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#