Let's solve each question step by step: --- ## **Question 9** **For any vector U, what is the value of \( |\text{Curl}(\text{Grad}U)| \)?** ### **Step-by-step Explanation:** - The gradient of a scalar function \( f \) is a vector: \( \nabla f \). - The curl of a vector field \( \mathbf{F} \) is another vector: \( \nabla \times \mathbf{F} \). - **A key vector calculus identity:** \[ \nabla \times (\nabla f) = 0 \] The curl of a gradient is always the **zero vector**, for any scalar function \( f \). - If \( U \) is a scalar function: \(\text{Curl}(\text{Grad}(U)) = 0 \). - If \( U \) is a vector field, then \(\text{Grad}(U)\) refers to the Jacobian matrix, and its "curl" isn't standard, but for most calculus contexts, the answer is **zero**. ### **Final Answer for Q9:** \[ \boxed{0} \] --- ## **Question 10** \[ \int_{0}^{4} \int_{0.5x^2}^{2x} dy\, dx \] Let's solve step by step: ### **Step 1: Integrate with respect to \( y \)** \[ \int_{0.5x^2}^{2x} dy = [y]_{0.5x^2}^{2x} = 2x - 0.5x^2 \] ### **Step 2: Integrate with respect to \( x \)** So, the integral becomes: \[ \int_{0}^{4} (2x - 0.5x^2) dx \] ### **Step 3: Integrate** \[ \int (2x - 0.5x^2) dx = \int 2x\, dx - \int 0.5x^2\, dx \] \[ = 2 \int x\, dx - 0.5 \int x^2\, dx \] \[ = 2 \left( \frac{x^2}{2} \right) - 0.5 \left( \frac{x^3}{3} \right) \] \[ = x^2 - \frac{0.5}{3} x^3 \] \[ = x^2 - \frac{1}{6} x^3 \] ### **Step 4: Evaluate limits from 0 to 4** \[ \left[ x^2 - \frac{1}{6} x^3 \right]_{0}^{4} \] Plug in \( x = 4 \): \[ 4^2 - \frac{1}{6} (4^3) = 16 - \frac{1}{6} (64) = 16 - \frac{64}{6} = 16 - 10.6667 = 5.3333 \] Plug in \( x = 0 \): \( 0^2 - \frac{1}{6}(0^3) = 0 \). So the result is: \[ \boxed{5.33} \] --- ## **Final Answers:** **Q9:** \( \boxed{0} \) **Q10:** \( \boxed{5.33} \)