Question

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

$\int {\int }_{\Omega }x{y}^{3}dA$, where $\Omega$is the region from Exercise 43.

Short answer

$\int {\int }_{\Omega }x{y}^{3}dA=8736$

Step-by-step solution

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

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

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.

Evaluate the double integral.

$$\begin{gathered} \int {\int }_{\Omega }\left(x{y}^3\right)dA=\frac{1}{2}{\int }_{u=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{u=3}{\int }_{v=3}^{v=27}\left(\frac{v^2}{u^2}\right)dvdu \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right)dA=\frac{1}{2}{\int }_{u=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{u=3}\frac{1}{u}\left({\int }_{v=3}^{v=27}\left(v^2\right)dv\right)du \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right)dA=\frac{1}{2}{\int }_{u=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}^{u=3}\frac{1}{u}\left({\left[\frac{v^3}{3}\right]}_3^{27}\right)du \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right)dA=3276{\int }_{u=1/3}^{u=3}\left(\frac{1}{u^2}\right)du \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right)dA=3276{\left[-\frac{1}{u}\right]}_{1/3}^3 \\ \int {\int }_{\Omega }\left(\frac{x^2}{y^2}+x^2{y}^2\right)dA=8736 \end{gathered}$$