Question #81a78

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Answer 1 · 2017-05-18T08:53:17

$1/3$

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:
$(1/6--(1/6))|_0^(pi/3)$
$=1/3$