Transformation to uv plane integral, Carnot engine

I'm curious how do you choose the u and v transformations here to be equal to the constants and why is the f(u,v)=1 for the area? I see that the area in xy is difficult to integrate because the sides are curved. is the transformation proven in a different book than Stewart? the problem: the work done by an ideal Carnot engine is equal to the area enclosed by two isotherms and adiabatic curves. xy is a hyperbola $\displaystyle xy=a, xy=b, xy^{1.4}=c, xy^{1.4}=d$$\displaystyle \begin{bmatrix}\frac{\partial x}{\partial u}\ & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} &\frac{\partial y}{\partial v}\\ \end{bmatrix}=\frac{5}{2v}, \int_{c}^{d}\int_{a}^{b}\frac{5}{2v}dudv=\frac{5}{ 2}(b-a)ln\frac{d}{c}$

by the way, how long does it take people to do these?

Re: Transformation to uv plane integral, Carnot engine

$\displaystyle x-y=0, 2, x+y=0,2, u=x-y, v=x+y, \int \int (x+y)e^{x^2+y^2}dA=\frac{1}{2}\int_{0}^{3}\int_{0} ^{2}ve^{uv}dudv=\frac{1}{4}(e^6-7)$

Re: Transformation to uv plane integral, Carnot engine

$\displaystyle trapezoidvertices (1,0)(2,0)(0,1)(0,2)\rightarrow (-2,2)(-1,1)(2,2)(1,1),I=\int \int cos(\frac{y-x}{y+x}), u=y+x, v=y-x, I=\frac{1}{2}\int_{1}^{2}\int_{-u}^{u}cos(\frac{v}{u})dvdu=\frac{3}{2}sin1$

it's interesting it doesn't matter if you reverse u=y-x and v=y+x when transforming to the uv plane

is it bad to just map the vertices instead of lines?