# Area in polar coordinates

• archaic
Yes, sloppy of me, sorry!In summary, the conversation discusses the concept of a negative differential, specifically with regards to the integral of a function with limits in an anticlockwise direction. The participants also consider different constructions and examples to better understand this concept.f

#### archaic

Homework Statement
Find the area of the region inside the circle ##r=-2\sin\theta## and outside the circle ##r=1##.
Relevant Equations
$$A=\frac 12\int_{\theta_1}^{\theta_2}(r^2-r^2_0)d\theta$$
-2\sin\theta=1\Leftrightarrow\theta=-\frac{\pi}{6},\,-\frac{5\pi}{6}\\ \begin{align*} \int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(4\sin^2\theta-1\right)d\theta &=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(1-2\cos2\theta\right)d\theta\\ &=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12d\theta-\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\cos2\theta d\theta\\ &=\frac 12\left[\theta-\sin2\theta\right]_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\\ &=-\frac{5\pi}{12}-\frac{\sqrt 3}{4}+\frac{\pi}{12}-\frac{\sqrt 3}{4}\\ &=-\frac{\pi}{3}-\frac{\sqrt 3}{2} \end{align*}\\ A=\frac{\pi}{3}+\frac{\sqrt 3}{2}

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That's what I got as well; is it supposedly wrong?

archaic
That's what I got as well; is it supposedly wrong?
Nope, was just checking.
Thank you!

etotheipi
Nope, was just checking.
Thank you!

Oh cool, that's a relief. Something sort of interesting also is that it seems the integral "sweeps out" areas in the anticlockwise direction, so it comes out positive if you put the limits in that order. I'm still not entirely sure why, but luckily I don't think it matters too much...

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Oh cool, that's a relief. Something sort of interesting also is that it seems the integral "sweeps out" areas in the clockwise direction, so it comes out positive if you put the limits in that order. I'm still not entirely sure why, but luckily I don't think it matters too much...

etotheipi

Yeah, it's fair to say it still confuses me!

Yeah, it's fair to say it still confuses me!
Well, if you think about it as a limit of a difference (as in ##dx=\lim_{n\to\infty}\Delta x(n)##), then if the LHS of the difference is less than the RHS, regardless of it being a limit, the result is surely negative!

Well, if you think about it as a limit of a difference, then if the LHS of the difference is less than the RHS, regardless of it being a limit, the result is surely negative!

Well sure, though what's to tell us which of the two options is being chosen? Like in this case, I think the resolution is that ##d\theta## is the limit of a positive difference and that would make the integral overall positive for anticlockwise ordered limits. But if we had ##d\theta## as a negative difference, we'd have a negated result - so there doesn't seem to be any way of telling?

Well sure, though what's to tell us which of the two options is being chosen? Like in this case, I think the resolution is that ##d\theta## is the limit of a positive difference and that would make the integral overall positive for anticlockwise ordered limits. But if we had ##d\theta## as a negative difference, we'd have a negated result - so there doesn't seem to be any way of telling?
In this example, I like to look at it as
$$\sum_{k=0}^\infty\frac 12r^2(\theta+kd\theta)d\theta$$
My inside function is positive, but the theta differential is negative.
It also depends on the function of course, if it is positive, then the integral with decreasing theta is negative, since it is the rate of change of a growing function, i.e ##F(\theta+kd\theta)-F(\theta)=dF<0##, hence ##\sum dF=\sum\frac{dF}{d\theta}d\theta=\sum f(\theta)d\theta<0##.

etotheipi
In this example, I like to look at it as
$$\sum_{k=0}^\infty\frac 12r^2(\theta+kd\theta)d\theta$$

I'm not entirely sure I understand your construction; do you mean ##\theta## to be the initial ##\theta = \theta_1##? As in $$I = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{2}r(\theta_{k})^{2}\delta \theta = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{2}r(\theta_{1} + k\delta \theta)^{2}\delta \theta$$ Though I think I understand your point. With ##\delta \theta = \frac{\theta_2 - \theta_1}{n}## if the limits are anticlockwise then we will have positive ##\delta \theta## and otherwise negative ##\delta \theta##.

archaic
I'm not entirely sure I understand your construction; do you mean ##\theta## to be the initial ##\theta = \theta_1##? As in $$I = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{2}r(\theta_{k})^{2}\delta \theta = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{2}r(\theta_{1} + k\delta \theta)^{2}\delta \theta$$ Though I think I understand your point. With ##\delta \theta = \frac{\theta_2 - \theta_1}{n}## if the limits are anticlockwise then we will have positive ##\delta \theta## and otherwise negative ##\delta \theta##.
Yes, sloppy of me, sorry!