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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.

Baby Rudin, chap 10, ex 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?

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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 $(u,v)\in \Bbb R^2$.

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You are right.

The affine transformation is

  • $\binom{x}{y}=T\binom{u}{v} = \binom{1}{1} + A_T\binom{u}{v}$ with $A_T = \begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix}$

Now, you have

$$\int_{T(S)} e^{x-y} d(x,y) = \int_{S}e^{x(u,v)-y(u.v)}\left|\det A_T \right| d(u,v)$$

So, you get

$$\alpha = \int_S e^{1+2u+v - (1+u+3v)}\left|\det\begin{pmatrix}2 & 1 \\ 1 & 3\end{pmatrix} \right|d(u,v) = 5\int_S e^{u-2v}d(u,v)$$

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