Analytical solution for an integral in polar coordinates?

• A
• derya
In summary, the conversation discusses finding open-form solutions to integrals involving known constants lambda and eta. The integrals involve trigonometric functions and can be expressed as Taylor series. The last integral may be a multiple of the complete elliptic integral of the second kind, which could be considered an analytical solution. The possibility of finding a special function for the logarithmic integral is uncertain.
derya
TL;DR Summary
Seeking the solution of an integral over polar coordinates.
Hi,

I am trying to find open-form solutions to the integrals attached below. Lambda and Eta are positive, known constants, smaller than 10 (if it helps). I would appreciate any help! Thank you!

As for 1st integral after squared most of the terms are easily integrated except
$$-2 \int \sqrt{\lambda^2 \cos^2\theta + \eta^2 \sin^2\theta} \ d\theta$$
$$=-2\lambda\int \sqrt{1+a\sin^2\theta}\ d\theta$$
where
$$a = \frac{\eta^2-\lambda^2}{\lambda^2}> -1$$
Say ##\lambda > \eta## you may express the integrand as Taylor series terms of which are easily integrated. Say ##\lambda < \eta## you can do similar thing for integrand ##\sqrt{1+b \cos^2\theta}##. Similar Taylor expansion seem to apply for the 2nd integral also.

Last edited:
anuttarasammyak said:
As for 1st integral after squared most of the terms are easily integrated except
$$-2 \int \sqrt{\lambda^2 \cos^2\theta + \eta^2 \sin^2\theta} \ d\theta$$
$$=-2\lambda\int \sqrt{1+a\sin^2\theta}\ d\theta$$
where
$$a = \frac{\eta^2-\lambda^2}{\lambda^2}> -1$$
Say ##\lambda > \eta## you may express the integrand as Taylor series terms of which are easily integrated. Say ##\lambda < \eta## you can do similar thing for integrand ##\sqrt{1+b \cos^2\theta}##. Similar Taylor expansion seem to apply for the 2nd integral also.
The last integral here looks like a multiple of the complete elliptic integral of the second kind. If a special function counts as an analytical solution, then you might be in luck. As for the logarithmic integral, I'm unclear if you can even find special functions.

anuttarasammyak

1. What is the formula for finding the analytical solution for an integral in polar coordinates?

The formula for finding the analytical solution for an integral in polar coordinates is:
∫∫f(r,θ)rdrdθ = ∫∫f(r,θ)rdθdr

2. How do you convert an integral in Cartesian coordinates to an integral in polar coordinates?

To convert an integral in Cartesian coordinates to an integral in polar coordinates, you can use the following substitutions:
x = rcosθ
y = rsinθ
dx dy = rdrdθ

3. Can you use the same limits of integration for both r and θ in an integral in polar coordinates?

No, the limits of integration for r and θ are different in an integral in polar coordinates. The limits for r depend on the shape of the region being integrated, while the limits for θ are typically from 0 to 2π.

4. What are the advantages of using polar coordinates in integration?

One advantage of using polar coordinates in integration is that it can simplify the integrand and make it easier to solve. Additionally, polar coordinates are often more useful for problems involving circular or symmetric regions.

5. Are there any special cases to consider when using polar coordinates in integration?

Yes, there are a few special cases to consider when using polar coordinates in integration. These include when the region being integrated has a hole in the middle, when the region is unbounded, and when the integrand contains terms that are not expressed in polar form.

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