Need help deducing the region for this double integration problem

[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?

 

[etotheipi]

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...

 

[PeroK]

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

 

[JC2000]

... 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 !

 

[etotheipi]

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 $r \in [1, \sqrt{e}]$.

 

[benorin]

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

 

[LuccaP4]

You can use inequalities in geogebra as well.

 

[haruspex]

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