Find the area of the intersection of the circle r=sin theta and r= 1/(3)^(1/2) cos theta

Answer 1

sinθ = cosθ/√3; tanθ = 1/√3; θ = π/6; Pole also common point (for diffeerent θ)

Area S = 1/2∫₀^π/6sin²θdθ + 1/2∫_π/6^π/21/3cos²θdθ = 1/4(θ - sin2θ/2)₀^π/6 + 1/12(θ + sin2θ/2)_π/6^π/2 =

= 1/4(π/6 - √3/4) + 1/12(π/2) - 1/12(π/6 + √3/4) = 5π/72 - √3/12 = 0.0738

Answer 2

r=sinT

r^2=rsinT

x^2+y^2 = y

x^2 +(y-1/2)^2 = 1/4

is a circle with radius 1/2, center (0, 1/2)

r = (1/sqr3)cosT =x/rsqr3

r^2 =x/sqr3

x^2+y^2 =x/sqr3

x^2 -x/sqr3 +y^2= 0

(x-1/2sqr3)^2 + y^2 = 1/12

is a smaller circle with center = (1/2sqr3, 0), radius 1/sqr2

Graph the 2 circles

They overlap in a small sliver in quadrant 1. Overlap area is very small, less than 0.5.

intersection points are (1/sqr2, 1/6) and ((sqr3)/4, 1/4)

the x coordinates of the intersection points are integral evaluation limits if integrated with respect to x

integrate with respect to y use the y coordinates as the limits

integrate (sqr(x/sqr3-x^2) -1/2 +sqr(.25- x^2))dx from (sqr3)/4 to (sqr3)/2