# Rudin Principles of Mathematical Analysis Chapter 10, Exercise 8

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?

## 2 Answers

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

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

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