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?


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.