Integrating Over Angles vs. Integrating Over a Surface: What's the Difference?

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Homework Help Overview

The discussion revolves around the integration of a function defined in terms of angles, specifically S(θ, φ) = U (c/2) cos²(θ), over specified ranges for φ and θ. Participants are exploring the implications of integrating over angles versus integrating over a surface parameterized by those angles.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning whether to perform the double integral directly or to consider a transformation involving a Jacobian for surface integration. There is discussion about the correct interpretation of the integration limits and the nature of the variables involved.

Discussion Status

Some participants have provided insights into the necessity of including the Jacobian when integrating over a surface, while others are clarifying the difference between treating angles as arbitrary variables versus integrating over a defined area. There is ongoing exploration of the correct setup for the integral.

Contextual Notes

There is mention of potential confusion regarding the notation for angles and the need for clarity in the problem statement. Participants are also reflecting on the expected outcome of the integral and the conditions under which it holds true.

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Homework Statement



Suppose I have a relation S(\theta, \phi) = U \frac{c}{2} \cos{^{2} \, \theta} and I want to integrate over \phi from 0 to 2 \pi and \theta from 0 to \frac{\pi}{2}. How do I do this double integral? Do I just do it normally (without any transformation), or do I use a total angle? I know that I'm supposed to get S = U \frac{c}{4} but I'm not sure how it works.

Here's an image I made:
S9VStKK.png


Angles are a little confusing for me so I would like some feedback.
 
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LoadedAnvils said:
Suppose I have a relation S(\theta, \phi) = U \frac{c}{2} \cos{^{2} \, \theta} and I want to integrate over \phi from 0 to 2 \pi and \theta from 0 to \frac{\pi}{2}. How do I do this double integral? Do I just do it normally (without any transformation), or do I use a total angle? I know that I'm supposed to get S = U \frac{c}{4} but I'm not sure how it works.
What you have literally stated as the problem is to find ##\int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} U \frac{c}{2} \cos{^{2} \, \theta} d\theta d\phi##. But that isn't going to give the desired answer.
What you mean, I suspect, is that you want to integrate over a surface parameterised by those angles. In that case you need to include the Jacobian. This represents the area of an element dA in terms of small changes in the two angles. See e.g. http://en.wikipedia.org/wiki/Spheri..._and_differentiation_in_spherical_coordinates.
(But make sure you get your θ and ϕ the right way around. Not everyone uses the same notation.)
 
What is the difference between integrating over the angles and integrating over the surface parameterised by these angles?

I shouldn't have put S as a function. If I do an integral on each side, I get the desired result if it is an integral over the angles.
 
LoadedAnvils said:
What is the difference between integrating over the angles and integrating over the surface parameterised by these angles?
Integrating over the angles doesn't take any regard of the fact that they are angles. it just treats them as arbitrary variables of integration. ∫∫(function)dθdϕ.
Integrating over a surface means ∫(function)dA, where dA is an element of area. In spherical polar, dA = r2 sin(θ)dθdϕ, where θ is the polar angle and ϕ is the azimuthal angle (which is the way you have used them).
I shouldn't have put S as a function. If I do an integral on each side, I get the desired result if it is an integral over the angles.
You mean ##\int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} U \frac{c}{2} \cos{^{2} \, \theta} d\theta d\phi##?
I get a π2 in the value of that. Please post your working.
OTOH, if I include the sin(θ), I still get a single factor of π, so still not the answer given in the OP.
Please post the problem exactly as given.
 

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