Find the area of the region that lies inside both curves. PLEASE HELP!

Question

Find the area of the region that lies inside both curves.
r = √3 cos θ, r = sin θ

 

Thank you in advance!

Answer 1

for 0 < θ < π/2  ,  these curves cross at    θ =arctan( 3^1/2 ) =  π/3  

The cosine curve is the upper one, so the desired area  is

 

  A = integral from 0 to π/3  of [ 3^1/2 cos(θ) - sin(θ)  ]

 

The anti-derivative of the expression in the square brackets is

  

     [ 3^1/2 sin(θ) + cos(θ) ]   

 

  So  A =   (3^1/2 sin(π/3) +cos(π/3) ) - (3^1/2 sin(0) + cos(0)  )

 

    Remembering that π/3  radians is the same as 60 degrees, this evaluates to

 

     (3/2 + 1/2) -1  =   1

 

This is actually the ares bounded by the two curves and the y axis.   There is another

intersection of the two curves  at θ = -2π/3 

Plugging this value in for the lower limit results in A =  2 - (-3/2 - 1/2)  = 4