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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!

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$dr$ is infinitesimal change in radius, since $dr$ itself is small, higher power of $dr$ are even insignificant, so it's usually ignored. –  Santosh Linkha Jan 13 '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 '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 '13 at 20:43

1 Answer 1

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:

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

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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 '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 '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 '13 at 21:26
    
@user825089 I agree! –  amWhy Jan 13 '13 at 22:08
    
+1 nice attempt. –  B. S. Feb 11 '13 at 7:48

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