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I'm in doubt on how to approach a problem of double integrals over a specific region.

I have to calculate $\int\int\limits_R e^x dA$, R being the region between $y=\frac{x}{2}$, $y=x$, $y=\frac{1}{x}$ and $y=\frac{2}{x}$. I am only interested in the first quadrant. That being said, the region is as follows:

enter image description here

And the points:

enter image description here

Where 1.414 is $\sqrt{2}$ and 0.707 is $\frac{\sqrt{2}}{2}$.

My approach, which I'm in doubt if it's a valid one, was the following:

Divide the region into 2 regions and consider each new region a "case 2" region and sum the integrals over each region to obtain the integral over the original region:

enter image description here

The division is made in order to obtain well-defined functions in each region. Is that a valid approach? If not, how should I approach this problem?

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Looks good. Another way to approach this problem would be through a change of variables (tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx). –  Adriano Jul 22 '13 at 23:42
@Adriano I never used change of variables with double integrals, I'll give it a try. A problem with the approach I was taking is that it leads to $\int e^\frac{1}{y} dy$, which I don't think should happen - i.imgur.com/sGDWrUa.jpg. –  Alex Jul 23 '13 at 0:10
@Alex: I have been sorry not to make you an easier way to solve the finial integrals. +1 –  B. S. Jul 24 '13 at 10:14
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2 Answers

Your approach looks just fine, is certainly valid, and it's clear you thought through the problem carefully. In a situation like this, another possible approach to consider would be using a change of variables.

Using a change of variables is one method we can use to integrate over one region, instead of breaking it into the sum of integrals over sub-regions. A good example to look through is the trapezoidal region discussed at the linked website, e.g.

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A leading hint +1 –  B. S. Jul 23 '13 at 1:33
Trying to suggest a potentially fruitful approach. The OP's approach and problem set up is picture-perfect, but leads to an unpleasant integral! –  amWhy Jul 23 '13 at 1:35
The same story is coming up when you use mine as well. :( –  B. S. Jul 23 '13 at 1:56
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Setting $$u=x/y,~~v=xy$$ you get easily that $$1\le u,v\le 2$$ and the main integrals will be changed to the following ones:

$$\int_1^2\int_1^2\exp(\sqrt{uv})|J|dudv$$ wherein $J$ is the Jacobian determinant $\frac{\partial(u,v)}{\partial(x,y)}=1/2u$. It seems that the associated indefinite integral cannot be expressed by elementary functions.

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+1 I think you're right! –  amWhy Jul 23 '13 at 13:16
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