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If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined as $f(x,y) = e^{x+y}, e^{x-y}$, what is the area of the image of $\{0<x,y<1\}$?

My approach: I try to use the this composition, $(x,y) \mapsto (x+y,x-y) \mapsto (e^{x+y},e^{x-y})$

For $(x,y) \mapsto (x+y,x-y)$, I find the Jacobian (of the change of variable to be 2). So this means $\int \int dx dy = 2\int d(x+y) d(x-y)$

Now I use this composition $(s,t) \mapsto (e^s,e^t)$ where $s = x + y, t = x - y$.

Now I claim that the image is a square (as there is no mixing of variable). So the area is product of range of $(e^{s_2} - e^{s_1}).(e^{t_2} - e^{t_1})$

So the final answer is $(e^{2} - 1).(e^{1}- e^{-1})$ But my notes suggest the answer is $(e^{2} - 1)$. So can correct my mistake/argument.

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ManishKumar Singh
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  • As I understand it, you're saying that the image is $(1,e^2) \times (e^{-1},e),$ but that's not so. If you solve $f(x,y) = (e^2, e),$ you get $(x,y) = (3/2,1/2),$ which is not in the domain. You're on the right track with the change of variables theorem, but you're not using it right. Hint: The area is the integral of 1 over the image. – saulspatz Jan 27 '18 at 20:10

1 Answers1

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The Jacobian of the map $g(x,y)= (e^{x+y},e^{x-y})$ is $-2e^{2x}$, as can easily be verified. Therefore, if $S$ denotes your square, then according to the change-of-variable formula: $$\int\int_{gS}1\,dx\,dy=\int\int_{S}1(g(x,y))\,|-2e^{2x}|\,dx\,dy=\int_0^1\int_0^12e^{2x}\,dx\,dy=e^2-1,$$ as in your notes.

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