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 integral from Exercise 33, but use the transformation given by $u=x+y$ and$v=\frac{x}{y}$

Short answer

(a). The region is shown below,

(b). The transformed region is shown below,

(c).$\int {\int }_{\Omega }{(x+y)}^{2}dA=2560$

Step-by-step solution

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,

role="math" $$\begin{gathered} AB:x+y=4 \\ AD:y=3x \\ CD:x+y=12 \\ BC:y=\frac{1}{3}x \end{gathered}$$

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

Consider the new set of variables defined as

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

After solving we get that,

$$\begin{gathered} \frac{u}{1+v}=y \\ \frac{uv}{1+v}=x \end{gathered}$$

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

$$\begin{gathered} AB:x+y=4\Rightarrow u=4 \\ AD:y=3x\Rightarrow v=\frac{1}{3} \\ CD:x+y=12\Rightarrow u=12 \\ BC:y=\frac{1}{3}x\Rightarrow v=3 \end{gathered}$$

Plot these limits on u v - plane.

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

Set up the double integral.

$$\begin{gathered} \int {\int }_{\Omega }{(x+y)}^{2}dA=2{\int }_{u=4}^{u=12}{\int }_{v=1/3}^{v=3}\frac{u^{3}v}{{(1+v)}^3}dvdu \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=2{\int }_{u=4}^{u=12}{u}^3\left({\int }_{v=1/3}^{v=3}\frac{v}{{(1+v)}^3}dv\right)du \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=2{\int }_{u=4}^{u=12}{u}^3\left({\int }_{v=1/3}^{v=3}\left({\left(1+v\right)}^{-2}-{(1+v)}^{-3}\right)dv\right)du \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=2{\int }_{u=4}^{u=12}{u}^3\left(\frac{1}{4}\right)du \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=\frac{1}{2}{\int }_{u=4}^{u=12}{u}^{3}du \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=\frac{1}{2}{\left[\frac{u^4}{4}\right]}_4^{12} \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=\frac{1}{2}\left(\frac{{12}^4}{4}-\frac{4^4}{4}\right) \\ \int {\int }_{\Omega }{(x+y)}^{2}dA=2560 \end{gathered}$$