Author Topic: fun integration in my hw (Read 2754 times) Tweet Share

kamil9876 · « **on:** August 10, 2009, 11:37:21 pm »

Just wanted to see if there is a more elegant solution than mine to this problem:

A sphere of radius $R$ has total charge $Q$ . The volume charge density( $C/m^3$ ) within the sphere at a distance r<R from the centre is:

$p=p_o (1-\frac{r}{R})$

Find an expression for $Q$ in terms of $R$ and $p_o$

Now i used this funny mathematical identity thing that was more or like of the form:

$\lim_{n \rightarrow \infty} \sum_{i=0}^{n-1}(f(\frac{Ri}{n}))(\frac{R}{n})^k = \int_0^R lim_{n \rightarrow \infty} (f(x)(\frac{R}{n})^{k-1}) dx = 0$ where $k$ is an integer greater than 1

Basically stated in a less rigorous but more comprehensible form(what I actually wrote down, the above was just for the mathematical audience to scrutenize the idea) I just did the limit of a riemann sum but with powers on the $\Delta x$ :

$\lim_{\Delta x \rightarrow 0} \sum f(x) \Delta x^k=\lim_{\Delta x \rightarrow 0} \sum f(x)\Delta x \Delta x^{k-1}=\lim_{\Delta x \rightarrow 0} \sum f(x)\Delta x\lim_{\Delta x \rightarrow 0} \Delta x^{k-1}=(\int _a ^b f(x) dx) (0) = 0$

I've used this identity before but I find it quite funny.

Anyways, any comment/more elegant ways of doing this will be appreciated

/0 · « **Reply #1 on:** August 11, 2009, 12:32:32 am »

:p I'll give it a go...
The charge densities are constant at a distance $r$ from the centre, i.e. hollow spheres.
The surface area of one of these spheres is $4\pi r^2$ , and the sphere has thickness $dr$ with charge $dQ$

*leap of faith* volume of the outside bit of one of these hollow spheres is $4 \pi r^2dr$
(my reasoning is that if $dr$ were 'big', the approximation would be very bad, but as $dr$ approaches zero, the accuracy of this expression increases)

$p_0(1-\frac{r}{R}) = \frac{dQ}{4\pi r^2dr}$

$4\pi r^2 p_0(1-\frac{r}{R})dr = dQ$

$Q = \int_0^R 4\pi r^2 p_0(1-\frac{r}{R})dr=\frac{R^3p_0\pi}{3}$

kamil9876 · « **Reply #2 on:** August 11, 2009, 12:57:34 am »

yes that is correct answer. In fact your leap of faith can be used to justify why the derivative of Volume is surface area(or analogously, why the derivative of area is circumfere of circle). I see you are very physicsy by using infinitesimal reasoning.

I basically did the same idea, split the thing into onion shells of thickness $\Delta r$ .

The volume of this onion shell at a distance r is:

$V(r)=\frac{4\pi}{3}(r+\Delta r)^3 - \frac{4\pi}{3}r^3$
$=\frac{4}{3}(3\pi r(\Delta r)^2 + 3\pi r^2\Delta r + \Delta r^3)$

Therefore the charge in the onion shell at distance r is:

$Q(r)= pV(r)$

total charge is:

$Q=lim_{\Delta r \rightarrow 0} \sum \frac{4}{3} (3p\pi r(\Delta r)^2 + 3p\pi r^2\Delta r + p\Delta r^3)$

$=\frac{4}{3} (lim_{\Delta r \rightarrow 0} \sum 3p\pi r(\Delta r)^2 + lim_{\Delta r \rightarrow 0} \sum 3p\pi r^2\Delta r + \frac{4}{3} lim_{\Delta r \rightarrow 0} \sum p\Delta r^3)$

Now by using the magical identities I wrote up in the first post I can make two of the terms dissapear:

$Q=\frac{4}{3} lim_{\Delta r \rightarrow 0} \sum 3p\pi r^2\Delta r$
$=4 \int_0^R p\pi r^2\ dr$

and the rest is spec maths.

/0 · « **Reply #3 on:** August 11, 2009, 01:45:02 am »

lol wow
I tried $\frac{4\pi}{3} ((r+\Delta r)^3-r^3)$ but I couldn't get anywhere with it
But yeah that's pretty sick how you can do that with advanced math

kamil9876 · « **Reply #4 on:** August 11, 2009, 11:28:22 am »

oh yeah, just realised two other ways perhaps:

Split the thing into onion layers of equal volume, $\Delta V$

The integral merely becomes:

$\int_0^{\frac{4}{3}\pi R^3} p dV$

And just get p in terms of V or make those dV=... substituions.

But i think I prefer my crazy riemann sums

/0 · « **Reply #5 on:** October 07, 2009, 03:11:42 am »

Sorry to necro lol, but a problem like this came up in an assignment recently and I think I had a more thorough 'think' (at 1:00am) about how to prove that $\rho = 4\pi r^2 dr$

I'm hoping this is a less physicsy (i know you meant waffly

) explanation

Basically, you have a thin shell with inner radius $r$ and outer radius $R$ . We cut the shell in half, and then we 'unfold' both halves of the shell so that they look like two frustum (3D trapezoid), with two parallel circular sides each. It's meant to be a bijective mapping (as if the jargon's gonna help), like shining a light on top of a riemann sphere and observing the shadow on the plane (it's in this "mobius transformations" video on youtube somewhere). The larger side is the outside surface of the halfshell and the inner side is the inside surface of the halfshell.

Now, take the larger side to be the base of a cylinder, and the smaller side to be the base of another cylinder, both with heights equal to the height of the frustum, $\Delta r$ .

Then we will have $V_{small\ cylinder} \leq V_{frustum} \leq V_{larger \ cylinder}$

$\Rightarrow 2 \pi r^2 \Delta r \leq V_{frustum} \leq 2\pi R^2 \Delta r$

Now, as we let $R \to r$ ,

$2 \pi r^2 dr \leq dV_{frustum} \leq 2 \pi r^2 dr$

$\therefore dV_{halfshell} = dV_{frustrum} = 2\pi r^2 dr$

$\therefore dV_{shell} = 4 \pi r^2dr$

woohooooooo

kamil9876 · « **Reply #6 on:** October 07, 2009, 06:21:57 pm »

I think your idea is the 3D analog of an argument I thought of when trying to prove $A_{circle}=\pi r^2$ back in yr 7-8. My idea was to think of the disk as made up of many coiled-circle strings. Then as you straighten them out you see that they form a trapezium since the circumference(and hence length) of each string is a linear function of the radius. This gives another formula for a ring, and by taking the inner radius as zero you get the circle.

The picture represents how each string is stretched out to make straight line segments, and the triangle's perimeter is basically the set of the endpoints of these line segments. Easily seeing the area is $A=\frac{1}{2}(r)(2\pi r)$

Lol but taking it to 3D is insane

, was pretty difficult to imagine but good work

TrueTears · « **Reply #7 on:** October 07, 2009, 06:27:52 pm »

Quote from: kamil9876 on October 07, 2009, 06:21:57 pm

I think your idea is the 3D analog of an argument I thought of when trying to prove $A_{circle}=\pi r^2$ back in yr 7-8. My idea was to think of the disk as made up of many coiled-circle strings. Then as you straighten them out you see that they form a trapezium since the circumference(and hence length) of each string is a linear function of the radius. This gives another formula for a ring, and by taking the inner radius as zero you get the circle.

The picture represents how each string is stretched out to make straight line segments, and the triangle's perimeter is basically the set of the endpoints of these line segments. Easily seeing the area is $A=\frac{1}{2}(2\pi r)$

Lol but taking it to 3D is insane , was pretty difficult to imagine but good work

I just learnt something new today.

kamil9876 · « **Reply #8 on:** October 07, 2009, 07:11:08 pm »

oh shit just realised i forgot the $(r)$ haha

mark_alec · « **Reply #9 on:** October 07, 2009, 08:30:20 pm »

If you use spherical coordinates, the $4\pi r^2 dr$ just pops out.

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