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I want to convert some integrals to use polar coordinates as my differentials, my problem is getting the limits.

So this is the first concept I am not understanding: If I have a circle in the xy-plane and want to represent it in polar coordinates with the restriction that the region is strictly everything on the right hand side of the line $x=1$ why would the lower limit be $1/\cos(\theta)$ if the circle I have is $ (x-1)^2 +y^2 = 1 $. I sort of understand how to get the upper limit but not the theory behind the lower. For the upper I simply expanded the $x$ terms to get $ r^2 = 2r*cos(\theta)$ and solved to get an upper limit of $2\cos(\theta)$

Please could someone explain the theory about the lower limit.

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the lower limit.... $x = r\cos\theta\\ y=r\sin\theta\\ x = 1\\ r\cos\theta=1\\ r = 1/\cos\theta$

I suggest until you have a solid intuition about these things, always sketch the curve. Draw in some line $r = \theta$ and think about the value of $r$ at the endpoints of that line.

One more step, what are your limits for $\theta$?

$1/\cos\theta = 2\cos\theta\\ 1/2 = \cos^2 \theta\\ \cos\theta = \sqrt2/2\\ \theta = -\pi/4,\pi/4$

How is it that we were able to discard $\cos\theta = -\sqrt2/2$? I will leave that to you to think about.

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It seems you want to describe the red semi circle:

The situation (Large Version)

This would mean $\theta \in [-\pi/4, \pi/4]$ and $r \in [\sqrt{2}, 2]$.

For the minimum $\theta=\pm \pi/4$ we have $\cos \theta = 1 / \sqrt{2} \iff \sqrt{2} = 1 / \cos \theta$.

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  • $\begingroup$ yes thats correct , I understand the theta restrictions, not the distance (r) restrictions $\endgroup$ – Teyash Arjun May 4 '16 at 19:49

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