| Posted on Tuesday, 27 July, 2004 - 01:29 pm: |
Hi, I've been trying to do the surface area of a sphere by integration using: A=4pò0r y dx,
| y= | _____ Ör2-x2 |
Substitute x=rsinf, and fiddle around with limits, ''dx''s and stuff, and you get: A=4pò0p/2r2cos2fdf
How do you evaluate this integral?
Please help, I need to prove to my Dad that 14 year old's can do integration!
Thanks,
Tristan
| Posted on Tuesday, 27 July, 2004 - 01:42 pm: |
Take out the factor of r squared and note that
2cos2 x=cos(2x)+1
Can you integrate it now?
-Phil.
| Posted on Tuesday, 27 July, 2004 - 01:48 pm: |
How do you mean? Should I have the cos2x thing and substitute again?
| Posted on Tuesday, 27 July, 2004 - 01:49 pm: |
Also, the formula for the surface area of a solid of revolution is 2pòy.ds where Using this makes it a lot easier and the integral simplifies greatly.
-Phil.
| Posted on Tuesday, 27 July, 2004 - 01:50 pm: |
True. But why doesn't my method work?
| Posted on Tuesday, 27 July, 2004 - 06:28 pm: |
OK. The simple answer is that it's to do with the curvature of the sphere. You seem to be integrating over the infinitesimal cylinders y dx, this is not what a sphere looks like locally. Imagine zooiming in to the edge of a circle, your intuition will tell you that it will not start looking like a zig-zag line, but that it should start looking like a straight diagonal line, so in this sense, the small bits we should be adding up should be truncated cones. In the infinitesimal limit, this turns into y ds. Where ds is the length of the diagonal, and is equal to what Phil says above. So, the integral should be 2pò0r y ds. For more verification of the phenomenon: draw a square (of sides 1), and split it into smaller squares. You will notice if you measure the zig-zag length from one corner to the opposite, the length always stays at 2, however small you make the zig-zag line. But if you measure the diagonal distance from the corner to the opposite, it is Ö2. Julian
| Posted on Tuesday, 27 July, 2004 - 07:13 pm: |
That's something along the lines that I was thinking, but I didn't want to make a fool out of myself by saying it and it being wrong
-Phil.
| Posted on Tuesday, 27 July, 2004 - 07:20 pm: |
So, what's the working then? I'm probably being stupid, but I can't see it!
Tristan.
| Posted on Tuesday, 27 July, 2004 - 07:35 pm: |
Use
| ds= | _________ Ö1+(dy/dx)2 | dx |
. In order to find dy/dx simply rearrange to give x2+y2=r2 then differentiate wrt x implicitly to get
| dy/dx=-x/y=-x/ | _____ Ör2-x2 |
Then you get
| ds/dx=x/ | _____ Ör2-x2 |
Use area =2pòy.ds and the big square roots cancel nicely giving 2pò-rr r.ds=4pr2 Phil
| Posted on Wednesday, 28 July, 2004 - 10:11 am: |
Phil,
Great, thanks!
Tristan
| Posted on Wednesday, 28 July, 2004 - 10:28 am: |
However... When finding the area of a circle with integration, you use the "rectangle rule", i.e. the ëlement of area" is ydx. When rotating a rectangle through 2p radians around the x-axis, you discover a cylinder, surface area 2py dx. So why doesn't my method work?
| Posted on Wednesday, 28 July, 2004 - 11:42 am: |
I've wondered about that one too. Strange.
-Phil.
| Posted on Wednesday, 28 July, 2004 - 04:33 pm: |
I don't want to get too much into the specifics of Riemann Integration (which is what you're trying to justify here), so analysts should turn off 'rigour mode'.
The problem in this case is linked heavily to the example I gave in my last post.
Consider the square which I talked about earlier (I'll try and get an image uploaded for you if you're having trouble picturing it). I'll try and explain it again, draw a square. Suppose we taking dx to be 1/2 then the area under the zig-zag will be 3/4. If we take dx = 1/4 then the area under the zig-zag is 3/4-1/8, and in general:
If dx = 1/2n , call the area under the zig-zag An . We have the equation:
An =An-1 -2-n-1 , and A1 =3/4
Analysts recognise this as a Cauchy Sequence, and will know that it converges to a certain value, we can call this A¥ , so let's consider
Sum the geometric series, and we get A¥ =A1 +1+1/2+1/4-2 = 1/2
Low and behold, if we draw a diagonal line on the large square from one corner to the next and calculate the area of the triangle underneath the area is 1/2. So we have concluded as dx tends to zero, the sum of the rectangles tends to the integral under the line.
*But*: as I wrote earlier, the length of the zig-zag line does *not* converge on the length of the diagonal line - no matter how small you go - since the length always stays the same at 2.
It is in this same sense that the sum of the areas of the rectangles converges to the area in a circle, but the sum of the dx's doesn't converge on the arc length of the curve.
In terms of doing things like this, because it can seem a big hand wavey you should always begin by writing down the infintesimal for what you're trying to find.
In this case, you want to find the surface area, call this S.
so S = integral dS
where dS is the infintesimal surface area element. You can then continue from there.
- Julian
| Posted on Wednesday, 28 July, 2004 - 04:53 pm: |
Julian,
Thanks! I almost understand it now, although I do need to give it some serious thought.
Tristan.
| Posted on Friday, 30 July, 2004 - 08:37 pm: |
PROBLEM!
Why does the original method work for finding the volume of a sphere (and the area of a circle), but not for the SA of a sphere? I'm confused!
| Posted on Saturday, 31 July, 2004 - 12:01 pm: |
Tristan, this is exactly what julian is trying to explain to you...
The question you should be asking is why does the method they gave you can't work for the volume
Yatir
| Posted on Monday, 02 August, 2004 - 10:22 am: |
Tristan, I'm sorry for the late reply - I can only access the internet in the weekdays.
Regards your question: you have to be *very* careful when dealing with infinite quantities, because they often behave very unpredictably.
In particular, if we have an infinite series of terms, it is often the case that the order in which we add them up changes the final value of the sum.
As an example, the sum 1-1/2+1/3-1/4+... converges on the value of log(2). But if we add it up using the order 1+1/3-1/2+1/5+1/7-1/4+... then this converges on the value of (3/2)log(2).
Both series are exhaustive in the sense that they contain the same terms.
The point I am trying to make to you is that you *have no guarantee* that the integral you have described will add up to the surface area of a sphere. I agree that it seems feasible that it *could*, but when adding an infinite number of infintesimal terms we have to be more rigourous.
As I've shown above on a slightly more simple example, the integral does *not* converge on the concept that we require.
In the case of your problem your thought process should always be focussed on what you're trying to integrate. (the below uses concepts beyond A-Level)
1. We're trying to find the surface area, S. So we want to integrate dS.
2. What is dS? Well, because we're considering a volume of revolution, we will use symmetry. It will be a small bit of the arc length of the function (in this case a circle) - call this ds, and we want to rotate this through a small bit of angle - call this dq.
So dS=dsdq.
3. So,
4. Now, what's ds? (remember we're only concerned with 2d things now, which is nice for us, because that's much simpler!)
ds2 =dx2 +dy2 by pythagoras' theorem.
So, ds = (1+(dy/dx)2 )1/2 dx
5. So,
with y being the equation of a circle.
(If we were going to use multiple integration, as above, there is a much simpler way to work out the surface area of a sphere!)
I hope this hasn't been too complicated. You'll find with a lot of mathematics at A-Level that the reason you're having a hard time understanding niggling details is because the explainations often require some higher level maths!
- Julian
| Posted on Monday, 02 August, 2004 - 09:49 pm: |
Thanks again! Sorry for the late reply also...
Thanks!
Tristan.