Let's solve **Question 11** step by step: --- ### 11. Evaluate the integral \[ \iint_R 2x \cos(y^3) \, dy \, dx \] where \( R = \{(x, y) \in \mathbb{R}^2 \mid 0 \leq y \leq 1 \text{ and } 0 \leq x \leq 3y\} \). --- #### **Step 1: Write the Integral with Limits** Given \( R \), the limits for \( y \) are from \( 0 \) to \( 1 \), and for each \( y \), \( x \) goes from \( 0 \) to \( 3y \): \[ \int_{y=0}^1 \int_{x=0}^{3y} 2x \cos(y^3) \, dx \, dy \] --- #### **Step 2: Integrate with Respect to \( x \) First** \[ \int_{y=0}^1 \left[ \int_{x=0}^{3y} 2x \cos(y^3) \, dx \right] dy \] Since \( \cos(y^3) \) is treated as a constant with respect to \( x \): \[ \int_{x=0}^{3y} 2x \cos(y^3) \, dx = \cos(y^3) \int_{x=0}^{3y} 2x \, dx \] Now, integrate \( 2x \): \[ \int_{x=0}^{3y} 2x \, dx = [x^2]_{x=0}^{3y} = (3y)^2 - (0)^2 = 9y^2 \] So, the inner integral is: \[ \cos(y^3) \cdot 9y^2 = 9y^2 \cos(y^3) \] --- #### **Step 3: Integrate with Respect to \( y \):** \[ \int_{y=0}^1 9y^2 \cos(y^3) \, dy \] Let’s use substitution: Let \( u = y^3 \implies du = 3y^2 dy \implies y^2 dy = \frac{du}{3} \) When \( y = 0, u = 0 \). When \( y = 1, u = 1 \). So, \[ \int_{y=0}^1 9y^2 \cos(y^3) \, dy = 9 \int_{y=0}^1 y^2 \cos(y^3) \, dy = 9 \int_{u=0}^{u=1} \cos(u) \cdot \frac{du}{3} \] \[ = 3 \int_{u=0}^{1} \cos(u) \, du \] \[ = 3 [\sin(u)]_{u=0}^{1} \] \[ = 3 (\sin(1) - \sin(0)) \] \[ = 3 \sin(1) \] --- ### **Final Answer:** \[ \boxed{3\sin(1)} \] --- If you would like the solution to Question 12 as well, let me know!

Let's work through **Question 12** step by step. --- ### 12. Given the region \( R = \{(x,y) \in \mathbb{R}^2 \mid y > 0 \text{ and } 4 < x^2 + y^2 < 9\} \): --- #### **(a) Sketch the region \( R \) and find the corresponding region \( S \) in the polar coordinates** **Step 1: Understanding the region \( R \)** - The inequalities \( 4 < x^2 + y^2 < 9 \) describe the area between two circles: - Inner circle: \( x^2 + y^2 = 4 \) (radius \( 2 \)) - Outer circle: \( x^2 + y^2 = 9 \) (radius \( 3 \)) - \( y > 0 \) restricts the region to the upper half-plane. **Step 2: Sketch** - Draw two circles with radii 2 and 3 centered at the origin. - Shade the region between these circles in the upper half-plane (above \( x \)-axis). **Step 3: Convert to polar coordinates** Recall: \[ x = r \cos \theta, \quad y = r \sin \theta \] In polar coordinates: - \( x^2 + y^2 = r^2 \) - The region between the circles: \( 4 < r^2 < 9 \Rightarrow 2 < r < 3 \) - Since \( y > 0 \), the angle \( \theta \) spans from \( 0 \) to \( \pi \). **Thus, the corresponding region \( S \) in the \( r\theta \)-plane is:** \[ S = \{ (r, \theta) \mid 2 < r < 3, \quad 0 < \theta < \pi \} \] --- ### **(b) Evaluate the double integral** \[ I = \iint_R 2 y \, dx \, dy \] **Step 1: Convert the integrand and limits to polar coordinates** - \( y = r \sin \theta \) - \( dx \, dy = r \, dr \, d\theta \) - The integrand becomes: \[ 2 y = 2 r \sin \theta \] **Step 2: Set up the integral** \[ I = \int_{\theta=0}^{\pi} \int_{r=2}^{3} 2 r \sin \theta \cdot r \, dr \, d\theta \] Note: \( dx \, dy = r \, dr \, d\theta \), so: \[ I = \int_{0}^{\pi} \int_{2}^{3} 2 r \sin \theta \cdot r \, dr \, d\theta = \int_{0}^{\pi} \int_{2}^{3} 2 r^2 \sin \theta \, dr \, d\theta \] --- ### **Step 3: Integrate with respect to \( r \)** \[ \int_{2}^{3} 2 r^2 \, dr = 2 \int_{2}^{3} r^2 \, dr = 2 \left[ \frac{r^3}{3} \right]_{2}^{3} = 2 \left( \frac{3^3}{3} - \frac{2^3}{3} \right) = 2 \left( \frac{27}{3} - \frac{8}{3} \right) = 2 \left( 9 - \frac{8}{3} \right) \] Simplify: \[ 2 \left( \frac{27}{3} - \frac{8}{3} \right) = 2 \times \frac{19}{3} = \frac{38}{3} \] --- ### **Step 4: Integrate with respect to \( \theta \)** \[ I = \frac{38}{3} \int_{0}^{\pi} \sin \theta \, d\theta \] \[ \int_{0}^{\pi} \sin \theta \, d\theta = [-\cos \theta]_{0}^{\pi} = -\cos \pi + \cos 0 = -(-1) + 1 = 1 + 1 = 2 \] --- ### **Final result:** \[ I = \frac{38}{3} \times 2 = \frac{76}{3} \] --- ### **Answer:** \[ \boxed{\frac{76}{3}} \] --- If you have further questions or need clarifications, feel free to ask!

Certainly! Here are the diagrams and the final answers for both parts: --- ### **Question 12: Diagram Description** - **Region \( R \):** The area between two circles centered at the origin with radii 2 and 3, restricted to the upper half-plane (above the x-axis). - **Polar coordinates:** \( r \) runs from 2 to 3, \( \theta \) from 0 to \( \pi \). *(Since I can't draw directly here, imagine:)* - Draw a circle of radius 2 and another of radius 3. - Shade the region between these two circles in the upper half-plane. --- ### **Final Answers** **Part (a):** **Region \( S \) in \( r\theta \)-plane:** \[ \boxed{ 2 < r < 3, \quad 0 < \theta < \pi } \] **Part (b):** **Value of the integral:** \[ \boxed{ \frac{76}{3} } \] Let me know if you'd like me to guide you on how to draw the diagram more specifically!