What is the area under the polar curve #f(theta) = 3theta^2+thetasin(4theta-(5pi)/12 ) +cos(2theta-(pi)/3)# over #[pi/8,pi/6]#?

1 Answer
Jun 21, 2016

0.455

Explanation:

Area under the polar curve for any function #f(theta)# is given by #int_a^b 1/2 r^2 d(theta)#

The function #f(theta)# represents the r or radius as the curve moves from #pi/8# to #pi/6#.

So all we have to do is plugin whole #f(theta)# for r.

That would give us,
#int_(pi/8)^(pi/6) (3(theta)^2 +theta*sin(4theta-((5pi)/12))+cos(2theta-pi/3))^2d(theta)#

Doing this integral by hand would be little tedious so you can just plug integral in your calculator and get the answer.

The answer would be 0.455