem:** Since \(\Sigma\) is a closed surface only if we include the disk \(D\) in the \(xy\)-plane (\(z=0\)) to close the hemisphere, and the orientation is downward, we can use the Divergence Theorem on the closed surface \(S = \Sigma + D\) with outward normal. The Divergence Theorem states: \[ \int_{S} \vec{F} \cdot d\vec{S} = \int_{V} \nabla \cdot \vec{F} \, dV \] where \(V\) is the volume enclosed by \(S\). 5. **Adjust for orientation:** The hemisphere \(\Sigma\) is oriented downward, which is inward for the volume \(V\) (since outward normal on the hemisphere points upward). Thus, \[ \int_{\Sigma} \vec{F} \cdot d\vec{S} = - \int_{\Sigma} \vec{F} \cdot d\vec{S}_{\text{outward}}. \] 6. **Calculate divergence:** \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2yz - 3xyz^2) + \frac{\partial}{\partial y}(x + 2yz - 2xz) + \frac{\partial}{\partial z}(2 + yz^3 - z^2) \] Calculate each term: - \(\frac{\partial}{\partial x}(2yz - 3xyz^2) = -3yz^2\) - \(\frac{\partial}{\partial y}(x + 2yz - 2xz) = 2z\) - \(\frac{\partial}{\partial z}(2 + yz^3 - z^2) = 3yz^2 - 2z\) Sum: \[ \nabla \cdot \vec{F} = -3yz^2 + 2z + 3yz^2 - 2z = 0 \] 7. **Interpretation:** The divergence is zero everywhere. 8. **Apply Divergence Theorem:** \[ \int_{S} \vec{F} \cdot d\vec{S} = \int_{V} 0 \, dV = 0 \] 9. **Split the closed surface integral:** \[ \int_{S} \vec{F} \cdot d\vec{S} = \int_{\Sigma} \vec{F} \cdot d\vec{S} + \int_{D} \vec{F} \cdot d\vec{S} = 0 \] 10. **Orientation of \(D\):** The disk \(D\) lies in the plane \(z=0\) with upward normal \(\vec{k}\) for the closed surface outward orientation. 11. **Calculate \(\int_D \vec{F} \cdot d\vec{S}\):** On \(D\), \(z=0\), so \[ \vec{F}(x,y,0) = (0) \vec{i} + (x + 0 - 0) \vec{j} + (2 + 0 - 0) \vec{k} = 0 \vec{i} + x \vec{j} + 2 \vec{k} \] The surface element is \(d\vec{S} = \vec{k} \, dA\) (upward). Dot product: \[ \vec{F} \cdot \vec{k} = 2 \] Area of disk radius 1 is \(\pi\). So, \[ \int_D \vec{F} \cdot d\vec{S} = \int_D 2 \, dA = 2 \pi \] 12. **Find \(\int_{\Sigma} \vec{F} \cdot d\vec{S}\):** From step 9, \[ \int_{\Sigma} \vec{F} \cdot d\vec{S} = - \int_D \vec{F} \cdot d\vec{S} = -2 \pi \] **Final answer:** \[ \boxed{\int_{\Sigma} \vec{F} \cdot d\vec{S} = -2 \pi} \]