# Evaluate the integral \iint_{R} 7(x + y)e^{x^{2} - y^{2}} dA, where R is the rectangle enclosed...

## Question:

Evaluate the integral

{eq}\displaystyle\; \iint_{R} 7\left(x + y\right)e^{x^{2} - y^{2}} \,dA {/eq},

where {eq}R {/eq} is the rectangle enclosed by the lines {eq}\;x - y = 0 {/eq}, {eq}\;\;x - y = 7 {/eq}, {eq}\;\;x + y = 0 {/eq},{eq}\; {/eq} and {eq}\;x + y = 3 {/eq}.

## Double Integral Using Jacobian Transformation:

If we have any functions {eq}x=g(u,v) {/eq} and {eq}y=h(u,v) {/eq} then the Jacobian of {eq}x \ and \ y {/eq} with respect to {eq}u \ and \ v {/eq} is given by:

{eq}\left \| J \right \| =\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix}\\ {/eq}

We can use the Jacobian to change variables in a double integral by virtue of the formula

{eq}\iint_{R}f(x,y)dy dx=\iint_{D}f(x(u,v), y(u, v))\left \| J \right \|dudv {/eq}

We have to evaluate the integral{eq}\displaystyle\; \iint_{R} 7\left(x + y\right)e^{x^{2} - y^{2}} \,dA {/eq},

where {eq}R {/eq} is the rectangle enclosed by the lines {eq}\;x - y = 0 {/eq}, {eq}\;\;x - y = 7 {/eq}, {eq}\;\;x + y = 0 {/eq},{eq}\; {/eq} and {eq}\;x + y = 3 {/eq}.

A simple coordinate transformation can be used to simplify the region of integration:

{eq}x-y=u \ and \ x+y=v\\ \Rightarrow x=\dfrac{u+v}{2} , y=\dfrac{u-v}{2} {/eq}

With this transformation, the region of integration is transformed into a new region bounded by lines:

{eq}u=0 , u=7, v=0, v=3 {/eq}

The Jacobian of transformation is given by:

{eq}\begin{align} \left \| J \right \| &=\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix}\\ &=\begin{vmatrix} \dfrac{1}{2} & \dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{-1}{2} \end{vmatrix}\\ &=\left |\dfrac{-1}{4}-\dfrac{1}{4} \right |\\ &=\dfrac{1}{2}\\ \end{align} {/eq}

Therefore, the double integral in the transformed coordinates is evaluated as:

{eq}\begin{align} \iint_{R} 7\left(x + y\right)e^{x^{2} - y^{2}} \,dA &= \iint_{D} 7\left(x + y\right)e^{(x+y)(x-y)} \,dA\\ &=\int_{v=0}^{v=3}\int_{u=0}^{u=7}7ve^{uv}\begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} du dv\\ &=\int_{v=0}^{v=3}\int_{u=0}^{u=7}7ve^{uv}\left ( \frac{1}{2} \right )dudv\\ &=\dfrac{7}{2}\int_{v=0}^{v=3}v\left (\int_{u=0}^{u=7}e^{uv}du \right )dv\\ &=\dfrac{7}{2}\int_{v=0}^{v=3}v\left ( \dfrac{e^{uv}}{v} \right )_{u=0}^{u=7} dv\\ &=\dfrac{7}{2}\int_{v=0}^{v=3}(e^{uv})_{u=0}^{u=7} dv\\ &=\dfrac{7}{2}\int_{v=0}^{v=3}(e^{7v}-e^{(0)v})dv\\ &=\dfrac{7}{2}\int_{v=0}^{v=3}(e^{7v}-1)dv\\ &=\dfrac{7}{2}\left ( \dfrac{e^{7v}}{7}-v \right )_{v=0}^{v=3}\\ &=\dfrac{7}{2}\left ( \dfrac{e^{7(3)}}{7}-3-\left ( \dfrac{e^{7(0)}}{7}-0 \right ) \right )\\ &=\dfrac{7}{2}\left ( \dfrac{e^{21}}{7}-3-\dfrac{1}{7} \right )\\ &= \dfrac{e^{21}-22}{2} \\\ \end{align} {/eq}