# Area under the curve of a Polar Graph

Hello all!

I'm just wanting a quick clarification on how finding the area under a polar graph works. Say we have the polar graph of ##r\left(\theta \right)=\frac{\arctan \left(2\theta \right)}{\theta }## as shown below:

I know that the area under the graph between ##0## and ##\frac{\pi }{2}## can be found with ##\int _0^{\frac{\pi }{2}}\left(\frac{1}{2}\left(\frac{\arctan \left(2\theta \right)}{\theta }\right)^2\right)d\theta ##. However, the thing I'm not quite sure about is if this integral finds only the red shaded area in the above image, or if it is both the red and blue shaded areas. Any help with this would be much appreciated :)

Anora

Orodruin
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The area element in polar coordinates is ##dA = r\, dr\, d\theta##. The area bounded by a given radial function ##r(\theta)## between ##\theta_1## and ##\theta_2## is therefore on the form
$$A(\theta_1,\theta_2) = \int_{\theta_1}^{\theta_2} \int_0^{r(\theta)} r\, dr \, d\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} r(\theta)^2 d\theta.$$

The area element in polar coordinates is ##dA = r\, dr\, d\theta##. The area bounded by a given radial function ##r(\theta)## between ##\theta_1## and ##\theta_2## is therefore on the form
$$A(\theta_1,\theta_2) = \int_{\theta_1}^{\theta_2} \int_0^{r(\theta)} r\, dr \, d\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} r(\theta)^2 d\theta.$$
Thank you for taking the time to reply :)

I know that to find the area enclosed under under the polar graph between ##\theta=0## and ##\theta={\frac{\pi }{2}}## I should use ##\int _0^{\frac{\pi }{2}}\left(\frac{1}{2}\left(\frac{\arctan \left(2\theta \right)}{\theta }\right)^2\right)d\theta##. However, the part which I'm confused about is how this is visually represented on the graph it self. Referring back to the image in the OP: would integrating between ##\theta=0## and ##\theta={\frac{\pi }{2}}## be visually represented as just the red shaded area? Or would it include the blue shaded area as well?

Orodruin
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What do you think based on the integrals in #2?

Hello all!

I'm just wanting a quick clarification on how finding the area under a polar graph works. Say we have the polar graph of ##r\left(\theta \right)=\frac{\arctan \left(2\theta \right)}{\theta }## as shown below:

View attachment 197680

I know that the area under the graph between ##0## and ##\frac{\pi }{2}## can be found with ##\int _0^{\frac{\pi }{2}}\left(\frac{1}{2}\left(\frac{\arctan \left(2\theta \right)}{\theta }\right)^2\right)d\theta ##. However, the thing I'm not quite sure about is if this integral finds only the red shaded area in the above image, or if it is both the red and blue shaded areas. Any help with this would be much appreciated :)
Hopefully I can give you the clear answer you asked for. The integral given should take the curve from 0 to π/2 separately from the rest of the graph, and evaluate the area underneath it as including the blue section with the red. I think that is what you are asking, but just to clarify, the blue section would not be counted additionally for the extra times that the graph passes through quadrant one. Hope that was helpful!

Mark44
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Hello all!

I'm just wanting a quick clarification on how finding the area under a polar graph works. Say we have the polar graph of ##r\left(\theta \right)=\frac{\arctan \left(2\theta \right)}{\theta }## as shown below:

View attachment 197680

I know that the area under the graph between ##0## and ##\frac{\pi }{2}## can be found with ##\int _0^{\frac{\pi }{2}}\left(\frac{1}{2}\left(\frac{\arctan \left(2\theta \right)}{\theta }\right)^2\right)d\theta ##. However, the thing I'm not quite sure about is if this integral finds only the red shaded area in the above image, or if it is both the red and blue shaded areas. Any help with this would be much appreciated :)
A major problem with your integral is that the integrand is undefined when ##\theta = 0##.

Another problem, I believe, is that the section of the curve in Quadrant I that is the farthest from the pole doesn't correspond to the interval ##[0, \pi/2]##. f(0) is undefined (as I already mentioned), and ##f(\pi/2) = \frac{\arctan(\pi)}{\pi/2} \approx 0.402##. Neither of these values of ##\theta## corresponds to any point on the upper arch of the curve in Q I.

If all you're interested in is finding out how polar integration works, try working with a simpler polar function, like r = 1, between 0 and ##\pi/2##.

Orodruin
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A major problem with your integral is that the integrand is undefined when ##\theta = 0##.
I do not agree. The integral is perfectly well defined. An integrand does not need to be well defined at every point to make the integral well defined. Furthermore, the extension to ##\theta = 0## is trivial by assuming continuity.

Also, I get ca 0.8 at ##\theta = \pi/2##. Did you miss the multiplication by 2?

Mark44
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I do not agree. The integral is perfectly well defined. An integrand does not need to be well defined at every point to make the integral well defined. Furthermore, the extension to ##\theta = 0## is trivial by assuming continuity.
I understand that the integrand doesn't need to be defined at every point in the interval of concern. Strictly speaking, the integrand is of the indeterminate form [0/0] at ##\theta = 0##, so it's not a simple matter of finding an antiderivative and plugging in the two endpoints.

Because the function is indeterminate at ##\theta = 0##, I considered the possibility that ##\lim_{\theta \to 0} f(\theta)## exists, but didn't make the effort to find this limit.
Orodruin said:
Also, I get ca 0.8 at ##\theta = \pi/2##. Did you miss the multiplication by 2?
Yes. I have the 2 written down as ##2 \frac{\arctan(\pi)}{\pi}## in my work, but neglected to incorporate the factor of 2 in the value I wrote earlier.

Orodruin
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I understand that the integrand doesn't need to be defined at every point in the interval of concern. Strictly speaking, the integrand is of the indeterminate form [0/0] at ##\theta = 0##, so it's not a simple matter of finding an antiderivative and plugging in the two endpoints.
But this does not really affect whether or not the integral is well defined or not. Also, that the function is of the form ##\theta/\theta## as ##\theta \to 0## does not mean that its anti-derivative will be ill defined in the same limit. This is not even the case when the function limit is diverging. Consider the integral
$$I = \int_0^1 \frac{dx}{\sqrt{x}}.$$
Although the integrand diverges as ##x \to 0##, the result can be perfectly well found as
$$I = 2[\sqrt{x}]_0^1 = 2\sqrt{1} - 2 \sqrt{0} = 2.$$

The limit as ##\theta \to 0## is rather trivial if you just remember that ##\arctan(x) = x + \mathcal O(x^2)##.

Mark44
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But this does not really affect whether or not the integral is well defined or not. Also, that the function is of the form ##\theta/\theta## as ##\theta \to 0## does not mean that its anti-derivative will be ill defined in the same limit. This is not even the case when the function limit is diverging. Consider the integral
$$I = \int_0^1 \frac{dx}{\sqrt{x}}.$$
Right, but a slightly different integral, ##\int_0^1 \frac{dx} x## has an antiderivative whose limit diverges at x = 0. For the OP's question, I didn't do the work of calculating the integral -- I was looking for some evidence that he had considered that his integrand wasn't defined at ##\theta = 0##, and that the interval he is considering, ##[0, \pi/2]## corresponds to the boundary arc in Quadrant I.
Orodruin said:
Although the integrand diverges as ##x \to 0##, the result can be perfectly well found as
$$I = 2[\sqrt{x}]_0^1 = 2\sqrt{1} - 2 \sqrt{0} = 2.$$

The limit as ##\theta \to 0## is rather trivial if you just remember that ##\arctan(x) = x + \mathcal O(x^2)##.

Orodruin
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Right, but a slightly different integral, ##\int_0^1 \frac{dx} x## has an antiderivative whose limit diverges at x = 0.
Yes, but in that case the integral also diverges and is not well defined. As long as the anti-derivative has a well defined limit, the definition of the integral
##
\int_0^a \ldots dx = \lim_{\epsilon \to 0^+} \int_\epsilon^a \ldots dx
##
is unambiguous.

I do not agree. The integral is perfectly well defined. An integrand does not need to be well defined at every point to make the integral well defined. Furthermore, the extension to ##\theta = 0## is trivial by assuming continuity.

Also, I get ca 0.8 at ##\theta = \pi/2##. Did you miss the multiplication by 2?
I disagree with your disagreement. Since the integral must be evaluated at both points in order to obtain the required solution, the equation must be continuous and defined on the given interval to be differentiable. My stance would be that the lower bound would need to be the 'limit as x approaches 0 from the positive direction' of x.

Hopefully I can give you the clear answer you asked for. The integral given should take the curve from 0 to π/2 separately from the rest of the graph, and evaluate the area underneath it as including the blue section with the red. I think that is what you are asking, but just to clarify, the blue section would not be counted additionally for the extra times that the graph passes through quadrant one. Hope that was helpful!
Yes thank you, that was what I was wondering. I had guessed that the area would be the total area under the curve between ##0## and ##pi/2## as that's how integration works with regular functions of x. The only thing that made me unsure with this question was the fact that polar graphs can be one:many and many:many, so I was just looking for clarification.

Orodruin
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I disagree with your disagreement. Since the integral must be evaluated at both points in order to obtain the required solution, the equation must be continuous and defined on the given interval to be differentiable. My stance would be that the lower bound would need to be the 'limit as x approaches 0 from the positive direction' of x.
Mathematics does not care about your stance. You can put an arbitrary function value at ##\theta=0## and still have the same integral. In fact, you could redefine the integrand to arbitrary values at a set of zero measure and still obtain the same integral. That the function is differentiable is completely irrelevant for its integrability. With your argumentation, a discontinuous function would not be integrable, which is nonsense.

However, the thing I'm not quite sure about is if this integral finds only the red shaded area in the above image, or if it is both the red and blue shaded areas. Any help with this would be much appreciated :)
The important thing is that arctan(x) has several branches. I had some trouble forcing WolframAlpha to show that but here goes.
Also the antiderivative of your function has several branches. When computing the integral, you get to pick the branch. If both endpoints are on the same branch, you get the red area or one of the blue ones.
If the two endpoints are on a different branch, it will count the red and blue areas and also the other quadrants so the result will not be very useful.
(If the two endpoints are on an adjacent branch, it might still make sense depending on the actual angles).

Orodruin
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The important thing is that arctan(x) has several branches. I had some trouble forcing WolframAlpha to show that but here goes.
Also the antiderivative of your function has several branches. When computing the integral, you get to pick the branch. If both endpoints are on the same branch, you get the red area or one of the blue ones.
If the two endpoints are on a different branch, it will count the red and blue areas and also the other quadrants so the result will not be very useful.
(If the two endpoints are on an adjacent branch, it might still make sense depending on the actual angles).
No, this is not the case. The curve shown in the OP is the polar curve for the principal branch of arctan (taking values between -pi/2 and pi/2). The fact that it has several radii in the same direction is due to the periodicity of the polar angle, not the branching of the arctan function. The OP therefore has very little to do with branch cuts of arctan.

The integral in the OP quite clearly goes from zero radius to the radius described by the polar function as described in #2.

SlowThinker