Find area of region that lies inside r1sin theta and outside

Solution

well here is solution You can solve this by solving double integral with the curves as limits in polar coordinates. The limits of integration for the radial coordinate are : 1 = r = 1 - sin(?) There is some region, in which inequality above cannot be satisfied, because the first curve, which is not really a circle, lies inside the circle of radius 1. To exclude this region you need to restrict the angle: 1 - sin(?) = 1 <=> sin(?) = 1 <=> (1/6)·p = ? = (5/6)·p We need to to integrate over the whole range of angles except region above, i.e. 0 = ? = (1/6)·p (5/6)·p = ? = 2·p Due to periodicity of the angles, (i.e. ?+2·p = ? ), you can rewrite this as one continuous region: (5/6)·p = ? = (13/6)·p With dA = dxdy = r drd?, you find the integral for the area as: A = ? dA = ? (5/6)·p?(13/6)·p ? (1/2)?(1-sin(?)) r drd? because ? r dr = (1/2)·r² A = ? (5/6)·p?(13/6)·p (1/2)·[ (1-sin(?))² - (1/2)² ]d? = ? (5/6)·p?(13/6)·p (1/2)·[ 1 - 2·sin(?) + sin²(?)) - (1/4) ] d? = ? (5/6)·p?(13/6)·p [ (3/8) - sin(?) + (1/2)·sin²(?))) ] d? because ? - sin(?) d? = cos(?) ? sin²(?) d? = (1/2)·(? - cos(?)·sin(?)) => ? (3/8) - sin(?) + (1/2)·sin²(?))) ] d? = (3/8)·? + cos(?) + (1/4)·(? - cos(?)·sin(?)) = (5/8)·? + cos(?) - (1/4)·cos(?)·sin(?) A = (5/8)·(13/6)·p + cos((13/6)·p) - (1/4)·cos((13/6)·p)·sin((13/6)·p) - (5/8)·(5/6)·p - cos((5/6)·p) + (1/4)·cos((5/6)·p)·sin((5/6)·p) = (65/48)·p + v(3/4) - (1/4)·v(3/4)·(1/2) - (25/48)·p - (- v(3/4) + (1/4)·(-v(3/4))·(1/2) = (40/48)·p + v(3/4)·(1 - (1/8) + 1 - (1/8)) = (5/6)·p + (7/8)·v3 ˜ 4.1335