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1. [-/0.15 Points] DETAILS SCALCET8 10.4.AE.001. 0/100 Submissions Used MY NOTES ASK YOUR TEACHER EXAMPLE 1 Find the area enclosed by one loop of the four-leafed rose r = 3 cos(20). SOLUTION The curve r = 3 cos(20) is sketched in the figure to the left. Notice from the figure that the region enclosed by the right loop is swept out by a ray that rotates from 0 = -7/4 to 0 = 1/4. Therefore this formula gives " 1, 2 de -7/4 2 de 0= -# de Video Example () OP ( ( Ob )50) + 1 / 7 16 - 173 / 4 N/O Jo Need Help? Read It 2. [-/0.15 Points] DETAILS SCALCET8 10.4.AE.002. 0/100 Submissions Used MY NOTES ASK YOUR TEACHER EXAMPLE 2 Find the area of the region that lies inside the circle r = 12 sin(0) and outside the cardioid r = 4 + 4 sin(0). SOLUTION The cardioid (in blue) and the circle (in red) are sketched in the figure. The value of a and b in this formula are determined by finding the points of intersection of the two curves. They intersect when 12 sin(0) = , which gives sin(0) = , so 0 = 1/6, 0 = 5n/6. The desired area can be found by subtracting the area inside the cardioid between 0 = 1/6, 57/6 from the area inside the circle from 7/6 to 5nt/6. Thus A = 2 ( 12 sin( 0) ) 2 20 - 2 ( 4 + 4 sin( 0 ) ) 2 do . Since the region is symmetric about the vertical axis 0 = 1/2, we can write A = 2 2 144 sin ? ( 0 ) de - 2 Ja/6 16 ( 1 + 2 sin ( 8 ) + sin ? ( 8) ) de Video Example () [128 sin (0) - 16 - de -64 cos ( 20 ) - sin ( 8) ) de [because sin2(0) = = (1 - cos(20))