Find the area enclosed by the polar curve: r = 90.40 On the interval o = = 7 and the straight line Segment between its ends Area : A= \"/71 To determine the area of the line segment between its enos we haveto find limits If 0 = 0 then r= 9 0.4 (0) = 9 IF D = then r= 9.0.4(* ) = 9.52926.. A = 81 0/ + In polar coordinates : X = rcoso A = 81 [20. 80 71/7 Y= rsino Lo.8 Jo At (r, 0) = ( 9, 0 ) we have 260.8) 0.8 ( /7 )_ 0 .8 ( 0) *= 9 cos ( 0)= 9 y = 9 sin (0 )= 0 A= 50.625 [0.12107 . . .] Thus the point is (9, 0) A = 6.129 At (r,)= (9.5292, 7 ) we have Area of triangle is given by *= 9.529 2cos (#)= 9.5292 y= 9.52qasim (7 )= 0.02375 Area A= 2 x (X-coordinate at 0 = 0) x ( yat Q=7) Thus the point is (9.5292, 0.023) Area A = 2 ( 9 ) x (0.02375...) Area A= 0.106 Areaof Polar Coordinate : 6. 129 Area of the enclosed polar curve is Area of triangle: 0.106 Area enclosed= Area of Polar coordinate- Area of triangle Area enclosed = 6.129- 0.106 Area enclosed= 6.0223Find the area enclosed by the polar curve 7' : 980.43 on the interval 0 g 6' g % and the straight line segment between its ends. Area =