You have to integrate 1/2 r^2 from theta = 0 to 2 pi.
1/2 r^2 = 1/2 [25 + 20 sin(theta)
+ 4 sin^2(theta)]
The sin(theta) term doesn't contribute to the integral and sin^2(theta) can be replaced by 1/2 [because it would yield the same contribution as cos^2(theta) and together they are equal to 1].
So, the area is given by:
pi[25 + 2 ] = 27 pi
Determine the area of the region enclosed by the limaçon
r=5+2sin(theta)
1 answer
1 answer