Ring integral qustion

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  • #2
Cyosis
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What you're calculating is the area of the disk. In your example they add up all the rings to get the area of the disk. Perhaps it's more insightful to write it like this, using cylindrical coordinates.

[tex]A=\int_S dA=\int_0^R \int_0^{2\pi}r d\theta dr=\int_0^R 2 \pi r dr[/tex]
 
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  • #3
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What you're calculating is the area of the disk. In your example they add up all the rings to get the area of the disk. Perhaps it's more insight full to write it like this, using cylindrical coordinates.

[tex]A=\int_S dA=\int_0^R \int_0^{2\pi}r d\theta dr=\int_0^R 2 \pi r dr[/tex]
i know that double integral meaning signs

what is the maning of this part
[tex]r d\theta dr[/tex]
r d\theta is the length of a small arc
what is the meaning of adding dr at the end
??
 
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  • #4
Cyosis
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Have you ever had calculus, Cartesian coordinates, cylindrical coordinates and spherical coordinates? Do you know what a Jacobian is? If not my explanation probably won't make a lot of sense.

The area of a disk in Cartesian coordinates is: [tex]A=\int_{-r}^r \int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}} dxdy =\int_0^R \int_0^{2\pi}r d\theta dr[/tex]

A coordinate transformation from Cartesian to cylindrical coordinates was performed here with a Jacobian r.
 
  • #5
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what words i need to search
in google
in order to understand your integral
 
  • #6
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specially this ring integral stuff
 
  • #7
Cyosis
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The integral is nothing special, just basic polar coordinates. If these are unfamiliar to you you're going to have serious issues with electromagnetism. I really suggest you pick up a calculus book and teach yourself integration, curvilinear coordinates and coordinate transformations. I am really not sure how I can explain this further to you without using any calculus.
 
  • #8
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i know polar coordinated
and i know how to calculate a double integral
i dont know how to get the Cartesian formula from the drawing?
 
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  • #9
Cyosis
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Cartesian coordinates:
(x,y,z)

Polar coordinates:
[tex]
\begin{align*}
& x=r\cos \theta
\\
& y=r\sin \theta
\end{align*}
[/tex]

Circle:
[tex]x^2+y^2=r^2[/tex]

Are these things familiar to you? Draw [itex]y=\sqrt{r^2-x^2}[/itex] what kind of line does it describe?
 
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  • #10
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its a formula of a circle
what are he meaning of the ranges you presented??(how it calculates the AREA)
 
  • #11
Cyosis
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It is a part of a circle, not a whole circle. What part of a circle does the line [itex]y=\sqrt{r^2-x^2}[/itex] represent and what does [itex]y=-\sqrt{r^2-x^2}[/itex] represent? Draw them in the same drawing. What is the resultant figure, what is its radius?
 
  • #12
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the integral is the sum of all the rings combining
the big rings.
this is the first integral
the second one sums the parts of each ring
correct?
 
  • #13
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a small part of the ring
i will call "ds"
so the whole small ring is
[tex]
\int_{0}^{dr}\int_{0}^{2\pi}ds
[/tex]
is it correct?
 
  • #14
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transgalactic, going back to your original question, an intuitive reason that pi*(r+dr)^2 - pi*(r)^2 is approximately 2*pi*r*dr is that (r+dr)^2 = r^2 +2*r*dr + (dr)^2, so that

pi*(r+dr)^2 - pi*(r)^2 = pi*[r^2 +2*r*dr + (dr)^2] - pi*(r)^2 = 2*pi*r*dr + (dr)^2

When dr is small, (dr)^2 is really small so it is ignored. That is, keep only the first power of dr, and discard higher powers.

Here's another explanation. Let R be the larger radius and r the smaller radius, so that the area is pi * (R^2 - r^2). Now (ignoring the pi), R^2-r^2=(R+r)*(R-r)=2*( (R+r)/2 )*(R-r).

Observe that (R+r)/2 is the average radius, the average of R and r. And R-r is the change in radius, delta r, or intuitively dr in the infinitesimal limit. Again, intuitively, in the infinitesimal limit, R and r are close, so the average radius (R+r)/2 is simply r.

For Google, don't use "ring." Use "cylindrical shells."
 
  • #15
Cyosis
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a small part of the ring
i will call "ds"
so the whole small ring is
[tex]
\int_{0}^{dr}\int_{0}^{2\pi}ds
[/tex]
is it correct?

This is not correct. What is correct is that ds is a small piece of a ring, but ds doesn't run from 0 to 2 pi. Evaluating your integral would give us 2 pi dr, which isn't equal to the area of a disk. Can you see why [itex]ds=r d\theta[/itex] and that you can then take the limits to run from 0 to 2 pi (full circle)? Secondly you can't put a differential in your integration limits. The second integral sums all the rings that lay in the interval 0 to R.

However we have strayed away from your original question quite a bit now. If you still want to know how I calculated the area of a disk using Cartesian coordinates feel free to ask. That said Billy Bob has done a great job explaining your original question so I suggest you read his post very carefully and try to understand it.
 
  • #16
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how you get [tex]\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}} [/tex]
what is y??
i cant see what each integral represents
??

what those roots represent?

the from r to -r
is it on x axes or y axes
??
 
  • #17
Cyosis
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To understand what I am doing you should start answering my questions. It is pointless to just jump to and fro all the time.

Answer this:

It is a part of a circle, not a whole circle. What part of a circle does the line [itex]y=\sqrt{r^2-x^2}[/itex] represent and what does [itex]y=-\sqrt{r^2-x^2}[/itex] represent? Draw them in the same drawing. What is the resultant figure, what is its radius?
 
  • #18
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http://img87.imageshack.us/img87/5584/55857037.gif [Broken]

each formula represents a different hemisphere of a cicrle with radius r.

but its not a range is two formulas
i cant say
"go from this formula to the other one"
it doesnt makes sense.
we need to give the integral an actual values
 
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  • #19
Cyosis
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You're correct both are different hemispheres with radius r. To find the area underneath one hemisphere you have to integrate it from -r to r.

Integrate: [tex]\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy[/tex]. You will see that this gives you a function that describes a full circle. The next step is to integrate both hemispheres from -r to r to get the area underneath both hemispheres. Adding them together gives you the are of a circle with radius r.
 
  • #20
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as i see it
"To find the area underneath one hemisphere you have to integrate it from -r to r."
i can do the same thing with
[tex]
\int_{0}^{\pi}\sqrt{r^2-x^2}
[/tex]
i cant see what from r to -r does
??
 
  • #21
Cyosis
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Have you actually calculated the integral you just posted yourself? If not I suggest you do, seeing is believing. I can tell you right now though that you won't get the area of a disk out of that integral.

If you want to calculate the area under some graph between a and b you take your integration limits to be a and b. In our case we want to know the area under the graph of a semicircle so -r<x<r.

I hope this picture makes it clear.

http://www.phys.uu.nl/~0362417/phfinal.jpg
 
  • #22
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ahh i forgot that its not a trigonometric function
and similarly i will find the bottom hemisphere and sum them both.

so when the double integral gets involved
??
 
  • #23
Cyosis
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Yes that's correct the second integral (first in the order of integration) is just the function 1 that is bound by a curve that represents a circle.

[tex]\int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy dx=\int_{-r}^{r} 2 \sqrt{r^2-x^2} dx=\pi r^2[/tex]
 
  • #24
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i cant understand your words
"Yes that's correct the second integral (first in the order of integration) is just the function 1 that is bound by a curve that represents a circle."

second or first of what

what function is bounded
by what?
 
  • #25
Cyosis
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That's because you make me guess all the time. We started out with a double integral of which you didn't understand the inner integral and its integration boundaries. After a while you understood that the boundaries represent two semicircles. After that you didn't understand the outer integral from -r to r. After a while you understood this too (I think). So by now I have explained the entire double integral, but now it seems you want to know when the double integral comes into play? Did we not start with that altogether explaining step by step what each term in the double integral means?

So the big question is what part of the following integral do you not understand? Make sure you formulate your question clear and precise or we are getting nowhere.

[tex]
\int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy dx=\int_{-r}^{r} 2 \sqrt{r^2-x^2} dx=\pi r^2
[/tex]
 
  • #26
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i understand perfectly the right side of the equation.
i cant understand the meaning of the double integral
i cant say "we calculate function f(x) on some interval."
i dont have a function in the double integral
i have two functions acting as intervals (functions does tell me a real number)
i am used to have a solid number interval like from r to -r

??
 
  • #27
Cyosis
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There is a function in the double integral it is the function f(y)=1. Can you calculate [itex]\int_0^1 dx[/itex] for me?
 
  • #28
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i solved the double integral
its equal to the one integral formula.

but how you got it??
how to write double integral for the ring
?
[tex]
\int_0^1 dx=1
[/tex]
 
  • #29
Cyosis
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How I got the formula was explained in post 9,11,17,19 and 21. You calculated the integral correctly yes. Why didn't you have any problems with this one regarding that there is "no function" in the integral?
 
  • #30
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because i know that its 1 turning to x
the big problem is with the intervals of the double integral
i cant see the "scanning process" like in a single integral
there are two simultaneous scannings going on
one from -r to r

and the other one has no regions no scanning interval
 
  • #31
Cyosis
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I have no idea what a scanning interval is but it works exactly the same as [itex]\int_0^1 dx[/itex]. Here goes:


[tex]
\begin{align*}
\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy & = \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} 1 dy
\\
& =\left[ y \right]_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}
\\
& =\sqrt{r^2-x^2}-(-\sqrt{r^2-x^2})
\\
& =2 \sqrt{r^2-x^2}
\end{align*}
[/tex]
 
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  • #32
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your latex code failed
 
  • #33
Cyosis
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Yes I know it should be fixed now.
 
  • #34
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i did that already
i know that it equals the single integral.

i want to know how you thought of it.
and how to think of a double integral to calculate the area of the ring
??
 
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  • #35
Cyosis
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My reasoning is explained in 9,11,17,19 and 21. A ring does not have an area so you can't calculate the area of a ring. We have gone so far off topic by now that we're running into a dead end. You ask a question I answer your question only to find out that the question you asked wasn't really the question you wanted answered.

So regarding your original question, which was to integrate a function (1/r^2) over the area of a disk. Have you been able to solve this or are you still stuck and if you're still stuck how far have you gotten?
 

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