Question

Question #81a78

Answer 1

`1/3`

Explanation:

To find an interval for a petal
Solve for `r=0`.

`0=sqrt(sin(3 theta)), theta= 0, pi/3`

The area of a polar curve Is determined by:

`int_a^b 1/2 r^2 d theta`
`1/2 int (sqrt(sin 3theta))^2 d theta`
`1/2 int sin(3 theta) d theta`

Substitute `u=3 theta`

`=1/6 int_0^(pi/3) 3sin(3 theta) d theta`
`=1/6 int_0^(pi/3) sin(u) du = (-cos(u))/6=(-cos(3 theta))/6=-cos(3 (pi/3))/6=1/6`

Also substitue `theta=0`.
`-cos(3(0))/6=-1/6`

Now compute the definite integral:
`(1/6--(1/6))|_0^(pi/3)`
`=1/3`