# Thread: Using Calculus to determine Luminous Flux

1. ## Using Calculus to determine Luminous Flux

I need to determine the total luminous flux of an LED lamp. I have readings of the intensity (I) at a range of given latitude and longitude angles ( $\theta, \phi$), but I do not know how to solve with the equation given for calculating total luminous flux:

$\int^{\pi2}_{\phi=0} \int^{\pi}_{\theta=0} I(\theta,\phi) \sin\theta d\theta d\phi$

My knowledge of calculus is quite limited as I have only this year started to learn it at school. I would really appreciate an explanation of how to apply this equation!

2. ## Re: Using Calculus to determine Luminous Flux

It would be nice if we knew what the function \displaystyle \begin{align*} I \left( \theta, \phi \right) \end{align*} is...

3. ## Re: Using Calculus to determine Luminous Flux

Hey boomhelm.

You will need to use numeric integration for this problem to get an approximate answer that will be better with more observations (the more observations in a fixed region, the better the approximation).

Have you covered numerical integration in class?

4. ## Re: Using Calculus to determine Luminous Flux

Thank you both for your replies!

I am learning pre-calc at school at the moment, but I have briefly started to touch on numerical integration through my own independent learning - although not massively confident with it yet!

The function $(I\theta, \phi$) is the intensity in candela at a given angle, where $\theta$ is the polar axis and $\phi$ is the azimuth axis.

For example, I have measured 407 candela at 90 degrees on the polar axis and 0 degrees on the azimuth. So I guess $I$ simply becomes 407. Is this correct?

Then I guess the next part $(sin\theta)$ becomes $sin 90$.

This is as far as I have got with it... It really is bugging me. Like what do I do where there are two definite integrals? Do I multiply them...?

I understand this may come across a bit amateur, but I am only 13.

Any further help would be really appreciated with this.

5. ## Re: Using Calculus to determine Luminous Flux

When you have multiple integrals (like a double one) you sum volumes or hypervolumes instead of areas.

Pick a dx and dy that are small and each new volume will be under normal Riemann approximation: I(x,y)*sin(x)*dx*dy for every x starting from the left limit and increasing in steps of size dx with the y doing the same thing (start at y limit and increase every dy size).

You then add all of these volumes together and you get the approximation for the integral.

This is the most basic of numerical integration and there are many different ways to do the same thing: each with advantages and dis-advantages of their own.