- #1

- 1,395

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, what I need is to find the area of the ring with is dr and the radius is r. So pi*(r+dr)^2 - pi*(r)^2 \mathsf{..} So why the dA is that simple expression??f

- #1

- 1,395

- 0

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

Homework Helper

- 1,495

- 5

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]

Last edited:

- #3

- 1,395

- 0

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

??

Last edited:

- #4

Homework Helper

- 1,495

- 5

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

- 1,395

- 0

what words i need to search

in google

in order to understand your integral

in google

in order to understand your integral

- #6

- 1,395

- 0

specially this ring integral stuff

- #7

Homework Helper

- 1,495

- 5

- #8

- 1,395

- 0

i know polar coordinated

and i know how to calculate a double integral

i don't know how to get the Cartesian formula from the drawing?

and i know how to calculate a double integral

i don't know how to get the Cartesian formula from the drawing?

Last edited:

- #9

Homework Helper

- 1,495

- 5

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?

Last edited:

- #10

- 1,395

- 0

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

- #11

Homework Helper

- 1,495

- 5

- #12

- 1,395

- 0

the big rings.

this is the first integral

the second one sums the parts of each ring

correct?

- #13

- 1,395

- 0

i will call "ds"

so the whole small ring is

[tex]

\int_{0}^{dr}\int_{0}^{2\pi}ds

[/tex]

is it correct?

- #14

- 392

- 0

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

Homework Helper

- 1,495

- 5

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

- 1,395

- 0

what is y??

i can't see what each integral represents

??

what those roots represent?

the from r to -r

is it on x axes or y axes

??

- #17

Homework Helper

- 1,495

- 5

Answer this:

- #18

- 1,395

- 0

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 can't say

"go from this formula to the other one"

it doesn't 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 can't say

"go from this formula to the other one"

it doesn't makes sense.

we need to give the integral an actual values

Last edited by a moderator:

- #19

Homework Helper

- 1,495

- 5

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

- 1,395

- 0

"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 can't see what from r to -r does

??

- #21

Homework Helper

- 1,495

- 5

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

- 1,395

- 0

and similarly i will find the bottom hemisphere and sum them both.

so when the double integral gets involved

??

- #23

Homework Helper

- 1,495

- 5

[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

- 1,395

- 0

"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

Homework Helper

- 1,495

- 5

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

- 1,395

- 0

i can't understand the meaning of the double integral

i can't say "we calculate function f(x) on some interval."

i don't 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

Homework Helper

- 1,495

- 5

- #28

- 1,395

- 0

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

Homework Helper

- 1,495

- 5

- #30

- 1,395

- 0

the big problem is with the intervals of the double integral

i can't 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

Homework Helper

- 1,495

- 5

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]

[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]

Last edited:

- #32

- 1,395

- 0

your latex code failed

- #33

Homework Helper

- 1,495

- 5

Yes I know it should be fixed now.

- #34

- 1,395

- 0

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

??

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

??

Last edited:

- #35

Homework Helper

- 1,495

- 5

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?

Share:

- Replies
- 6

- Views
- 223

- Replies
- 7

- Views
- 545

- Replies
- 1

- Views
- 328

- Replies
- 7

- Views
- 569

- Replies
- 10

- Views
- 1K

- Replies
- 5

- Views
- 709

- Replies
- 5

- Views
- 396

- Replies
- 8

- Views
- 508

- Replies
- 3

- Views
- 140

- Replies
- 9

- Views
- 227