- #1

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the with is dr and the radius is r

so pi*(r+dr)^2 - pi*(r)^2

so why the dA is that simple expression??

http://www.freeimagehosting.net/image.php?cce537487f.gif

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- Thread starter transgalactic
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- #1

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the with is dr and the radius is r

so pi*(r+dr)^2 - pi*(r)^2

so why the dA is that simple expression??

http://www.freeimagehosting.net/image.php?cce537487f.gif

- #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]

[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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i know that double integral meaning signsWhat 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]

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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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

in google

in order to understand your integral

- #6

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specially this ring integral stuff

- #7

Cyosis

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- #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?

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?

(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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what are he meaning of the ranges you presented??(how it calculates the AREA)

- #11

Cyosis

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- #12

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the big rings.

this is the first integral

the second one sums the parts of each ring

correct?

- #13

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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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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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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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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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Answer this:

- #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

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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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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"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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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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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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[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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"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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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]

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