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

In summary, the conversation discusses the process of integrating over a surface parameterized by two angles, ϕ and θ, using a relation S(θ, ϕ) = U * c * cos^2(θ). It is noted that integrating over the angles without taking into account the fact that they are angles will not give the desired result. To properly integrate over the surface, the Jacobian must be included, which represents the area of an element dA in terms of small changes in the angles. The conversation also clarifies the notation used for the two angles and the difference between integrating over the angles and integrating over the surface. The desired result for the integral is S = U * c / 4.
  • #1
LoadedAnvils
37
0

Homework Statement



Suppose I have a relation [itex]S(\theta, \phi) = U \frac{c}{2} \cos{^{2} \, \theta} [/itex] and I want to integrate over [itex]\phi[/itex] from [itex] 0 [/itex] to [itex] 2 \pi [/itex] and [itex] \theta [/itex] from [itex] 0 [/itex] to [itex] \frac{\pi}{2} [/itex]. 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 [itex] S = U \frac{c}{4} [/itex] 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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  • #2
LoadedAnvils said:
Suppose I have a relation [itex]S(\theta, \phi) = U \frac{c}{2} \cos{^{2} \, \theta} [/itex] and I want to integrate over [itex]\phi[/itex] from [itex] 0 [/itex] to [itex] 2 \pi [/itex] and [itex] \theta [/itex] from [itex] 0 [/itex] to [itex] \frac{\pi}{2} [/itex]. 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 [itex] S = U \frac{c}{4} [/itex] 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.)
 
  • #3
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.
 
  • #4
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.
 

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

What is "Integration over all angles"?

"Integration over all angles" is a mathematical concept that involves calculating the area under a curve of a function over all possible angles. It is used to determine the total amount of a quantity that is distributed over a range of angles.

Why is "Integration over all angles" important?

Integration over all angles is important because it allows us to accurately calculate the total amount of a quantity that is distributed over a range of angles. This is especially useful in fields such as physics, where understanding the distribution of forces or particles is crucial.

How is "Integration over all angles" different from regular integration?

The main difference between "Integration over all angles" and regular integration is that regular integration involves calculating the area under a curve over a specific interval, while "Integration over all angles" involves calculating the area under a curve over all possible angles. This requires using specialized techniques and formulas.

What are some real-world applications of "Integration over all angles"?

"Integration over all angles" has many real-world applications, such as determining the total amount of light emitted by a light source, calculating the total amount of force exerted by a rotating object, and determining the total amount of radiation received by an object from all directions.

What are some common techniques used for "Integration over all angles"?

There are various techniques used for "Integration over all angles", including the use of trigonometric identities, change of variables, and specialized integration formulas such as the Jacobi-Anger expansion. It is also common to use computer software or numerical methods to perform the calculations.

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