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Evaluate the double integral by re-writing them in polar coordinates:

$\displaystyle\iint\limits_{R}\frac{y^2}{x^2}\ dA$, where $R$ is part of the annulus (ring) $9\leq x^2+y^2\leq 25$ lying in the first quadrant and below the line $y=x$.

So from this, I gather (assuming I understood correctly) that $R=\{(x,y)\mid9\leq x^2+y^2\leq 25,\ 0\leq y\leq x\}$.

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There are two circles, one where $r=5$ and one where $r=3$. I would guess that what I'm looking for is the larger circle minus the smaller circle, but doing that the only way I can think of yields the following:

\begin{gather} (x^2+y^2)-(x^2+y^2)=25-9\\ 0=16 \end{gather}

I'm clearly way off track here; how do I look at this so I can define the bounds of the double integral in polar coordinates?

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1 Answer 1

up vote 3 down vote accepted

You can use polar coordinates, then you have

$$ 9\leq x^2+y^2\leq 25 \implies 3\leq r\leq 5, $$

and

$$ 0\leq \theta \leq \frac{\pi}{4},\quad \rm{since}\quad y=x. $$

$$ \displaystyle\iint\limits_{R}\frac{y^2}{x^2}\ dA = \int_{0}^{\pi/4}\int_{3}^{5}\frac{\sin^{2}(\theta)}{\cos^2(\theta)}rdrd\theta $$

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Ah; it had not crossed my mind to specify $3\leq r\leq 5$. That makes perfect sense. All of my questions to date were about solid circles, not rings. –  agent154 Aug 2 '13 at 4:09

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