Question

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

$\int {\int }_R(3x^2-5xy-2y^2)dA$, where R is the parallelogram from Exercise 41

Short answer

$\int {\int }_R(3x^2-5xy-2y^2)dA=-\frac{343}{4}$

Step-by-step solution

Draw the region

Step 1: 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,

style="max-width: none;">

Plot these limits on u v plane,

Find the integral

Use the antiderivative to evaluate all integrals,

$$\begin{gathered} \int {\int }_R(3x^2-5xy-2y^2)dA=\frac{7}{2}{\int }_{-7}^{0}udu \\ =\frac{7}{2}{\left[\frac{u^2}{2}\right]}_{-7}^0 \\ =\frac{7}{2}\left(-\frac{7^2}{2}\right) \\ =-\frac{343}{4} \end{gathered}$$