Question

Evaluate the integral by making an appropriate change of variables.

\(\iint_{R}{{}}\cos \left( \frac{y-x}{y+x} \right)dA\)where R is the trapezoidal region with vertices\((1,0),(2,0),(0,2)\) and \((0,1)\)

Short answer

The value of the integral \({\iint }_{R}\cos \left( \frac{y-x}{y+x} \right)dA\) is\(\frac{3}{2}\sin (1)\)

Step-by-step solution

Concept

Jacobian transformation,

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

Find the partial derivatives

Firstly, we need partial derivatives:

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

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

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

Substituting the values and simplify the final calculation

Now, we know the substitution we are using is:

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

With

\((x,y) \in R\:{\rm{and}}\:;(u,v) \in D, - v \le u \le v\:{\rm{and}}\:1 \le v \le 2\)

Therefore:

\(\begin{aligned}\iint_{R}{{}}\cos \left( \frac{y-x}{y+x} \right) & \;=\iint_{D}{{}}\cos \left( \frac{u}{v} \right)\cdot \frac{1}{2}dA \\{} & \;=\int_{1}^{2}{{}}\int_{-v}^{v}{{}}\cos \left( \frac{u}{v} \right)\cdot \frac{1}{2}dudv \\{} & \;=\frac{1}{2}\int_{1}^{2}{{}}\left( v\sin \frac{u}{v} \right)_{-v}^{v}dv \\{} & \;=\frac{1}{2}\int_{1}^{2}{{}}2\sin (1)vdv \\{} & \;=\sin (1)\int_{1}^{2}{{}}vdv \\{} & \;=\sin (1)\left( \frac{{{v}^{2}}}{2} \right)_{1}^{2} \\{} & \;=\frac{3}{2}\sin (1) \\\end{aligned}\)

Thus, the value of the integral \({\iint }_{R}\cos \left( \frac{y-x}{y+x} \right)dA\) is\(\frac{3}{2}\sin (1)\)