Let's solve **both problems step-by-step** as requested: --- ## **1. Change of Variables for the Integral** \[ \int_0^1 \int_y^{y+2} \sqrt{x-y} \, dx \, dy \] Hint: Try \( u = x-y, \ v = y \). ### **Step 1: Change of Variables** Let - \( u = x-y \) - \( v = y \) Then, - \( x = u + v \) - \( y = v \) The Jacobian determinant (\( J \)) for the transformation is: \[ \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = (1)(1) - (1)(0) = 1 \] So, the Jacobian is 1. --- ### **Step 2: Change the Limits** Original limits: - \( y \) goes from \( 0 \) to \( 1 \) - For each \( y \), \( x \) goes from \( y \) to \( y+2 \) In \( (u,v) \) coordinates: - \( v = y \) goes from \( 0 \) to \( 1 \) - For each \( v \), \( u = x-y \) goes from \( 0 \) (when \( x = y \)) to \( 2 \) (when \( x = y+2 \)) --- ### **Step 3: Substitute and Integrate** The integrand \( \sqrt{x-y} = \sqrt{u} \). So, the new integral is: \[ \int_{v=0}^1 \int_{u=0}^2 \sqrt{u} \cdot 1 \ du \ dv \] --- ### **Step 4: Compute the Integral** First, integrate with respect to \( u \): \[ \int_{u=0}^2 \sqrt{u} \ du = \int_0^2 u^{1/2} du = \left[ \frac{2}{3} u^{3/2} \right]_0^2 = \frac{2}{3} (2)^{3/2} - 0 = \frac{2}{3} \cdot 2 \sqrt{2} = \frac{4\sqrt{2}}{3} \] Now, integrate with respect to \( v \) from \( 0 \) to \( 1 \): \[ \int_{v=0}^1 \frac{4\sqrt{2}}{3} dv = \frac{4\sqrt{2}}{3}[v]_0^1 = \frac{4\sqrt{2}}{3} (1-0) = \frac{4\sqrt{2}}{3} \] --- ### **Final Answer (Problem 1):** \[ \boxed{\frac{4\sqrt{2}}{3}} \] --- ## **2. Region Bounded by Lines and Change of Variables** The region \( R \) is bounded by: - \( x + y = 4 \) - \( x + y = 1 \) - \( x - 2y = 0 \) - \( x - 2y = -4 \) We need to compute: \[ \iint_R 3xy \ dA \] --- ### **Step 1: Choose Substitutions** Let: - \( u = x + y \) - \( v = x - 2y \) Then, - \( x = \frac{2u + v}{3} \) - \( y = \frac{u - v}{3} \) --- ### **Step 2: Find the Jacobian** \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{vmatrix} = \left( \frac{2}{3} \cdot -\frac{1}{3} \right) - \left( \frac{1}{3} \cdot \frac{1}{3} \right) = -\frac{2}{9} - \frac{1}{9} = -\frac{1}{3} \] So, \( |J| = \frac{1}{3} \). --- ### **Step 3: Express Integrand in Terms of \( u, v \)** \[ xy = x y = \left(\frac{2u + v}{3}\right)\left(\frac{u - v}{3}\right) = \frac{(2u + v)(u - v)}{9} = \frac{2u(u-v) + v(u-v)}{9} = \frac{2u^2 - 2uv + vu - v^2}{9} = \frac{2u^2 - uv - v^2}{9} \] --- ### **Step 4: Set Up the Bounds** Given: - \( x + y = u \), from 1 to 4 - \( x - 2y = v \), from -4 to 0 So \( u \) goes from 1 to 4, and \( v \) goes from -4 to 0. --- ### **Step 5: Set Up the Integral** \[ \iint_R 3xy \ dA = \iint_{u=1}^{4} \int_{v=-4}^{0} 3 \left( \frac{2u^2 - uv - v^2}{9} \right) \cdot \left| -\frac{1}{3} \right| dv du \] \[ = \iint_{u=1}^{4} \int_{v=-4}^{0} \frac{3}{9}(2u^2 - uv - v^2) \cdot \frac{1}{3} dv du \] \[ = \iint_{u=1}^{4} \int_{v=-4}^{0} \frac{1}{9}(2u^2 - uv - v^2) dv du \] --- ### **Step 6: Integrate with Respect to \( v \)** \[ \int_{v=-4}^{0} (2u^2 - uv - v^2) dv = \int_{v=-4}^{0} 2u^2 dv - \int_{v=-4}^{0} uv dv - \int_{v=-4}^{0} v^2 dv \] \[ = 2u^2 [v]_{-4}^0 - u [v^2/2]_{-4}^0 - [v^3/3]_{-4}^0 \] \[ = 2u^2 (0 - (-4)) - u \left(0 - \frac{16}{2}\right) - \left(0 - \frac{(-64)}{3}\right) \] \[ = 2u^2 (4) - u (-8) + \frac{64}{3} \] \[ = 8u^2 + 8u + \frac{64}{3} \] --- ### **Step 7: Integrate with Respect to \( u \)** So the integral is: \[ \int_{u=1}^{4} \frac{1}{9} \left(8u^2 + 8u + \frac{64}{3}\right) du \] \[ = \frac{1}{9} \left( \int_{u=1}^{4} 8u^2 du + \int_{u=1}^{4} 8u du + \int_{u=1}^{4} \frac{64}{3} du \right) \] Compute each: 1. \( \int_{1}^{4} 8u^2 du = 8 \left[ \frac{u^3}{3} \right]_{1}^{4} = 8 \left( \frac{64}{3} - \frac{1}{3} \right) = 8 \times \frac{63}{3} = 8 \times 21 = 168 \) 2. \( \int_{1}^{4} 8u du = 8 \left[ \frac{u^2}{2} \right]_{1}^{4} = 8 \times \left( \frac{16}{2} - \frac{1}{2} \right) = 8 \times \frac{15}{2} = 4 \times 15 = 60 \) 3. \( \int_{1}^{4} \frac{64}{3} du = \frac{64}{3} (4-1) = \frac{64}{3} \times 3 = 64 \) Sum: \( 168 + 60 + 64 = 292 \) --- ### **Step 8: Multiply by \( \frac{1}{9} \)** \[ \frac{1}{9} \times 292 = \frac{292}{9} \] --- ### **Final Answer (Problem 2):** \[ \boxed{\frac{292}{9}} \] --- ## **Summary of Answers** 1. \(\boxed{\frac{4\sqrt{2}}{3}}\) 2. \(\boxed{\frac{292}{9}}\) If you need any clarification, let me know!