Integrating a circle in polar form

Find the area enclosed by the curve $r = acos \theta$.

$A = \frac{1}{2} \int_0^{2\pi} a^2cos^2\theta d\theta$

$A = \frac{a^2}{4} \int_0^{2\pi} cos2\theta + 1 d\theta = \frac{a^2}{4}[\frac{1}{2}sin2\theta + \theta]_0^{2\pi} = \frac{a^2 \pi}{2}$

The answer is half of mine which can be obtained easily using geometry, my question is; where did I miss a half in my integral?

I'd be inclined to go from -pi/2 to pi/2, and avoid the negative r's.

But as davros said. Going 0 to 2pi is covering the curve twice.

All hail the Daleks!

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One more little thing, how do I find the limits which will provide a single loop?

Clearly for theta=0, r > 0, then just work your way round positive/negative theta until you have r=0. Then you have your limits.

In general, when I have some more complex cases, is it safe to say that I should determine any angles at which r=0 in the original function and also along with these would be the points from this:

$y = 0 = rsin\theta \Rightarrow \theta = 0 , \pi, 2\pi, ... ,n\pi$

...and from here I could decide on the limits?

Can't see any errors in your working for this answer.

And checked it on my graph package - your answer is correct.

Mind me asking what software you use?

http://www.padowan.dk

It's free, and you can make a donation.

Handles standard y=f(x), parametric, and polar, as well as inequalities and relations.

Can supposedly handle complex numbers as well, but I've never taken the time to suss it out on that score.