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Ring integral qustion

  1. May 2, 2009 #1
  2. jcsd
  3. May 2, 2009 #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]
     
    Last edited: May 2, 2009
  4. May 2, 2009 #3
    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
    ??
     
    Last edited: May 2, 2009
  5. May 2, 2009 #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.
     
  6. May 2, 2009 #5
    what words i need to search
    in google
    in order to understand your integral
     
  7. May 2, 2009 #6
    specially this ring integral stuff
     
  8. May 2, 2009 #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.
     
  9. May 2, 2009 #8
    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?
     
    Last edited: May 2, 2009
  10. May 2, 2009 #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?
     
    Last edited: May 2, 2009
  11. May 2, 2009 #10
    its a formula of a circle
    what are he meaning of the ranges you presented??(how it calculates the AREA)
     
  12. May 2, 2009 #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?
     
  13. May 3, 2009 #12
    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?
     
  14. May 3, 2009 #13
    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?
     
  15. May 3, 2009 #14
    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."
     
  16. May 3, 2009 #15

    Cyosis

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    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.
     
  17. May 4, 2009 #16
    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
    ??
     
  18. May 4, 2009 #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:

     
  19. May 4, 2009 #18
    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
     
    Last edited by a moderator: May 4, 2017
  20. May 4, 2009 #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.
     
  21. May 4, 2009 #20
    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
    ??
     
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