- #1

JC2000

- 186

- 16

- 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?

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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