Set up the integral that calculates the area of the region enclosed by the circle r = 6 and the liner =3 v2 sec(0) in the right half plane. 14 2 (36 - 18 asc (0) ) do film 2 (36 - 18 CSC (0) de P 7x/4 o * / 4 (36 - 18 sec- (0)) de (36 - 18 csc (0)) de (36 - 18 sec. (0)) de (36 - 18 sec (0) de O None of the above.Question number 7. Which of the following represents the area inside the inner loop of r = 8 + 16 cos(0)? TT /3 2 T / 3 (8 + 16 cos(0)) de 47/3 (8 + 16 cos(0))2 de 2T / 3 71/6 5T / 6 (8 + 16 cos(0)) de 57/3 AT / 3 (8 + 16 cos(0)) de 47/3 O (8 + 16 cos(0))2 de 7 / 3Question number 8. Which of the following represents the area inside one petal of r = 9 sin (3 0)? 27/3 O (9 sin(3 0)) de TT/ 4 O (9 sin(3 0))2 de 7T/3 O (9 sin(3 0) )2 de . 7/3 O (9 sin(30)) de o (9 sin(30))2 deQuestion number 9. Which of the following represents the area interior to both r = 11 - 11 sin(0) and r = 11 sin(@)? 2 (Jo ?(11 - 11 sin(0)? do + (11 sin()" de 0 2 2(11 sin(0))? de + 2 (11 - 11 sin(0))? de PTT /2 (11 - 11 sin(0)) de + / 5 (11 sin(@))2 de 0 2 (11 sin(@)) de + 16 2(11 -11 sin(0) de . 7/6 1 0 2 (11 - 11 sin(@)) de + 7 (11 sin(@))? de . 7/2 1