Questions: This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Evaluate the given integral by making an appropriate change of variables. ∬ over R (x-2y)/(3x-y) dA, where R is the parallelogram enclosed by the lines x-2y=0, x-2y=2, 3x-y=1, and 3x-y=2 Step 1 Let u=x-2y and v=3x-y. Then in terms of u and v, x=(2/5)v- square and y=(1/5)v- square . Therefore, ∂(x, y)/∂(u, v) = (∂x/∂u) (∂x/∂v) (∂y/∂u) (∂y/∂v) = (-1/5) (2/5) (-3/5) (1/5) = square

Solution Steps

To solve the given integral, we need to perform a change of variables using the provided transformations $u = x - 2y$ and $v = 3x - y$. First, express $x$ and $y$ in terms of $u$ and $v$. Then, compute the Jacobian determinant of the transformation to adjust the differential area element $dA$. Finally, substitute these into the integral and evaluate it over the new region in the $uv$-plane.

We start with the transformations defined by: $$u = x - 2y$$ $$v = 3x - y$$ Solving these equations for $x$ and $y$ gives: $$x = -\frac{u}{5} + \frac{2v}{5}$$ $$y = -\frac{3u}{5} + \frac{v}{5}$$

Next, we compute the Jacobian matrix of the transformation: $$\frac{\partial(x, y)}{\partial(u, v)} = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ -\frac{3}{5} & \frac{1}{5} \end{bmatrix}$$ Calculating the determinant of this matrix yields: $$\text{det} = \left(-\frac{1}{5}\right) \left(\frac{1}{5}\right) - \left(-\frac{3}{5}\right) \left(\frac{2}{5}\right) = \frac{1}{5}$$

Final Answer