Find the area of the region enclosed by one loop of the curv

Solution

here i show you for y = 4sn(3) you similarly proceed for y = 4sin(7)

r = 4 sin(3)
first, in order to determine the extent of one loop, solve:
4 sin(3) = 0
sin(3) = 0 (3) = 0
= 0 and (3) = = (/3)
thus 0 and (/3) are two consecutive zeros of your function,
which moreover is positive over the interval [0, /3] (since sinx is positive over [0, ]) (this statement is crucial, because any area must be positive);
thus the required area is:

A = [0 to (/3)] 4sin(3) d
let\'s find the antiderivative, first:

4sin(3) d = 4 sin(3) d = 4(1/3)[- cos(3)] + c = (- 4/3) cos(3) + c

then, evaluating the definite integral from 0 to (/3), you get:

{(- 4/3) cos[3(/3)]} - {(- 4/3) cos[3(0)]} =

(- 4/3) cos() + (4/3) cos(0) =

(- 4/3) (-1) + (4/3) (1) =

(4/3) + (4/3) = 8/3

the required area is (8/3)