Question

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

$\int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA$, where $\Omega$ is the region from Exercise 45.

Short answer

$\int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA=\frac{247}{192}$

Step-by-step solution

Draw the region and name the vertices  

The region $\Omega$is bounded by,

$y=2x,y=2x^2,y=x,y=x^2$

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

Consider the new set of variables defined as

$$\begin{gathered} u=\frac{y}{x} \\ v=\frac{y}{x^2} \end{gathered}$$

After solving ee get that,

$$\begin{gathered} \frac{u}{v}=x \\ \frac{u^2}{v}=y \end{gathered}$$

Determine the equation of each boundary in terms of u and v.  

We have,

$$\begin{gathered} \frac{u}{v}=x \\ \frac{u^2}{v}=y \end{gathered}$$

Use these equations to determine the equation of each boundary of the region.

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

Plot these limits on u v plane.

Evaluate the double integral. 

Set up the double integral,

$$\begin{gathered} \int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA={\int }_{u=1}^{u=2}{\int }_{v=1}^{v=2}\frac{u^2(u+v)}{v^3}dvdu \\ \int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA={\int }_{u=1}^{u=2}{u}^2\left({\int }_{v=1}^{v=2}\left(\frac{v}{v^3}+\frac{u}{v^3}\right)dv\right)du \\ \int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA={\int }_{u=1}^{u=2}{u}^2\left({\left[-\frac{1}{v}-\frac{u}{2v^2}\right]}_1^2\right)du \\ \int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA=\frac{1}{2}{\int }_{u=1}^{u=2}\left(\frac{u^2}{2}+\frac{3u^3}{8}\right)du \\ \int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA=\frac{1}{2}{\left[\frac{u^3}{6}+\frac{3u^4}{32}\right]}_1^2 \\ \int {\int }_{\Omega }\left(\frac{y+xy}{x^2}\right)dA=\frac{247}{192} \end{gathered}$$