To solve the given integral using the specified transformation, we will follow these steps:

Problem Statement

Evaluate: $\iint_R 2 \frac{x - 4y}{5x - y} \, dA,$ where $R$ is the parallelogram enclosed by the lines $x - 4y = 0$, $x - 4y = 1$, $5x - y = 1$, and $5x - y = 2$. We use the transformation: $u = x - 4y, \quad v = 5x - y.$

Solution

1. Express $x$ and $y$ in Terms of $u$ and $v$: We have: $u = x - 4y \quad \text{and} \quad v = 5x - y.$ Solve this system for $x$ and $y$ in terms of $u$ and $v$: $5u = 5x - 20y \quad \Rightarrow \quad v - 5u = 19y \quad \Rightarrow \quad y = \frac{v - 5u}{19}.$ Substitute $y$ back into $u = x - 4y$ to solve for $x$: $x = u + 4\left(\frac{v - 5u}{19}\right) = \frac{9u + 4v}{19}.$

2. Compute the Jacobian Determinant $\frac{\partial (x, y)}{\partial (u, v)}$: We have: $x = \frac{9u + 4v}{19}, \quad y = \frac{v - 5u}{19}.$ Find the partial derivatives: $\frac{\partial x}{\partial u} = \frac{9}{19}, \quad \frac{\partial x}{\partial v} = \frac{4}{19},$ $\frac{\partial y}{\partial u} = -\frac{5}{19}, \quad \frac{\partial y}{\partial v} = \frac{1}{19}.$ Then the Jacobian determinant is: $J = \begin{vmatrix} \frac{9}{19} & \frac{4}{19} \\ -\frac{5}{19} & \frac{1}{19} \end{vmatrix} = \frac{9 \cdot 1 - (-5) \cdot 4}{19^2} = \frac{9 + 20}{361} = \frac{29}{361}.$

3. Transform the Integral: Using the transformation, the integral becomes: $\iint_{R'} 2 \frac{u}{v} \cdot \frac{29}{361} \, du \, dv.$ Simplifying the integrand: $\iint_{R'} \frac{58u}{361v} \, du \, dv.$

4. Set up the New Limits for $u$ and $v$: The region $R$ is bounded by $u = 0$, $u = 1$, $v = 1$, and $v = 2$. So the integral becomes: $\int_1^2 \int_0^1 \frac{58u}{361v} \, du \, dv.$

5. Evaluate the Integral:

   • Integrate with respect to $u$: $\int_1^2 \int_0^1 \frac{58u}{361v} \, du \, dv = \int_1^2 \frac{58}{361v} \left[ \frac{u^2}{2} \right]_0^1 \, dv = \int_1^2 \frac{58}{361v} \cdot \frac{1}{2} \, dv = \int_1^2 \frac{29}{361v} \, dv.$
   • Integrate with respect to $v$: $\int_1^2 \frac{29}{361v} \, dv = \frac{29}{361} \left[ \ln v \right]_1^2 = \frac{29}{361} (\ln 2 - \ln 1) = \frac{29}{361} \ln 2.$

Final Answer:

$\iint_R 2 \frac{x - 4y}{5x - y} \, dA = \frac{29 \ln 2}{361}.$

Would you like further details on any step, or have any questions?

Related Questions

1. How would the solution change if a different transformation was used?
2. What is the significance of the Jacobian in transformations?
3. How can we verify the limits for the new region $R'$?
4. What are the general steps to transform a double integral?
5. Why do we apply the logarithmic property in the final integration step?

Tip

Always verify the transformed region’s limits by plugging boundary conditions into the transformation equations.