A circle has a radius of 14 cm. Another circle has a radius of 7 cm. The centre of the second circle lies on the circumference of the first. Find The common area for both circles.
(P.S The answer might not be elegant)
Calculus solution using polar coordinates
Origin at center of small circle.
Small circle: r=7
Large circle: r=2Rcos(t) where R is radius of big circle, t is angle from origin to a point on the circle (r,t)
or, r=28 cos(t)
Consider 2 segments. For the first segment, imagine rotating a radius from origin, of length 7 from t=0 up to where the 2 circles intersect. The second segment is from that angle up to pi/2. Then double the segments.
The 2 circles intersect when r=7=28 cos(t). ie cos(t)=1/4. or when t=arccos(1/4). Same as arcsin(sqrt(15)/4).
First segment (doubled): 2*integral from 0 to arccos(1/4) of (49/2) dt which is= to 49 arccos(1/4)
Second segment (doubled): is double integral of r dr dt ; the limits on the angle t are from arccos(1/4) up to pi/2. The limits on r are from 0 up to the equation for the big circle. so:
2* integral from arccos(1/4) to pi/2 integral from 0 to 28cos(t) of r dr dt. The r dr, when integrated become r^2/2. Since the upper limit of r is a function of t, we get a slightly more complicated function of t to integrate. so:
2* integral from arccos(1/4) to pi/2 [cos^2(t)]/2 dt
integral of cosine squared of t is .5*(t + sin(t)*cos(t))
Exact answer is:
49 arccos(1/4) + 2*(14^2)[pi/2  arccos(1/4)]  2*(14^2)*sqrt(15)/16
= 68.7502555446012 the same as others found.

Posted by Larry
on 20040609 01:40:09 