Calc 3, multiple integral and variable change in $\iint y^{-1}\,dx\,dy$.

Question

Sketch the domain $D$ bounded by $y=x^2,\ y=(1/2)x^2,$ and $y=2x$. Use a change of variables with the map $x=uv,\ y=u^2$ to calculate: $$\iint_D y^{-1}\,dx\,dy.$$

I believe the Jacobian is $v-2(u^2)$ and the integrand is the Jacobian multiplied by $u^{-2}$. I don't know how to find the bounds. Some tries resulted in the $u$ bounds being $0$ to $2v$ and the $v$ bounds being $1$ to $\sqrt{2}$ but I could not solve the integral that way. Any help at all would be appreciated.

Answer 1

$$\begin{array}{rcl} x &=& uv \\ y &=& u^2 \\ \dfrac{\partial(x, y)}{\partial(u,v)} &=& \begin{pmatrix}v&u\\2u&0\end{pmatrix} \\ \det \dfrac{\partial(x, y)}{\partial(u,v)} &=& -2u^2 \\ \mathrm dx \mathrm dy &=& 2u^2 \ \mathrm du \mathrm dv \\ \displaystyle \int \int_D \dfrac1y \mathrm dx \mathrm dy &=& \displaystyle \int_{v=1}^{v=\sqrt2} \int_{u=0}^{u=2v} \dfrac1{u^2} (2u^2)\ \mathrm du \mathrm dv \\ &=& \displaystyle 2 \int_{v=1}^{v=\sqrt2} \int_{u=0}^{u=2v} \mathrm du \mathrm dv \\ &=& \displaystyle 2 \int_{v=1}^{v=\sqrt2} 2v \mathrm dv \\ &=& \displaystyle 2 \\ \end{array}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ We can perform an explicit evaluation without a change of variables !!!.

Hereafter, $\ds{\bracks{\cdots}}$ are Iverson Brackets. Namely, $\ds{\bracks{P} = 1}$ whenever $\ds{P}$ is true and $\ds{0}$ otherwise. They are very convenient to handle 'region definitions'.

\begin{align} \mc{A} & \equiv \iint_{\large\mathbb{R}^{2}} {\bracks{y < 2x}\bracks{y < x^{2}}\bracks{y > x^{2}/2} \over y}\,\dd x\,\dd y = \iint_{\large\mathbb{R}^{2}} {\bracks{x^{2}/2 < y < \min\braces{2x,x^{2}}} \over y}\,\dd x\,\dd y \\[5mm] & = \int_{-\infty}^{\infty}\braces{ \overbrace{\bracks{2x < x^{2}}\bracks{{x^{2} \over 2} < 2x}} ^{\ds{\bracks{2 < x < 4}}}\ \int_{x^{2}/2}^{2x}{\dd y \over y}\ +\ \overbrace{\bracks{x^{2} < 2x}\bracks{{x^{2} \over 2} < x^{2}}} ^{\ds{\bracks{0 < x < 2}}}\ \int_{x^{2}/2}^{x^{2}}{\dd y \over y}}\dd x \\[5mm] & = \int_{2}^{4}\ln\pars{4 \over x}\,\dd x + \int_{0}^{2}\ln\pars{2}\,\dd x = \bbx{\ds{2}} \end{align}