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?