A 2414 DHW 9 (10.3 - 10.4) - $22 - X Bb Announcements - CS 1136.106 - x C Find The Area Of The Region Tha x + V X > C a webassign.net/web/Student/Assignment-Responses/submit?dep=29024201&tags=autosave#question3798389_27 20. [-/4 Points] DETAILS SCALCET8 10.4.AE.001. MY NOTES PRACTICE ANOTHER EXAMPLE 1 Find the area enclosed by one loop of the four-leafed rose r = 5 cos(20). SOLUTION The curve r = 5 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 = -It/4 to 0 = 1/4. Therefore this formula gives A = 1,2 de de 1 / 4 T/ 4 = 25 de Video Example () TT / 4 = 25 - (1 + cos(40)) de TE/ 4 = 25 Need Help? Read It 21. [1/1 Points] DETAILS PREVIOUS ANSWERS SCALCET8 10.4.019. MY NOTES PRACTICE ANOTHER Type here to search O 83.F Sunny ~ 9 /4x ENG 4:33 PM US 3/28/2022(- ) C i webassign.net/web/StudenUAssignmentResponses/submit?dep:29024201Mags:autosave#question3798389_27 IE F i I] . E 3V3 +7: A 4 i Nuanelp? ii 24. [13 Points] SCALCET8 10.4.038. Find all points of intersection of the given curves. (Assume 0 s 9 C webassign.net/web/Student/Assignment-Responses/submit?dep=29024201&tags=autosave#question3798389_27 25. [-/7 Points] DETAILS SCALCET8 10.4.AE.002. MY NOTES PRACTICE ANOTHER EXAMPLE 2 Find the area of the region that lies inside the circle r = 15 sin(0) and outside the cardioid r = 5 + 5 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 15 sin(0) = which gives sin(0) = , so 0 = 1/6, 0 = 51/6. The desired area can be found by subtracting the area inside the cardioid between 0 = 1/6, 5x/6 from the area inside the circle from 7/6 to 5n/6. Thus 5Tt/6 A = 2 16 (15 sin( 8) ) 2 de - Z Ju/6 (5 + 5 sin ( 0) )2 de . Since the region is symmetric about the vertical axis 0 = 1/2, we can write Video Example () A = 2 2 2 225 sin 2(0 ) do - 25 1:/ 2 (1 + 2 sin(@) + sin2(@)) de [200 sin (0) - 25- de It/ 2 - 100 cos(20) - |sin(8) ) de = JT/6 [because sin2(0) = =(1 - cos(20)) TE /2 = T/6 Need Help? Read It Type here to search O 83.F Sunny ~ 3 6 4X ENG 4:33 PM US 3/28/2022