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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 ...
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)
$\begingroup$ 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..
$\begingroup$ 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 $\sin^{-1}\frac{|AB|}{2r}$ $\endgroup$ –
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.
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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
We can also split the triangle into three smaller triangles using $\frac{1}{2} ab \sin C$.Therefore $\Delta ABC$ equals:
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Get the equation of a circle through the points $(1,1), (2,4), (5,3) $. I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?