find the area of the region within one loop of?

Question

r(theta) = 2 cos(3theta)

thanks for your time

Answer 1

In polar coordinates area is

A = ½∫r² dθ

The area of the sector bounded by the curve r = f(θ) and the two radial lines defined by θ1 and θ2 is

½∫[θ = θ1 to θ2] f(θ)² dθ

In your problem f(θ) = 2 cos 3θ. The curve starts when θ = 0 at (2,0) and continues. It first meets (0,0) when 2 cos 3θ = 0, or θ = π/6 making a region of half a loop.

So, area of one loop =

2 × ½∫[θ = 0 to π/6] ( 2 cos 3θ )² dθ

= 4∫[θ = 0 to π/6] cos² 3θ dθ

= 4∫[θ = 0 to π/6] ½ ( 1 + cos 6θ ) dθ

= 2 { ∫[θ = 0 to π/6] dθ + ∫[θ = 0 to π/6] cos 6θ dθ }

= 2 { [θ = 0 to π/6: θ] + 0 } because the area over the range 0 to π/6 of cos 6θ cancels to 0

= π / 3

Answer 2

First of all, one loop is swept out by -pi/3 Now, to find out what expression to integrate, just remember that in the limit, a sector with polar angle d(theta) and length r(theta) has area approximately (1/2)*r(theta)*r(theta)*sin(d(theta)), which is approximately 1/2*r(theta)^2*d(theta), and that is the expression to integrate.
That is, the area is 2*int(1/2*4cos(3theta)^2,theta,0,pi/3)=4int(cos(3theta)^2,theta,0,pi/3)=4/3*Int(cos(x)^2,x,0,pi)=4/3*1/2*(pi+cos(pi)sin(pi)-0-cos(0)sin(0))=2pi/3

Answer 3

One loop will be from 0 to pi, or theta = 0 to pi/3. So to get the area integrate from 0 to pi/3.
AREA=integral[0-->pi/3]2cos(3theta)dtheta

Answer 4

integrate it