# Assignment #1

## Tip for dealing with integrals over the volume, i.e.:

$Q_{encl} = \int_0^a \rho(r) d\tau$

### Short version of my tip:

If the the function you’re integrating over (e.g. charge density $$\rho(r)$$) only depends on $$r$$ then don’t integrate over both $$\phi$$ and $$\theta$$, just say

$Q_{encl} = \int_0^a \rho(r) 4\pi r^2 dr$

### Long version of my tip:

Lots of you said (on #6 on the HW):

$Q_{encl} = \int_0^a\int_0^\pi \int_0^{2\pi} \rho(r) r^2 \sin\theta d\theta d\phi dr$

which is absolutely correct, but you totally don’t need to go through that trouble.

For spherically symmetry functions

$d\tau = 4\pi r^2 dr ~~~~~{\rm For~spherically~symmetric~functions.}$

$$4\pi r^2 dr$$ is a perfect, differentially small volume element, as long as whatever you’re integrating over only depends upon $$r$$ and not on $$\theta$$ and $$\phi$$. In other words, as long as your problem has spherical symmetry.

Another version I saw was

$Q_{encl} = \int_0^a \rho(r) dr ~~~~~{\rm this~is~wrong}$

That version is actually incorrect. $$\rho(r)$$ is a charge per volume, so in order to get charge you need to integrate over the volume. You can get a clue about whether you’re integrating over something reasonable by looking at its units. $$dr$$ only has units of length, so it can’t be a volume. $$4 \pi r^2 dr$$ has units of length cubed, so it’s a volume.

Whenever I’m “setting up integrals” which I think is physics speak for “making stuff up that looks like an integral and will hopefully give me the answer I want” I look carefully at the units of the thing. The units have to work out in the integral for the units to work out in the problem. $$dr$$ has units of length.