How can iterated integrals be used to find the area of a ring?

[Cyosis]

I have no idea what a scanning interval is but it works exactly the same as $\int_0^1 dx$. Here goes:$$\begin{align*}<br /> \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy & = \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} 1 dy<br /> \\<br /> & =\left[ y \right]_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}<br /> \\<br /> & =\sqrt{r^2-x^2}-(-\sqrt{r^2-x^2})<br /> \\<br /> & =2 \sqrt{r^2-x^2}<br /> \end{align*}$$

 

[transgalactic]

your latex code failed

 

[Cyosis]

Yes I know it should be fixed now.

 

[transgalactic]

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
??

 

[Cyosis]

My reasoning is explained in 9,11,17,19 and 21. A ring does not have an area so you can't calculate the area of a ring. We have gone so far off topic by now that we're running into a dead end. You ask a question I answer your question only to find out that the question you asked wasn't really the question you wanted answered.

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?

 

[transgalactic]

i divide the ring into small squeres0 dA=dxdy
how to do double integral using that

 

[Mark44]

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 = x^1/2.

$$\int_{y = 0}^{1} \int_{x = y^2}^{y} 1 dx dy$$
The same region can be represented with the opposite order of integration.
$$\int_{x = 0}^{1} \int_{y = \sqrt{x}}^{x} 1 dy dx$$

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.