Question

Evaluate the integral by making an appropriate change of variables.

\(\iint_{R}{{}}\frac{x-2y}{3x-y}dA\), where R is the parallelogram enclosed by the lines \(x-2y=0,x-2y=4,3x-y=1\) and \(3x-y=8\) .

Short answer

The value of the integral \(\iint_{R}{{}}\frac{x-2y}{3x-y}dA\) is \(\frac{8}{5}\ ln (8)\)

Step-by-step solution

Concept

Jacobian transformation,

\(\begin{aligned}\frac{\partial (x,y)}{\partial (u,v)}&=\left| \begin{aligned}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\\end{aligned} \right| \\&=\frac{\partial x}{\partial u}\cdot \frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\cdot \frac{\partial y}{\partial u} \\\end{aligned}\)

Find the partial derivatives

Firstly, we need partial derivatives:

\(\begin{aligned}{} & \begin{aligned}& \frac{\partial x}{\partial u}=\frac{-1}{5} \\& \frac{\partial x}{\partial v}=\frac{2}{5} \\ \end{aligned} \\{} & \begin{aligned}& \frac{\partial y}{\partial u}=\frac{-3}{5} \\ & \frac{\partial y}{\partial v}=\frac{1}{5} \\ \end{aligned} \\\end{aligned}\)

And by plugging them in the formula for the Jacobian we get:

\(\begin{aligned}{{}{}}{\frac{{\partial (x,y)}}{{\partial (u,v)}}}&{\: = \left| {\begin{aligned}{{}{}}{\frac{{ - 1}}{5}}&{\frac{2}{5}}\\{\frac{{ - 3}}{5}}&{\frac{1}{5}}\end{aligned}} \right|}\\{}&{\: = \frac{1}{5}}\end{aligned}\)

Substituting the values and simplifying the final calculation

Now, we know the substitution we are using is:

\(\begin{aligned}{{}{}}{}&{x = \frac{{2v - u}}{5}}\\{}&{y = \frac{{v - 3u}}{5}}\end{aligned}\)

Therefore:

\(\begin{aligned}\iint_{R}{{}}\frac{x-2y}{3x-y}dA & \;=\int_{0}^{4}{{}}\int_{1}^{8}{{}}\frac{u}{v}\frac{1}{5}dvdu \\{} & \;=\int_{0}^{4}{{}}\left( \frac{1}{5}u\ln v \right)_{1}^{8}du \\{} & \;=\int_{0}^{4}{{}}\frac{1}{5}u\ln 8du \\{} & \;=\left( \frac{1}{5}\frac{{{u}^{2}}}{2}\ln 8 \right)_{0}^{4} \\{} & \;=\frac{8}{5}\ln 8 \\\end{aligned}\)

Thus, the value of the integral \(\iint_{R}{{}}\frac{x-2y}{3x-y}dA\) is \(\frac{8}{5}\ ln (8)\)