Sloppy math in the flux calculation?

[Mondo]

Hi,

I have stumbled upon following formula for flux colculation:

Some physical context - we want to calculate how much of the field defined by [imath]1/r^2[/imath] penetrates a sphere surface. The element of a sphere area is defined as [math]da = r^2 sin(\theta)d\theta d\phi[/math]. So in order to get the total flux we need to integrate this function over entire sphere - this is clear. However:

1. The integral in the middle of the equation above figure 2.15 is indefinite yet the [imath]sin(\theta)[/imath] function vanishes - how is that possible?
2. It is a single integral yet he integrates over two independent variables at the same time - is it legal? Shouldn't that be a double integral instead?

Thank you.

 

[logistic_guy]

Hi Mondo.


\(\displaystyle \large \int \frac{1}{4\pi \epsilon_0}\left(\frac{q}{r^2}\bold{\hat{r}}\right) \cdot r^2\sin \theta \ d\theta \ d\phi \ \bold{\hat{r}} = \int\int \frac{1}{4\pi \epsilon_0}\left(\frac{q}{r^2}\right) r^2\sin \theta \ d\theta \ d\phi \)


\(\displaystyle \large = \frac{q}{4\pi \epsilon_0}\int_{0}^{2\pi}\int_0^{\pi} \sin \theta \ d\theta \ d\phi = \frac{q}{4\pi \epsilon_0}\int_{0}^{2\pi} -\cos\theta\bigg |_{0}^{\pi} \ d\phi\)


\(\displaystyle \large = \frac{q}{4\pi \epsilon_0}\int_{0}^{2\pi} -(-1 - 1) \ d\phi = \frac{q}{4\pi \epsilon_0}\int_{0}^{2\pi} 2\ d\phi = \frac{q}{2\pi \epsilon_0}\int_{0}^{2\pi} \ d\phi\)


\(\displaystyle \large = \frac{q}{2\pi \epsilon_0} \phi \bigg |_{0}^{2\pi} = \frac{q}{2\pi \epsilon_0} (2\pi - 0) = \frac{q}{\epsilon_0}\)

 

[blamocur]

The answer to both of you questions is the same: sloppy notation.

1. The integral in the middle of the equation above figure 2.15 is indefinite yet the sin(θ)sin(\theta)sin(θ) function vanishes - how is that possible?

It is not indefinite: ranges for [imath]\theta[/imath] and [imath]\phi[/imath] should be [imath][0, \pi][/imath] and [imath][0, 2\pi][/imath] respectively.

2. It is a single integral yet he integrates over two independent variables at the same time - is it legal? Shouldn't that be a double integral instead?

You are right, the integral should be a double integral.

Having said that, I've seen people using notation like [imath]\int_S ... dS[/imath] for brevity.

 

[Mondo]

@logistic_guy, thanks for your insight. For the first line of your calculations, can you please tell what allows you to replace [imath]\hat{r}[/imath] with another integral?
Likewise, in the second line you all of the sudden introduced integral limits [imath]\pi [/imath] and [imath]2\pi[/imath]. I am not saying they are wrong, after all I know those are the correct limits for the spherical geometry, however it is a "knowledge" that doesn't follow from your previous calculation and that is my problem.

@blamocur, thanks. As I said above in my response to @logistic_guy , I know what sphere limits are but what I don't get here is how author can start with a single indefinite integral and then magically transform it to two definite integrals. Seems like we both agreed, he does shortcuts in notation. Yes, I like [imath]\int_{S} = [..] dS[/imath] more, however in this case it doesn't apply - we need to write two definite integrals with two limits [imath]\pi [/imath] and [imath]2\pi[/imath]

 

[logistic_guy]

For the first line of your calculations, can you please tell what allows you to replace [imath]\hat{r}[/imath] with another integral?

I did not replace \(\displaystyle \large \bold{\hat{r}}\) with another integral. This symbol \(\displaystyle \large \bold{\hat{r}}\) is a unit vector and the dot product of the same unit vectors is: \(\displaystyle \large \bold{\hat{r}} \cdot \bold{\hat{r}} = 1\)

\(\displaystyle \large\oint \bold{E} \cdot d\bold{a} = \int_{a}E\bold{\hat{r}} \cdot da\bold{\hat{r}} = \int_{a} (\bold{\hat{r}} \cdot \bold{\hat{r}})E \ da = \int_{a} (1)E \ da = \int_{a} E \ da\)


Some authors use this notation: \(\displaystyle \large \int_{a} E \ da\)


And some other authors us this notation: \(\displaystyle \large \iint\limits_a E \ da\)

where \(\displaystyle \large a\) is the surface area and \(\displaystyle \large da = \large r^2\sin \theta \ d\theta \ d\phi\) in this case.

Therefore, it does not matter if you will prefer one notation on the other as long as you are consistent.

The author wrote \(\displaystyle \large \int_{a} E \ da = \int_{a} E \ r^2\sin \theta \ d\theta \ d\phi \). When you look at this for the first time, it looks ambiguous but once you know there are different notations, you will understand what the author meant by that.

Likewise, in the second line you all of the sudden introduced integral limits [imath]\pi [/imath] and [imath]2\pi[/imath]. I am not saying they are wrong, after all I know those are the correct limits for the spherical geometry, however it is a "knowledge" that doesn't follow from your previous calculation and that is my problem.

You can introduce the limits of integration whenever you want, but I prefer to introduce them after I do some cleaning. As you can see I put a bunch of stuff outside the integral once I introduced the limits. Before that I just wanted to show you what was the next step of the author which he did not show!

 

[Mondo]

Thank you @logistic_guy , well explained