Question

In Exercises 11-16, use the indicated change of variables to evaluate the double integral. \begin{array}{l} \int_{R} \int 4(x+y) e^{x-y} d A \\ x=\frac{1}{2}(u+v) \\ y=\frac{1}{2}(u-v) \end{array}

Short answer

The transformed integral is \(\int \int u e^u du dv\), to be evaluated over the region \(R\) in \(uv\)-plane.

Step-by-step solution

Change of Variables

Substitute the given change of variables into the equation. This results in: \[ \int \int 4\left(\frac{1}{2}(u+v)+\frac{1}{2}(u-v)\right) e^{\frac{1}{2}(u+v)-\frac{1}{2}(u-v)} du dv \].

Simplify the Integral

Simplify the integral. This results in: \[ \int \int 2u e^u du dv \].

Compute the Jacobian

Differentiate x and y with respect to u and v to get Jacobian determinant. This results in: \[ J=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \]. Calculate the determinant of J: |J| = -(\frac{1}{2}*\frac{1}{2}) - (\frac{1}{2}*\frac{1}{2}) = -\frac{1}{2}. Since we only need the absolute value for our new integral, we will use \(\frac{1}{2}\).

Substitute the Jacobian into the Integral

Substitute |J| into the double integral. This results in: \[ \int \int \frac{1}{2} * 2u e^u du dv = \int \int u e^u du dv \].

Evaluate the Integral

Evaluate the integral by integrating \(u e^u\) with respect to \(u\), and then with respect to \(v\). The double integral becomes equivalent to evaluating the single integral \(e^u (u-1)\), and the area of integration for \(u\) and \(v\) would be determined by the region \(R\) in \(uv\)-plane, which is not specified in the problem.