Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite
1

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!

share|improve this question
1  
$dr$ is infinitesimal change in radius, since $dr$ itself is small, higher power of $dr$ are even insignificant, so it's usually ignored. – experimentX Jan 13 at 20:26
2  
Do you understand why the change of variables formula works for integrals? Here you want to write the integral with respect to $r$, and since $V(r) = (4/3)\pi{r}^3$, $dV = 4\pi{r}^2dr$ in the integrand. – KCd Jan 13 at 20:32
Thank you! That makes sense, of course. What I am asking is why the book has it in terms of change in $V$ instead of a simple differential $dV$ – user825089 Jan 13 at 20:43

1 Answer

up vote 3 down vote accepted

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:

$$Q = \int_{0}^{R}\rho\, 4\pi{r}^2 \,dr$$

share|improve this answer
Thank you! That is true, of course. My question though: is it a valid trick to write out the change in a function $V(r)$ instead of a differential $dV$ and then ignore the higher order $dr$s. Is it mathematically correct or the book is wrong? – user825089 Jan 13 at 20:53
I'm not clear what your book was doing: it seems evident that $dV/dr = 4\pi r^2$, that is, $dV = 4\pi r^2\,dr$. $dr$ is infinitesimal change in radius, so higher powers of $dr$ are even less significant, so can be ignored: I'm assuming that's what is being conveyed by your text's solution. – amWhy Jan 13 at 21:05
I think they should've used some different notation, e.g. $\Delta V$. Because $dV$ and the $dV$ in the book are clearly not the same thing. – user825089 Jan 13 at 21:26
@user825089 I agree! – amWhy Jan 13 at 22:08
+1 nice attempt. – Babak S. Feb 11 at 7:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.