I'm working on exercises of chapter 10 in Baby Rudin.
I refer to R. Cooke's solutions manual to Baby Rudin while I'm solving those exercises.(https://minds.wisconsin.edu/handle/1793/67009)
But I think there is a wrong solution for Chap 10, exercise 8.
the wrong part of a solution for chap 10, ex 8
Using Theorem 10.9 in Baby Rudin, which is about change of variables on a multiple integral, I think we should represent a integrand on the right side with a mapping T, not an inverse of T.
Could you guys check if I'm right or that solution is right?
Best Answer
If $S$ is the square with vertices $(0,0),\, (1,0),\, (0,1)$ and $(1,1)$ then $TS=H$, thus
$$\int_{TS} f(x,y) d(x,y)=\int_S(f\circ T)(u,v)|\det\partial T(u,v)| d(u,v)$$
where $\partial T(u,v)$ is the derivative of $T$ at the point $(u,v)\in S$. So yes, you are right, the stated solution is wrong.
Now using Fubini's theorem you can transform this integral over $S$ to a double integral on $[0,1]$ and $[0,1]$.
P.S.: in this case $T$ is an affine map, so it have the form $T=r+A$ for some $2\times 2$ matrix $A$ and some constant $r\in\Bbb R^2$. Hence $\partial T(u,v)=A$ for any chosen $(u,v)\in \Bbb R^2$.