$\begingroup$ What would you set the limits if you need to calculate the area of an infinitesimal ring in cartesian coordinates i.e. $\int dx \int dy $.. where you only want to integrate on the infinitesimal ring..

Aug 28, 2012 · does not represent an area because the integration is not bounded (also, a constant is missing on the RHS). An area should be for something with bounds (limits). However, the formula you mentioned is used in what is known as Onion proof for area of the circle (please do a find on 'onion'). This proof divides the circle into rings as explained ...

Hint: The circle has radius $1$. By symmetry (join the centre of the circle to the other corners of the square) the shaded part of the circle is one-quarter of the whole circle.

Dec 8, 2015 · $\begingroup$ Those data are not enough to find the area of the triangle. If you have a certain cord of the circle, then the opposite angle will be the same no matter where on the circle the third corner of the triangle is -- namely $\sin^{-1}\frac{|AB|}{2r}$ $\endgroup$ –

Hence we can find area of between chord AB and BC by multiplying the area of a circle with 1/6 i.e.πr^2/6 (because 60 degree/360 degree=1/6) We can subtract the area of triangle ( √3/4 * side^2) from it to find the area of the curved part. In all there are four such congruent parts. Then we can again add the area of the triangle (twice)

Mar 5, 2013 · The radius (half the length of the diameter, which is the length of a side of the cube) is 1 and the area of a circle is $\pi r^2$, so the area of the four half-circles is $2 ( \pi * (1)^2)$, which is $2\pi$. The square can only hold an area of $4$ and all its space is used, so the excess area of the circles' area must be contained in overlap.

$\begingroup$ I think it would be more clearly an Answer if you affirmed that the OP's approach is correct and leads to the familiar area of a circle. I'm sure you mean to say so with your brief note, but many Readers may jump to your conclusion without seeing …

Your question quotes a text on Archimedes: 'Archimedes used the method of exhaustion as a way to compute the area inside a circle by filling the circle with a polygon of a greater area and greater number of sides.' Regardless of how he originally did it, here is how it can be done now for the unit circle using more modern techniques.

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The area of the triangle is not how you represent it, you've given the points on the circle, i.e. $(r,\theta)$ and $(r,\theta+d\theta)$, so you are not integrating a differential area. Share Cite