# Use the change of variables to evaluate the integral R ( x + y ) sin ( x - y ) d A where R is...

## Question:

Use the change of variables to evaluate the integral {eq}\int \int_R (x+y)\sin(x-y)dA {/eq} where {eq}R {/eq} is the region enclosed by {eq}y=x,y=x-2,y=-x {/eq} and {eq}y=-x+1 {/eq} (Hint: Use {eq}u=x+y {/eq} and {eq}v=x-y {/eq})

## Double Integral:

We will solve the problem where we will find the points in the uv plane and the region will be the rectangle and then we will solve the integral with respect to u and v.

To solve the problem we will use the variable change form:

{eq}x+y=u\\ x-y=v\\ \int \int u\sin vdudv\\ x=\frac{u+v}{2}\\ y=\frac{u-v}{2} {/eq}

Now let us find the Jacobian:

{eq}\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}{2} &\frac{1}{2} \\ \frac{1}{2} & \frac{-1}{2} \end{bmatrix}\\ =\frac{-1}{2} {/eq}

The integral will be:

{eq}=\frac{-1}{2}\int_{0}^{2}\int_{0}^{1}u\sin vdudv\\ =\frac{1}{4}\left [ \cos v \right ]\\ =\frac{1}{4}(\cos 2-1) {/eq}