Let’s go step-by-step: Given: \[ u = \frac{y^3}{3} + 2x - x^2 y \] \[ v = x y^2 - 2y - \frac{x^3}{3} \] We are to check **continuity** (incompressibility) and **irrotationality**. --- ## **Step 1: Continuity Equation (Incompressibility)** For 2D incompressible flow: \[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \] Calculate \(\frac{\partial u}{\partial x}\): \[ u = \frac{y^3}{3} + 2x - x^2y \] \[ \frac{\partial u}{\partial x} = 0 + 2 - 2x y = 2 - 2x y \] Calculate \(\frac{\partial v}{\partial y}\): \[ v = x y^2 - 2y - \frac{x^3}{3} \] \[ \frac{\partial v}{\partial y} = x \cdot 2y - 2 - 0 = 2x y - 2 \] Add: \[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = (2 - 2x y) + (2x y - 2) = 0 \] **Conclusion:** The continuity equation is satisfied. --- ## **Step 2: Irrotationality Condition** For 2D flow, irrotationality requires: \[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 0 \] Calculate \(\frac{\partial v}{\partial x}\): \[ v = x y^2 - 2y - \frac{x^3}{3} \] \[ \frac{\partial v}{\partial x} = y^2 - x^2 \] Calculate \(\frac{\partial u}{\partial y}\): \[ u = \frac{y^3}{3} + 2x - x^2 y \] \[ \frac{\partial u}{\partial y} = y^2 - x^2 \] Now, \[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = (y^2 - x^2) - (y^2 - x^2) = 0 \] **Conclusion:** The flow is irrotational. --- ## **Final Answer** **Both continuity and irrotational flow conditions are satisfied for the given velocity field.**