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??
#36
transgalactic
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i divide the ring into small squeres0 dA=dxdy
how to do double integral using that
Transgalactic,
Some of the questions you have asked in this thread lead me to believe that you are not very knowledgeable about iterated integrals, either in cartesian coordinates or in polar coordinates. For one thing, you seem to have a problem understanding limits of integration if they are not constants.
Here is an iterated integral that represents the area between the curves y = x and y = x1/2.
[tex]\int_{y = 0}^{1} \int_{x = y^2}^{y} 1 dx dy[/tex]
The same region can be represented with the opposite order of integration.
[tex]\int_{x = 0}^{1} \int_{y = \sqrt{x}}^{x} 1 dy dx[/tex]
In the limits of integration, I have added "x = ..." or "y = ..." to give you more information that a limit of integration is from one x value to another or from one y value to another. I have also added a "1" as the integrand, since you had some problem understanding what the integrand was when there didn't appear to be one.
You also don't seem to understand what dA is in iterated polar integrals - dA = r dr dtheta, not dr dtheta.