Q. 37 - Calculus

Question

For each double integral in Exercises 33–38, (a) sketch the region, (b) use the specified transformation to sketch the transformed region, and (c) use the transformation to evaluate the integral. 

Evaluate the double integral from Exercise 36, using the transformation given by u = x + y and v = 2y − x. 

Step-by-Step Solution

Answer

(a). The sketch of the region is shown below,

(b). The sketch of the transformed region is shown below,

(c). $\int {\int }_{\Omega }xy dA=\frac{399}{8}$

Part (a): Step 1: Draw the region

The given region $\Omega$is said to be trapezoid with vertices at $(1, 3), (3, 1), (9, 3), (3, 9)$ .

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

In the region $\Omega$, the equations of the boundary are,

$$\begin{gathered} AB: 2x-y=1 \\ BC: x+y=8 \\ CD: -x+2y=1 \\ DA: x+y=5 \end{gathered}$$

Part (b): Step 1: Draw the transformed region.

Consider the new set of variables defined as

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

After solving we get that,

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

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

$$\begin{gathered} AB: 2x-y=1\Rightarrow u-v=1 \\ BC: x+y=8\Rightarrow u=8 \\ CD: -x+2y=1\Rightarrow v=1 \\ DA: x+y=5\Rightarrow u=5 \end{gathered}$$

Plot these limits on u v- plane.

Part (c): Step 1: Evaluate the double integral.

Set up the double integral.

$$\begin{gathered} \int {\int }_{\Omega }xy dA=\frac{1}{27}{\int }_{u=5}^{u=8}{\int }_{v=1}^{v=u-1}\left(2u^2+uv-v^2\right)dvdu \\ \int {\int }_{\Omega }xy dA=\frac{1}{27}{\int }_5^8\left(2u^2{\int }_1^{u-1}dv+u{\int }_1^{u-1}vdv-{\int }_1^{u-1}{v}^{2}dv\right)du \\ \int {\int }_{\Omega }xy dA=\frac{1}{162}{\int }_5^8(13u^3-24u^2-6u+4)du \\ \int {\int }_{\Omega }xy dA=\frac{1}{162}\left[\frac{13u^4}{4}-8u^3-3u^2+4u\right] \\ \int {\int }_{\Omega }xy dA=\frac{399}{8} \end{gathered}$$