Let's solve **Problem 1** step by step: ### **Step 1: Understanding the Region \( D \)** Given curves: - \( xy = 1 \) - \( xy = 2 \) - \( x^2y = 1 \) - \( x^2y = 4 \) These curves bound the region \( D \). ### **Step 2: Substitution** Let’s define: - \( u = xy \) (so \( u \) goes from 1 to 2) - \( v = x^2y \) (so \( v \) goes from 1 to 4) Now, solve for \( x \) and \( y \) in terms of \( u \) and \( v \): From \( v = x^2y \) and \( u = xy \): - \( v = x \cdot (xy) = x \cdot u \implies x = v/u \) - \( y = u/x = u/(v/u) = u^2/v \) So, - \( x = v/u \) - \( y = u^2/v \) ### **Step 3: Find the Jacobian** The Jacobian determinant is: \[ J = \left| \frac{\partial(x, y)}{\partial(u, v)} \right| \] Let's compute: - \( x = v/u \) - \( y = u^2/v \) So, \[ \frac{\partial x}{\partial u} = -\frac{v}{u^2}, \quad \frac{\partial x}{\partial v} = \frac{1}{u} \] \[ \frac{\partial y}{\partial u} = \frac{2u}{v}, \quad \frac{\partial y}{\partial v} = -\frac{u^2}{v^2} \] So the Jacobian is: \[ J = \begin{vmatrix} -\frac{v}{u^2} & \frac{1}{u} \\ \frac{2u}{v} & -\frac{u^2}{v^2} \\ \end{vmatrix} = \left(-\frac{v}{u^2}\right)\left(-\frac{u^2}{v^2}\right) - \left(\frac{1}{u}\right)\left(\frac{2u}{v}\right) \] \[ = \frac{v}{u^2} \cdot \frac{u^2}{v^2} - \frac{2u}{uv} = \frac{1}{v} - \frac{2}{v} = -\frac{1}{v} \] Take the absolute value (since areas are positive): \[ |J| = \frac{1}{v} \] ### **Step 4: Change the Integral** Original integral: \[ \iint_D x^2 y^2 \, dA \] Express \( x^2 y^2 \) in terms of \( u \) and \( v \): - \( x^2 y^2 = (xy)^2 = u^2 \) So the integral becomes: \[ \int_{u=1}^{2} \int_{v=1}^{4} u^2 \cdot \frac{1}{v} \, dv \, du \] ### **Step 5: Evaluate the Integral** First, integrate with respect to \( v \): \[ \int_{v=1}^{4} \frac{1}{v} dv = \ln v \Big|_{1}^{4} = \ln 4 - \ln 1 = \ln 4 \] Now, integrate with respect to \( u \): \[ \int_{u=1}^{2} u^2 \, du = \left[ \frac{u^3}{3} \right]_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \] So, the total integral is: \[ \left(\frac{7}{3}\right) \ln 4 \] ### **Step 6: Final Answer** \[ \boxed{ \iint_D x^2 y^2 \, dA = \frac{7}{3} \ln 4 } \] --- #### **Region in the \( u, v \)-plane:** - \( 1 \leq u \leq 2 \) - \( 1 \leq v \leq 4 \) --- If you need the region sketch or help with Problem 2, let me know!