I'm doing a problem in physics, but it's the math part I'm curious about:
Charge density is defined by $\rho = \frac{dQ}{dV}$, then $Q = \int_{V}^{} \rho \text{d}V$
The problem is dealing with a sphere and the answer book says $dV = V(r + dr) - V(r) = \frac{4}{3}\pi(r+dr)^{3} - \frac{4}{3}\pi r^{3}= 4\pi \cdot dr \cdot r^2 + 4\pi \cdot(dr)^{2} \cdot r + \frac{4}{3}\pi \cdot (dr)^{3}$
Then the integral becomes $Q = \int_{0}^{R} \rho \cdot 4\pi \cdot r^2 \text{d}r$
But only the first term of $dV$ is included here. Can we simply ignore the terms with higher powers of $dr$?
I hope someone could explain this. Thank you!
This is a matter of using a change-in-variables:
integrating with respect to volume $\implies $ integrating with respect to $r$:
We have $$\;Q = \int_{V}^{} \rho\,dV\tag{1}$$
We know $V$ as a function of $r$: $$\;V(r) = \tfrac43 \pi{r}^3\,;\;\text{ so}\;\;dV = 4\pi{r}^2\,dr\tag{2}$$
Now replace $dV$ in $(1)$ by its equivalent in $(2)$, and we get:
$$Q = \int_{0}^{R}\rho\, 4\pi{r}^2 \,dr$$