Question

\(\iint_{R}{{{e}^{x+y}}}dA,\) where R is given by the inequality\(|x|+|y|\le 1\)

Short answer

The transformation is \(\int_{ - 1}^1 {\int_{ - 1}^1 {{e^u}} } \left| {\frac{1}{2}} \right|{\rm{d}}u\;{\rm{d}}v = e - \frac{1}{e}\)

Step-by-step solution

Define Jacobian matrix

The name "Jacobian" is frequently used to refer to both the Jacobian matrix and determinants, which are both defined for finite numbers of functions with the same number of variables. With respect to the variables, each row contains the first partial derivative of the same function.

Any type of Jacobian matrix can be used. It could be a square matrix (with the same number of rows and columns) or a rectangle matrix (the number of rows and columns are not equal).

Step 2: Find the transformation of Jacobian.

Let's start by substituting \(u = x + y\)and \(v = y - x.\)

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

The sketch of the R will be

Step 3: Now evaluate the integral.

Now we are ready to find the Jacobian.

\(\begin{aligned}\frac{{\partial (x,y)}}{{\partial (u,v)}} &= \left| {\begin{array}{*{20}{c}}{1/2}&{ - 1/2}\\{1/2}&{1/2}\end{array}} \right|\\ &= \frac{1}{2}\end{aligned}\)

Now based on the sketch we can easily evaluate the integral.

\(\begin{align}\iint_{R}{{{e}^{x+y}}}~dA&=\int_{1}^{1}{\int_{1}^{1}{{{e}^{u}}}}\left| \frac{1}{2} \right|du~dv \\ & =\left. \frac{1}{2}\int_{1}^{1}{{{e}^{u}}} \right|_{-1}^{1}dv \\ & =\frac{1}{2}\cdot \left( ev-\frac{v}{e} \right)_{-1}^{1} \\ & =\frac{1}{2}\cdot \left( e-\frac{1}{e}+e-\frac{1}{e} \right) \\ & =e-\frac{1}{e} \end{align}\)

Thus,the transformation is \(\int_{ - 1}^1 {\int_{ - 1}^1 {{e^u}} } \left| {\frac{1}{2}} \right|{\rm{d}}u\;{\rm{d}}v = e - \frac{1}{e}\)