Question

Evaluate the double integrals in Exercises 39–48. Use suitable transformations as necessary.

$\int {\int }_R\frac{2y-x}{3x+y+1}dA$, where R is the parallelogram with vertices (0, 0), (2, 1), (1, 4), and (−1, 3) .

Short answer

$\int {\int }_R\frac{2y-x}{3x+y+1}dA=\frac{21}{2}\ln \left(2\right)$

Step-by-step solution

Draw the region

Plot the given points to form the region and name the vertices.

Consider the new set of variables defined as,

$$\begin{gathered} u=x-2y \\ v=3x+y \end{gathered}$$

From the above equations,

$$\begin{gathered} \frac{u+2v}{7}=x \\ \frac{v-3u}{7}=y \end{gathered}$$

Plot these limits on u v plane,

Find the integration

Use the antiderivative to evaluate the integrals,

$$\begin{gathered} \int {\int }_R\frac{2y-x}{3x+y+1}dA=-\frac{3}{7}\ln \left(2\right){\int }_{-7}^{0}udu \\ =-\frac{3}{7}\ln \left(2\right){\left[\frac{u^2}{2}\right]}_{-7}^0 \\ =-\frac{3}{7}\ln \left(2\right)\left[-\frac{49}{2}\right] \\ =\frac{21}{2}\ln \left(2\right) \end{gathered}$$