We will address this problem in steps.

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Step 1: Show that the mapping $(u, v)$ maps $\mathcal{D}$ to $\mathcal{R} = [2, 3] \times [0, 4]$

The mapping is defined as: $u = xy, \quad v = x - y.$

Constraints:

1. $2 \leq xy \leq 3$: This directly maps to $2 \leq u \leq 3$.
2. $0 \leq x - y \leq 4$: This directly maps to $0 \leq v \leq 4$.
3. $x \geq 0, \, y \geq 0$: These ensure $x$ and $y$ remain in the first quadrant.

Thus, the region $\mathcal{D}$ defined by these constraints maps to the rectangle $[2, 3] \times [0, 4]$ in the $(u, v)$-plane.

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Step 2: Compute $\frac{\partial(x, y)}{\partial(u, v)}$ by first finding $\frac{\partial(u, v)}{\partial(x, y)}$

Compute $\frac{\partial(u, v)}{\partial(x, y)}$

We have: $u = xy, \quad v = x - y.$

Take partial derivatives: $\frac{\partial u}{\partial x} = y, \quad \frac{\partial u}{\partial y} = x,$ $\frac{\partial v}{\partial x} = 1, \quad \frac{\partial v}{\partial y} = -1.$

Thus, the Jacobian determinant is:

\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} y & x \\ 1 & -1 \end{vmatrix}.$$ Compute the determinant: $$\frac{\partial(u, v)}{\partial(x, y)} = y(-1) - x(1) = -y - x = -(x + y).$$ #### Find $$\frac{\partial(x, y)}{\partial(u, v)}$$ The Jacobian $$\frac{\partial(x, y)}{\partial(u, v)}$$ is the reciprocal of $$\frac{\partial(u, v)}{\partial(x, y)}$$: $$\frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{-(x + y)}.$$ --- ### Step 3: Express $$(x, y)$$ in terms of $$(u, v)$$ From the definitions: $$x = \frac{u}{v + 1}, \quad y = \frac{uv}{v + 1}.$$ Substitute these into $$- (x + y)$$: $$x + y = \frac{u}{v + 1} + \frac{uv}{v + 1} = \frac{u(1 + v)}{v + 1} = u.$$ Thus: $$\frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{-u}.$$ --- ### Step 4: Use the Change of Variables Formula The integral becomes: $$I = \int \int_{\mathcal{D}} (x^2 - y^2) \, dx \, dy.$$ Under the change of variables $$(u, v)$$, the integral transforms as: $$I = \int \int_{\mathcal{R}} f(u, v) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| \, du \, dv,$$ where $$f(u, v) = v$$. Substitute $$\left| \frac{\partial(x, y)}{\partial(u, v)} \right| = \frac{1}{u}$$: $$I = \int_2^3 \int_0^4 v \cdot \frac{1}{u} \, dv \, du.$$ --- ### Step 5: Evaluate the Integral Split the integral: $$I = \int_2^3 \frac{1}{u} \left( \int_0^4 v \, dv \right) du.$$ Evaluate the inner integral: $$\int_0^4 v \, dv = \left[ \frac{v^2}{2} \right]_0^4 = \frac{4^2}{2} - \frac{0^2}{2} = 8.$$ Thus: $$I = \int_2^3 \frac{8}{u} \, du.$$ Evaluate the remaining integral: $$\int_2^3 \frac{8}{u} \, du = 8 \int_2^3 \frac{1}{u} \, du = 8 \ln u \Big|_2^3 = 8 (\ln 3 - \ln 2) = 8 \ln\left(\frac{3}{2}\right).$$ --- ### Final Answer: $$I = 8 \ln\left(\frac{3}{2}\right).$$ --- Would you like more details or assistance with related topics? --- ### Related Questions: 1. What happens if we modify the bounds of $$\mathcal{D}$$? 2. Can this problem be solved geometrically without a change of variables? 3. How do you determine when a change of variables is necessary? 4. What other transformations could map $$\mathcal{D}$$ to a rectangle? 5. How does the Jacobian determinant affect the evaluation of an integral? --- ### Tip: When performing a change of variables, always verify that the transformation is invertible and maps the original region correctly.