Find the area?

Question

Find the area of the region enclosed by the polar curve r=sin(2θ) for 0≤θ≤pi/2.

 

Answer: pi/8

 

Please show all your work.

Answer 1

The area of a very thin circular sector with central angle dθ radians
is (pi*r^2)*(dθ/(2pi)) = (1/2)*(r^2)*dθ.

Add them up with an integral:

A = ∫{0,pi/2} (1/2)*(r^2) dθ

 

Substitute given r = sin(2θ):
A = ∫{0,pi/2} ((1/2)*(sin(2θ))^2) dθ

 

Substitute sin(2θ))^2 = (1/2)*(1-cos(4θ)):

 

A = ∫{0,pi/2} ((1/2)*((1/2)*(1-cos(4θ))) dθ

 

= (1/4) ∫{0,pi/2} (1-cos(4θ)) dθ

 

= (1/4) [θ - (sin(4θ))/4]{0,pi/2}

 

= (1/4) [pi/2 - (sin(4*pi/2))/4 - 0 + (sin(4*0))/4]

 

= (1/4) [pi/2 - (sin(2*pi))/4 + 0]

 

= (1/4) [pi/2 - 0]

 

= pi/8

P.S.: This is why you studied identities in Trig; so you could use them to change integrands into something you can integrate.

Answer 2

Hi Sun,

 

Areas under polar graphs are solved using a special integral for polar functions.  The function is:

A = 1/2 ∫ r² dθ

 

This integral is based on the fact that the area of a sector of a circle of radius r and angle θ is (1/2)r²θ.

 

In your case,

 

A = 1/2 ∫ sin²(2θ) dθ  (from θ=0 to θ=π/2)

 

Since sin²β = 1/2 * (1 - cos(2β) ) (one of the so-called "double-angle formulas"), this integral becomes:

 

A = 1/2 ∫ 1/2 * (1 - cos (4θ) ) dθ (from θ=0 to θ=π/2)

 

A = 1/4 ∫ 1 - cos (4θ) dθ (from θ=0 to θ=π/2)

 

A = 1/4 ( θ - (1/4) sin(4θ) ) (evaluated from θ=0 to θ=π/2)

 

A = π/8.

 

***

PS: I'm gonna throw in a bit of "math-propaganda" here and say that though I used π in the process above to solve the problem, I think τ, the ratio of a circle's circumference to its radius, is a more natural choice for that beautiful mathematical constant. τ = 2π ≈ 6.283... for more info about why, check out the "Tau Manifesto", http://tauday.com/tau-manifesto

Answer 3

    π/2                               π/2

   ∫   Sin 2θ dθ  = - Cos 2θ l      = - ( cos π - cos 0 ) = - ( -1 - 1) = 2

   0                                   0