# Ring integral qustion

## Answers and Replies

Cyosis
Homework Helper
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.

$$A=\int_S dA=\int_0^R \int_0^{2\pi}r d\theta dr=\int_0^R 2 \pi r dr$$

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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.

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

what is the maning of this part
$$r d\theta dr$$
r d\theta is the length of a small arc
what is the meaning of adding dr at the end
??

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Cyosis
Homework Helper
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: $$A=\int_{-r}^r \int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}} dxdy =\int_0^R \int_0^{2\pi}r d\theta dr$$

A coordinate transformation from Cartesian to cylindrical coordinates was performed here with a Jacobian r.

what words i need to search
in google
in order to understand your integral

specially this ring integral stuff

Cyosis
Homework Helper
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.

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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Cyosis
Homework Helper
Cartesian coordinates:
(x,y,z)

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

Circle:
$$x^2+y^2=r^2$$

Are these things familiar to you? Draw $y=\sqrt{r^2-x^2}$ what kind of line does it describe?

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

Cyosis
Homework Helper
It is a part of a circle, not a whole circle. What part of a circle does the line $y=\sqrt{r^2-x^2}$ represent and what does $y=-\sqrt{r^2-x^2}$ represent? Draw them in the same drawing. What is the resultant figure, what is its radius?

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?

a small part of the ring
i will call "ds"
so the whole small ring is
$$\int_{0}^{dr}\int_{0}^{2\pi}ds$$
is it correct?

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."

Cyosis
Homework Helper
a small part of the ring
i will call "ds"
so the whole small ring is
$$\int_{0}^{dr}\int_{0}^{2\pi}ds$$
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 $ds=r d\theta$ 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.

how you get $$\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}$$
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
??

Cyosis
Homework Helper
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 $y=\sqrt{r^2-x^2}$ represent and what does $y=-\sqrt{r^2-x^2}$ represent? Draw them in the same drawing. What is the resultant figure, what is its radius?

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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Cyosis
Homework Helper
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: $$\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy$$. 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.

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
$$\int_{0}^{\pi}\sqrt{r^2-x^2}$$
i cant see what from r to -r does
??

Cyosis
Homework Helper
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

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
??

Cyosis
Homework Helper
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.

$$\int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy dx=\int_{-r}^{r} 2 \sqrt{r^2-x^2} dx=\pi r^2$$

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?

Cyosis
Homework Helper
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.

$$\int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy dx=\int_{-r}^{r} 2 \sqrt{r^2-x^2} dx=\pi r^2$$