Let's break down the problem step by step. ### **Given:** - \(\mathcal{S}\): Half-cylinder \(x^2 + y^2 = 1, \; x \geq 0, \; 0 \leq z \leq 1\) - \(\mathbf{F}\) is horizontal (\(F_z = 0\)), \(\mathbf{F}(0, y, z) = zy^2 \mathbf{i}\) - \(\mathcal{W}\): Solid region enclosed by \(\mathcal{S}\) - \(\iiint_{\mathcal{W}} \mathrm{div}(\mathbf{F})\, dV = 4\) - **Asked:** Find the flux of \(\mathbf{F}\) through the curved side of \(\mathcal{S}\). --- ## **Step 1: Gauss' (Divergence) Theorem** \[ \iint_{\partial \mathcal{W}} \mathbf{F} \cdot d\mathbf{A} = \iiint_{\mathcal{W}} \nabla \cdot \mathbf{F} \, dV = 4 \] \(\partial \mathcal{W}\) is the boundary of the half-cylinder, which consists of: 1. The curved side (\(\mathcal{S}_\text{curved}\)) — what we want! 2. The flat base at \(z=0\) (\(\mathcal{S}_\text{base}\)) 3. The top at \(z=1\) (\(\mathcal{S}_\text{top}\)) 4. The flat face at \(x=0\) (\(\mathcal{S}_\text{face}\)) So, \[ \text{Flux through all sides} = \text{Flux through curved} + \text{Flux through base} + \text{Flux through top} + \text{Flux through face} \] Let’s denote: - \(\Phi_\text{curved}\): Flux through the curved side (**what we want**). - \(\Phi_\text{base}, \Phi_\text{top}, \Phi_\text{face}\): Flux through base, top, and face. \[ \Phi_\text{curved} + \Phi_\text{base} + \Phi_\text{top} + \Phi_\text{face} = 4 \] --- ## **Step 2: Compute Fluxes Through Other Sides** ### **Base (\(z=0\)):** At \(z=0\), \(\mathbf{F} = 0\), so \[ \Phi_\text{base} = 0 \] ### **Top (\(z=1\)):** At \(z=1\), \(\mathbf{F}(0, y, 1) = y^2 \mathbf{i}\), but in general, on the top face (\(z=1\)), \(\mathbf{F}(x, y, 1)\). But we're told only the value at \(x=0\). But recall: \(\mathbf{F}\) has no \(z\)-component, so \(\mathbf{F} \cdot \mathbf{\hat{n}}\) (where \(\mathbf{\hat{n}}\) is in the \(z\)-direction) is 0. So, \[ \Phi_\text{top} = 0 \] ### **Face at \(x=0\):** The face at \(x=0, y^2 \leq 1\), \(0 \leq z \leq 1\): Outward normal is \(-\mathbf{\hat{i}}\), so \[ \Phi_\text{face} = \iint_{y^2 \leq 1, \; 0 \leq z \leq 1} \mathbf{F}(0, y, z) \cdot (-\mathbf{\hat{i}}) \, dA \] But \(\mathbf{F}(0, y, z) = zy^2 \mathbf{i}\), so dot with \(-\mathbf{\hat{i}}\): \[ \Phi_\text{face} = \iint_{y^2 \leq 1, \; 0 \leq z \leq 1} -zy^2 \, dy \, dz \] Calculate: \[ \int_{z=0}^1 \int_{y=-1}^1 -z y^2 \, dy \, dz \] But \(y^2\) is even, so: \[ \int_{y=-1}^1 y^2 dy = 2 \int_0^1 y^2 dy = 2 \left[ \frac{y^3}{3} \right]_0^1 = \frac{2}{3} \] So, \[ \Phi_\text{face} = \int_{z=0}^1 -z \cdot \frac{2}{3} dz = -\frac{2}{3} \int_0^1 z dz = -\frac{2}{3} \cdot \frac{1}{2} = -\frac{1}{3} \] --- ## **Step 3: Solve for the Desired Flux** Recall: \[ \Phi_\text{curved} + 0 + 0 + (-1/3) = 4 \implies \Phi_\text{curved} = 4 + \frac{1}{3} = \boxed{\frac{13}{3}} \] --- ### **Final Answer:** \[ \boxed{\frac{13}{3}} \] is the flux of \(\mathbf{F}\) through the curved side of \(\mathcal{S}\).