Evaluate $\iint_{R} x y \,d x \,d y$

Question

Using the change of variable evaluate $\iint_{R} x y\ dx\ dy,$ when the region $R$ is bounded by the curves $x y=1, x y=3, y=3 x, y=5 x$ in the $1^{\text {st }}$ quadrant.

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Then transformation domain is $\mathrm{D}=\{(\mathrm{u}, \mathrm{v}): 1 \leq u \leq 3,3 \leq v \leq 5\}$ $\therefore \mathrm{y}=\sqrt{u v}, x=\sqrt{\frac{u}{v}}$ Jacobian of the transformation is

$$ \begin{split} \mathrm{J} &= \left|\begin{array}{ll}\frac{1}{\sqrt{v}} \cdot \frac{1}{2 \sqrt{u}} & \sqrt{u}\left(-\frac{1}{2} \cdot \frac{1}{v \sqrt{v}}\right) \\ \sqrt{v} \cdot \frac{1}{2 \sqrt{u}} & \sqrt{u} \cdot \frac{1}{2 \sqrt{v}}\end{array}\right| =\frac{1}{4 v}+\frac{1}{4} \cdot \frac{1}{v}=\frac{1}{2 v}\\ \mathrm{I} &= \int_3^5 \int_1^3 u \cdot \frac{1}{2 v}\ du\ dv = \left[\left[\frac{u^2}{4}\right]_1^3[\ln v]_3^5 = \log \left(\frac{25}{9}\right)\right. \end{split} $$

It's the answer I got from a solution book. But I am unable to understand that. Can anybody help me out.TIA.

Answer 1

Since $u=xy$ while $v=y/x$, the first two curves are the bound $1\le u\le 3$, while the last two are $3\le v\le5$. Now we just need to rewrite $\underbrace{xy}_u\underbrace{dxdy}_{Jdudv}$, with $J=\frac{1}{2v}$ the Jacobian determinant.

Answer 2

You are told to integrate over a certain region $R$ in the $(x,y)$-plane. This region $R$ is bounded by the hyperbolic arcs $xy=1$ and $xy=3$, and by segments on the lines $ y=3x$ and $y=5x$.

Draw a figure!

These data suggest to consider the new variables $$u:= xy,\qquad v:={y\over x}\ .\tag{1}$$ These variables are bound by $$1\leq u\leq3,\qquad 3\leq v\leq5\ ,$$ hence have a rectangular domain $R'=[1,3]\times[3,5]$. Furthermore we get from $(1)$ that $$x=\sqrt{u\over v},\qquad y=\sqrt{uv}\qquad(1\leq u\leq3, \ 3\leq v\leq5)\ .\tag{2}$$ The equations $(2)$ define a function $f:\>(u,v)\mapsto(x,y)$ that maps the domain $R'$ bijectively onto $R$. The latter fact is "obvious" from the figure. It implies that $$Q:=\int_R x y\>{\rm d}(x,y)=\int_{R'}\sqrt{u\over v}\>\sqrt{uv}\>J_f(u,v)\>{\rm d}(u,v)=\ldots$$ You have computed $J_f(u,v)={1\over2v}$. When this is correct we therefore obtain $$Q=\int_1^3\int_3^5{u\over 2v}\>dv\>du=\ldots=\log{25\over9}\ .$$