# Need help deducing the region for this double integration problem

• JC2000
In summary, we are converting to a polar integral and integrating the function ##f(x, y)=\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}## over the region ##1 \leq x^{2}+y^{2} \leq e##, which can be simplified to a rectangular region in the first quadrant with boundaries ##x=1## and ##x=\sqrt{e}## for ##x## and ##y=1## and ##y=\sqrt{e}## for ##y##. This can also be interpreted as the radial coordinate ##r## ranging from ##1##f

#### JC2000

Homework Statement
Not sure how to use latex formatting for the question so I have moved it to the main body ....
Relevant Equations
\begin{array}{c}
1 \leq x^{2}+y^{2} \leq e \\
1 \leq x^{2} \leq e \quad 1 \leq y^{2} \leq e \\
1 \leq x \leq \sqrt{e} \quad 1 \leq y \leq \sqrt{e} \\
x=1 \\
x=\sqrt{e}
\end{array}

*Similarly for y which gives a rectangular region in the first quadrant.
Converting to a polar integral : Integrate ##$$f(x, y)=$$ $$\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}$$## over the region ##$$1 \leq x^{2}+y^{2} \leq e$$##

So,

\begin{array}{c}
1 \leq x^{2}+y^{2} \leq e \\
1 \leq x^{2} \leq e \quad 1 \leq y^{2} \leq e \\
1 \leq x \leq \sqrt{e} \quad 1 \leq y \leq \sqrt{e} \\
x=1 \\
x=\sqrt{e}
\end{array}

*Similarly for y which gives a rectangular region in the first quadrant.

Is the above simplification the correct way to proceed?

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What is the shape described by ##x^2 + y^2 = 1##? And what about ##x^2 + y^2 =e##?

JC2000
What is the shape described by ##x^2 + y^2 = 1##? And what about ##x^2 + y^2 =e##?

Oof! Didn't think of it that way, not sure why I made the inference that I did...
Although this must somehow be self-evident, I would appreciate it if you could walk me through the reasoning for it...

... I'm not sure that the equation of a circle needs further explanation.

... I'm not sure that the equation of a circle needs further explanation.
I was referring to the interpretation of the inequality. I guess I simply misinterpreted the inequality and need to brush up on basic algebra !

I was referring to the interpretation of the inequality. I guess I simply misinterpreted the inequality and need to brush up on basic algebra !

Well the inequality ##1 \leq x^{2}+y^{2} \leq e## is really two inequalities: ##1 \leq x^2 + y^2## and ##x^2 + y^2 \leq e##. Perhaps more simply, you can say the radial coordinate ##r \in [1, \sqrt{e}]##.

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JC2000 and Delta2
Read this (assuming you'll have more questions to post on this site).

JC2000
You can use inequalities in geogebra as well.

JC2000
Converting to a polar integral : Integrate ##$$f(x, y)=$$ $$\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}$$##
Is that supposed to be ##f(x, y)=\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}## ?

JC2000
Is that supposed to be ##f(x, y)=\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}## ?
Yes