Let's solve **Question 1** step by step: --- ### **Question 1** Let \( D \) be the region in the first quadrant of the \( xy \)-plane given by \( 1 \leq x^2 + y^2 \leq 4 \). Set up and evaluate a double integral of the function \( f(x, y) = xy \) over the region. --- #### **Step 1: Describe the Region \( D \) in Polar Coordinates** - The region is between two circles: radius 1 (\( x^2 + y^2 = 1 \)) and radius 2 (\( x^2 + y^2 = 4 \)), and only in the first quadrant. - In polar coordinates: - \( x = r\cos\theta \) - \( y = r\sin\theta \) - The limits are: - \( r \) goes from 1 to 2 - \( \theta \) goes from 0 to \( \frac{\pi}{2} \) (first quadrant) --- #### **Step 2: Write the Double Integral in Polar Coordinates** - \( x y = r\cos\theta \cdot r\sin\theta = r^2 \cos\theta \sin\theta \) - Differential area in polar coordinates is \( r dr d\theta \) - So the integral becomes: \[ \iint_D xy \; dA = \int_{0}^{\frac{\pi}{2}} \int_{1}^{2} r^2 \cos\theta \sin\theta \cdot r \; dr d\theta \] \[ = \int_{0}^{\frac{\pi}{2}} \int_{1}^{2} r^3 \cos\theta \sin\theta \; dr d\theta \] --- #### **Step 3: Separate and Integrate** - \( r^3 \) with respect to \( r \) from 1 to 2 - \( \cos\theta \sin\theta \) with respect to \( \theta \) from 0 to \( \frac{\pi}{2} \) So, \[ \int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta \; d\theta \int_{1}^{2} r^3 dr \] First, compute the \( r \) integral: \[ \int_{1}^{2} r^3 dr = \left[ \frac{r^4}{4} \right]_{1}^{2} = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \] Now, the \( \theta \) integral: \[ \int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta d\theta \] Recall that \( \sin\theta \cos\theta = \frac{1}{2} \sin 2\theta \): \[ \int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin 2\theta d\theta = \frac{1}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_{0}^{\frac{\pi}{2}} \] \[ = -\frac{1}{4} \left[ \cos(\pi) - \cos(0) \right] = -\frac{1}{4} [(-1) - (1)] = -\frac{1}{4} \times (-2) = \frac{1}{2} \] --- #### **Step 4: Combine Results** Multiply the two parts: \[ \frac{15}{4} \times \frac{1}{2} = \frac{15}{8} \] --- ### **Final Answer** \[ \boxed{\frac{15}{8}} \] This is the value of the double integral of \( xy \) over the described region \( D \).