Q. 43 - Calculus

Question

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

$\int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA$ where $\Omega$ is the following region:

Step-by-Step Solution

Answer

$\int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=\frac{160}{3}+6552 \ln \left(3\right)$

Step 1: Draw the region and name the vertices

The region $\Omega$ is bounded by,

$y=3x, xy=27, y=\frac{1}{3}x, xy=3$

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

Consider the new set of variables defined as,

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

After solving, We get that

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

Step 2: Determine the equation of each boundary in terms of u and v.

We have,

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

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

$$\begin{gathered} AB: y=3x\Rightarrow u=\frac{1}{3} \\ BC: xy=27\Rightarrow v=27 \\ CD: y=\frac{1}{3}x\Rightarrow u=3 \\ DA: xy=3\Rightarrow v=3 \end{gathered}$$

Plot these limits on a u v- plane.

Step 3: Evaluate the double integral.

$$\begin{gathered} \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=\frac{1}{2}{\int }_{u=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{u=3}{\int }_{v=3}^{v=27}\left(\frac{u^2+v^2}{u}\right)dvdu \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=\frac{1}{2}{\int }_{u=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{u=3}\frac{1}{u}\left({\int }_{v=3}^{v=27}\left(u^2+v^2\right)dv\right)du \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=\frac{1}{2}{\int }_{u=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{u=3}\frac{1}{u}\left({\left[u^{2}v+\frac{v^3}{3}\right]}_3^{27}\right)du \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=12{\int }_{u=1/3}^{u=3}\left(u+\frac{273}{u}\right)du \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=12{\left[\frac{u^2}{2}+273 \ln \left(u\right)\right]}_{1/3}^3 \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right) dA=\frac{160}{3}+6552 \ln \left(3\right) \end{gathered}$$