Calculating Area in Polar Coordinates

In summary: 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.
  • #1
archaic
688
214
Homework Statement
Find the area of the region inside the circle ##r=-2\sin\theta## and outside the circle ##r=1##.
Relevant Equations
$$A=\frac 12\int_{\theta_1}^{\theta_2}(r^2-r^2_0)d\theta$$
$$-2\sin\theta=1\Leftrightarrow\theta=-\frac{\pi}{6},\,-\frac{5\pi}{6}\\
\begin{align*}
\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(4\sin^2\theta-1\right)d\theta
&=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12\left(1-2\cos2\theta\right)d\theta\\
&=\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\frac 12d\theta-\int_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\cos2\theta d\theta\\
&=\frac 12\left[\theta-\sin2\theta\right]_{-\frac{\pi}{6}}^{-\frac{5\pi}{6}}\\
&=-\frac{5\pi}{12}-\frac{\sqrt 3}{4}+\frac{\pi}{12}-\frac{\sqrt 3}{4}\\
&=-\frac{\pi}{3}-\frac{\sqrt 3}{2}
\end{align*}\\
A=\frac{\pi}{3}+\frac{\sqrt 3}{2}$$
 

Attachments

  • Capture.PNG
    Capture.PNG
    9.9 KB · Views: 138
Last edited:
Physics news on Phys.org
  • #2
That's what I got as well; is it supposedly wrong?
 
  • Like
Likes archaic
  • #3
etotheipi said:
That's what I got as well; is it supposedly wrong?
Nope, was just checking. 😵
Thank you!
 
  • Like
Likes etotheipi
  • #4
archaic said:
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...
 
Last edited by a moderator:
  • #5
etotheipi said:
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...
Hint: your post about ##dx## being negative. :)
 
  • Like
Likes etotheipi
  • #6
archaic said:
Hint: your post about ##dx## being negative. :)

Yeah, it's fair to say it still confuses me!
 
  • #7
etotheipi said:
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!
 
  • #8
archaic said:
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?
 
  • #9
etotheipi said:
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##.
 
  • Informative
Likes etotheipi
  • #10
archaic said:
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##.
 
  • Like
Likes archaic
  • #11
etotheipi said:
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!
 

Related to Calculating Area in Polar Coordinates

1. What is the formula for calculating area in polar coordinates?

The formula for calculating area in polar coordinates is A = 1/2 ∫ab r2 dθ, where r is the distance from the origin and θ is the angle.

2. How is the concept of area in polar coordinates different from Cartesian coordinates?

In polar coordinates, the area is calculated by integrating over a sector of a circle, while in Cartesian coordinates, the area is calculated by finding the product of the length and width of a rectangle.

3. Can you give an example of finding the area in polar coordinates?

Yes, for example, if we want to find the area of the region bounded by the curve r = 2cos(θ) and the line θ = π/4, we would set up the integral as A = 1/2 ∫0π/4 (2cos(θ))2 dθ and solve for the area.

4. How is the concept of area in polar coordinates useful in real-world applications?

The concept of area in polar coordinates is useful in real-world applications such as calculating the area of a circular region, finding the area of a sector in a polar graph, and determining the area of a region bounded by polar curves.

5. Is there a specific method for converting between polar and Cartesian coordinates when calculating area?

Yes, when converting between polar and Cartesian coordinates, we use the formula x = rcos(θ) and y = rsin(θ) to find the corresponding x and y values for a given polar point. These values can then be used to calculate the area in Cartesian coordinates.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
433
  • Calculus and Beyond Homework Help
Replies
4
Views
388
  • Calculus and Beyond Homework Help
Replies
3
Views
956
  • Calculus and Beyond Homework Help
Replies
3
Views
670
  • Calculus and Beyond Homework Help
Replies
9
Views
380
  • Calculus and Beyond Homework Help
Replies
1
Views
333
  • Calculus and Beyond Homework Help
Replies
3
Views
480
  • Calculus and Beyond Homework Help
Replies
3
Views
867
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
679
Back
Top