Let's break down the problem step by step: --- ### **Given Data** - **Velocity field:** \(\vec{V} = ax\hat{i} + by\hat{j}\), with \(a = b = 1\, \text{s}^{-1}\) - **Density:** \(\rho = 1\, \text{kg/m}^3\) - **Shaded area:** Rectangle in the \(y-z\) plane at \(x=5\, \text{m}\), with sides \(3\, \text{m}\) (in \(z\)) and \(4\, \text{m}\) (in \(y\)) - **Coordinates:** All in meters --- ### **1. **Volume Flow Rate (\(Q\)) Through the Shaded Area** #### **Step 1: Area Vector** The **shaded area** is perpendicular to the \(x\)-axis, so the **area vector** is: \[ d\vec{A} = dA\,\hat{i} \] Area \(A = 3\, \text{m} \times 4\, \text{m} = 12\, \text{m}^2\) ---- #### **Step 2: Velocity at \(x=5\) (on the area)** On the area, - \(x = 5\) - \(y\) ranges from 0 to 4 - \(z\) ranges from 0 to 3 So, at any point on the area: \[ \vec{V} = a x \hat{i} + b y \hat{j} = (1 \cdot 5)\hat{i} + (1 \cdot y)\hat{j} = 5\hat{i} + y\hat{j} \] --- #### **Step 3: Volume Flow Rate Expression** The volume flow rate through the area: \[ Q = \iint_A \vec{V} \cdot d\vec{A} \] Here, \(d\vec{A} = dA\,\hat{i}\), so \(\vec{V} \cdot d\vec{A} = (5\hat{i} + y\hat{j}) \cdot \hat{i} dA = 5\,dA\) So, \[ Q = \iint_A 5\, dA = 5 \iint_A dA = 5 \times \text{Area} = 5 \times 12 = \boxed{60\ \text{m}^3/\text{s}} \] --- ### **2. **Momentum Flux Through the Shaded Area** The momentum flux (in the \(x\)-direction): \[ \text{Momentum flux} = \iint_A \rho V_x \vec{V} \cdot d\vec{A} \] But since \(d\vec{A}\) is in the \(x\)-direction and \(\rho = 1\): \[ \text{Momentum flux} = \iint_A V_x (\vec{V} \cdot \hat{i})\, dA = \iint_A V_x V_x\, dA = \iint_A (V_x)^2 dA \] Here, \(V_x = 5\) (from earlier), so: \[ \text{Momentum flux} = \iint_A 5^2\, dA = 25 \iint_A dA = 25 \times \text{Area} = 25 \times 12 = \boxed{300\ \text{kg}\cdot\text{m}/\text{s}^2} \] --- ## **Final Answers** 1. **Volume flow rate through the shaded area:** \[ \boxed{60\ \text{m}^3/\text{s}} \] 2. **Momentum flux through the shaded area (\(\rho = 1\)):** \[ \boxed{300\ \text{kg}\cdot\text{m}/\text{s}^2} \] --- **Let me know if you want the full integration steps or any clarifications!**

The volume flow rate is obtained by integrating the normal component of velocity over the cross-sectional area, representing the volumetric fluid passing per unit time. The momentum flux is the rate at which momentum crosses the area, calculated by integrating the product of density, velocity component, and velocity over that area.