Let's break down each part as requested, **step by step**. --- ## (a) **Compute the Jacobian** ### **Step 1: Find Partial Derivatives** Given: - \( u = x^2 - y^2 \) - v = xy \) partial derivatives: \[ \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y \] \[ \frac{\partial v}{\partial x} = 2y, \quad \frac{\partial v}{\partial y} = 2x \] The Jacobian matrix is: \[ J = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix} \] **Explanation:** We calculated each partial derivative to fill out the Jacobian matrix. --- ### **Step 2: Compute Determinant** \[ \text{det}(J) = (2x)(2x) - (-2y)(2y) = 4x^2 + 4y^2 = 4(x^2 + y^2) \] So, \[ \left| \frac{\partial(u,v)}{\partial(x,y)} \right| = 4(x^2 + y^2) \] **Explanation:** The determinant is the product of the diagonal minus the product of the off-diagonal terms. This gives the scaling factor for the area transformation. --- ### **Final Answer (a):** \[ \boxed{ \left| \frac{\partial(u,v)}{\partial(x,y)} \right| = 4(x^2 + y^2) } \] --- ## (b) **Describe the image region \(\mathcal{R}^*\) in the \(uv\)-plane** ### **Step 1: Rewrite Region in Terms of \(u, v\)** Given \(x^2 + y^2 = r^2\), the region is \[ 1 \leq x^2 + y^2 \leq 4 \] and \(x > \), \(\frac{1}{2}x \leq y \leq x\). Express \(u, v\) in polar coordinates (\(x = r\cos\theta, y = r\sin\theta\)): \[ u = x^2 - y^2 = r^2(\cos^2\theta - \sin^2\theta) = r^2\cos 2\theta \] \[ v = 2xy = 2r^2\cos\theta\sin\theta = r^2\sin 2\theta \] So, \(u + iv = r^2 e^{i2\theta}\). **Explanation:** The transformation "squares" the complex number \(z = x + iy = re^{i\theta}\), mapping it to \(z^2 = r^2 e^{i2\theta}\). --- ### **Step 2: Describe the Image Region** - \(1 \leq r^2 \leq 4\) ⟹ \(1 \leq \sqrt{u^2 + v^2} \leq 4\) - For \(x > \), \(\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})\) - \(\frac{1}{2}x \leq y \leq x\) ⟹ \(\tan\theta \in [1/2, 1]\) ⟹ \(\theta \in [\arctan(1/2), \arctan(1)]\) But since \(2\theta\) is mapped, the region in the \(uv\)-plane is: - \(1 \leq \sqrt{u^2 + v^2} \leq 4\) - \(2\theta \in [2\arctan(1/2), 2\arctan(1)]\) **Explanation:** The region in the \(uv\)-plane is an annular sector, with radii from 1 to 4, and angles doubled from the original \(\theta\) range. --- ### **Final Answer (b):** \[ \boxed{ \mathcal{R}^* = \left\{ (u, v) : 1 \leq \sqrt{u^2 + v^2} \leq 4,\quad 2\arctan(1/2) \leq \arg(u+iv) \leq 2\arctan(1) \right\} } \] --- ## (c) **Evaluate \(\iint_\mathcal{R} (x^2 + y^2) dA\) in \((u,v)\)-coordinates** ### **Step 1: Set Up the Integral in \((u,v)\)** From part (a), \(\left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac{1}{4(x^2 + y^2)}\). Let \(x^2 + y^2 = r^2\), and \(u^2 + v^2 = r^4\) ⟹ \(r^2 = \sqrt{u^2 + v^2}\). The integrand \((x^2 + y^2) dA = r^2 dA\). So, \[ \iint_\mathcal{R} (x^2 + y^2) dA = \iint_{\mathcal{R}^*} (x^2 + y^2) \left| \frac{\partial(x,y)}{\partial(u,v)} \right| du dv \] \[ = \iint_{\mathcal{R}^*} r^2 \cdot \frac{1}{4r^2} du dv = \iint_{\mathcal{R}^*} \frac{1}{4} du dv \] **Explanation:** The Jacobian cancels the \(r^2\) in the numerator, leaving a constant integrand. --- ### **Step 2: Integrate Over \(\mathcal{R}^*\) (Annular Sector)** The area of the annular sector in the \(uv\)-plane is: - Radii: \(1 \leq \rho \leq 4\) - Angles: \(\phi_1 = 2\arctan(1/2)\), \(\phi_2 = 2\arctan(1)\) The area is: \[ \int_{\phi_1}^{\phi_2} \int_{1}^{4} \frac{1}{4} \rho d\rho d\phi \] Integrate in \(\rho\): \[ \int_1^4 \frac{1}{4} \rho d\rho = \frac{1}{4} \left[ \frac{\rho^2}{2} \right]_1^4 = \frac{1}{4} \left( \frac{16}{2} - \frac{1}{2} \right) = \frac{1}{4} \cdot \frac{15}{2} = \frac{15}{8} \] Integrate in \(\phi\): \[ (\phi_2 - \phi_1) \cdot \frac{15}{8} \] Where: - \(\phi_1 = 2\arctan(1/2)\) - \(\phi_2 = 2\arctan(1) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}\) **Explanation:** We converted the region to polar coordinates and calculated the area of the sector, multiplying by the constant integrand. --- ### **Final Answer (c):** \[ \boxed{ \iint_\mathcal{R} (x^2 + y^2) dA = \frac{15}{8} \left( \frac{\pi}{2} - 2\arctan(1/2) \right) } \] --- **Let me know if you'd like a more detailed explanation or the numeric value!**

To solve all parts of the problem step by step, we will follow the structure provided while ensuring clarity and conciseness. --- # Problem Breakdown ## (a) **Compute the Jacobian** ### **Step 1: Find Partial Derivatives** Given the transformation: - \( u = x^2 - y^2 \) - \( v = 2xy \) Calculate the partial derivatives: \[ \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y \] \[ \frac{\partial v}{\partial x} = 2y, \quad \frac{\partial v}{\partial y} = 2x \] The Jacobian matrix \( J \) is: \[ J = \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix} \] **Explanation:** We computed the partial derivatives to construct the Jacobian matrix, which describes how the transformation affects the area. --- ### **Step 2: Compute Determinant** Calculate the determinant of the Jacobian: \[ \text{det}(J) = (2x)(2x) - (-2y)(2y) = 4x^2 + 4y^2 = 4(x^2 + y^2) \] Thus, the absolute value of the Jacobian determinant is: \[ \left| \frac{\partial(u,v)}{\partial(x,y)} \right| = 4(x^2 + y^2) \] **Explanation:** The determinant is derived from the formula for the determinant of a 2x2 matrix, indicating how area scales under the transformation. --- ### **Final Answer (a):** \[ \boxed{ \left| \frac{\partial(u,v)}{\partial(x,y)} \right| = 4(x^2 + y^2) } \] --- ## (b) **Describe the Image Region \(\mathcal{R}^*\) in the \(uv\)-plane** ### **Step 1: Rewrite Region in Terms of \(u, v\)** The region \( \mathcal{R} \) in the \(xy\)-plane is defined by: \[ 1 < x^2 + y^2 < 4, \quad x > 0, \quad \frac{1}{2}x < y < x \] Using polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \): The conditions become: - For \( x^2 + y^2 \): \[ 1 < r^2 < 4 \quad \Rightarrow \quad 1 < r < 2 \] - For the angles: \[ \frac{1}{2} x < y < x \quad \Rightarrow \quad \frac{1}{2} \cos \theta < \sin \theta < \cos \theta \] This leads to: \[ \tan \theta > \frac{1}{2} \quad \text{and} \quad \tan \theta < 1 \quad \Rightarrow \quad \theta \in \left( \arctan\left(\frac{1}{2}\right), \frac{\pi}{4} \right) \] ### **Step 2: Map to \(uv\)-coordinates** The transformation gives: \[ u = r^2 \cos(2\theta), \quad v = r^2 \sin(2\theta) \] Thus, the image region in the \(uv\)-plane is an annular sector: - \(1 < \sqrt{u^2 + v^2} < 4\) - Angles: \( 2 \arctan\left(\frac{1}{2}\right) < \arg(u + iv) < \frac{\pi}{2} \) **Explanation:** The mapping transforms the circular region in polar coordinates to a squared sector in the \(uv\)-plane. --- ### **Final Answer (b):** \[ \boxed{ \mathcal{R}^* = \left\{ (u, v) : 1 < \sqrt{u^2 + v^2} < 4,\quad 2 \arctan\left(\frac{1}{2}\right) < \arg(u+iv) < \frac{\pi}{2} \right\} } \] --- ## (c) **Evaluate \(\iint_{\mathcal{R}} (x^2 + y^2) dA\) by Transforming to \((u,v)\)-coordinates** ### **Step 1: Set Up the Integral in \((u,v)\)** From part (a), we know: \[ \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac{1}{4(x^2 + y^2)} \] Using \(x^2 + y^2 = r^2\), we have: \[ \iint_{\mathcal{R}} (x^2 + y^2) dA = \iint_{\mathcal{R}^*} r^2 \cdot \frac{1}{4r^2} du dv = \iint_{\mathcal{R}^*} \frac{1}{4} du dv \] **Explanation:** We simplified the integral by noting that the Jacobian cancels the \(r^2\) in the numerator, leaving a constant integrand. --- ### **Step 2: Integrate Over \(\mathcal{R}^*\) (Annular Sector)** The area of the annular sector is found through: \[ \text{Area} = \int_{\phi_1}^{\phi_2} \int_{1}^{2} \frac{1}{4} \rho d\rho d\phi \] Calculating the \(\rho\) integral: \[ \int_{1}^{2} \frac{1}{4} \rho d\rho = \frac{1}{4} \left[ \frac{\rho^2}{2} \right]_{1}^{2} = \frac{1}{4} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8} \] Calculating the \(\phi\) integral, where: - \(\phi_1 = 2 \arctan\left(\frac{1}{2}\right)\) - \(\phi_2 = \frac{\pi}{2}\) The area becomes: \[ \text{Area} = \left(\frac{\pi}{2} - 2 \arctan\left(\frac{1}{2}\right)\right) \cdot \frac{3}{8} \] Thus, the final integral evaluates to: \[ \iint_{\mathcal{R}} (x^2 + y^2) dA = \frac{3}{8} \left(\frac{\pi}{2} - 2\arctan\left(\frac{1}{2}\right)\right) \] **Explanation:** We computed the area of the annular sector and multiplied by the constant integrand. --- ### **Final Answer (c):** \[ \boxed{ \iint_{\mathcal{R}} (x^2 + y^2) dA = \frac{3}{8} \left(\frac{\pi}{2} - 2\arctan\left(\frac{1}{2}\right)\right) } \] --- This completes the solution for all parts of the problem. Let me know if you need further clarifications or detailed calculations!