Question

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

$\int {\int }_R{(x-y)}^{3}dA$, where R is the parallelogram with vertices (0, 0), (3, 0), (5, 2), and (2, 2).

Short answer

$\int {\int }_R{(x-y)}^{3}dA=\frac{81}{2}$

Step-by-step solution

Draw the region

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

In the region R, the equations of boundary curves are,

$$\begin{gathered} AB:y=0 \\ BC:x-y=3 \\ CD:y=2 \\ DA:x-y=0 \end{gathered}$$

Consider the new set of variables defined as,

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

Add the above equations,

$u+v=x$

Use these equations to determine the equation of each boundary of the region in terms of variables u and v.

$$\begin{gathered} AB:v=0 \\ BC:u=3 \\ CD:v=2 \\ DA:u=0 \end{gathered}$$

Plot these points on UV plane,

Find the integral

Set up the double integral for the region R' between u = 0 and u = 3.

$$\begin{gathered} \int {\int }_R{(x-y)}^{3}dA={\int }_0^3\left(u^3{\int }_0^{2}dv\right)du \\ ={\int }_0^3\left(u^3{\left[v\right]}_0^2\right)du \\ =2{\int }_0^3{u}^{3}du \\ =2{\left[\frac{u^4}{4}\right]}_0^3 \\ =2\times \frac{81}{4} \\ =\frac{81}{2} \end{gathered}$$