area of polar graphs help??

Question

Find the area of a common interior.

r= 3(1 + sin(θ))
r= 3(1-sin(θ))

Answer 1

By symmetry we can integrate over θ from 0 to π/2.

4∫{1/2r²}dθ

= 4∫{1/2[3(1 - sinθ)]²}dθ = 2∫{9(1 - 2sinθ + sin²θ)}dθ

= 18∫[1 - 2sinθ + (1 - cos2θ)/2]dθ

= 18∫[3/2 - 2sinθ - (cos2θ)/2]dθ

= 9∫[3 - 4sinθ - cos2θ]dθ

= 9[3θ + 4cosθ - (sin2θ)/2] | [Evaluated from 0 to π/2]

= 9[3π/2 + 0 - 0] - 9[0 + 4 - 0] = 9(3π/2 - 4) ≈ 6.4115008

Answer 2

r=5 has cartesian equation x^2+y^2=25 and r=-10sin(theta) is r^2=-10rsin(theta) or x^2+y^2=-10y, that's circle x^2+(y+5)^2=25. the two curves are circles of radius 5 and centres (0,0) and (0,-5). enable the circles intersect at A and B with their centres at C and D. From the symmetry of the parent AC=CD=advert so attitude ACD=pi/3 and attitude ACB=2pi/3. the component of the international bertween radii CA and CB is (a million/2)(5)^2(2pi/3)=25pi/3 . the component of triangle ACB = (a million/2)5^2sin(2pi/3)= 25sqrt(3)/4 and the section on one component of AB=25pi/3-25sqrt(3)/4=25(pi/3-sqrt(3)/4) so the mandatory section is double this =50(pi/3-sqrt(3)/4) or equivalent. the answer relies upon on the component of a sector of a circle radius and attitude t in radians = (a million/2)(r^2)t and the component of an isosceles triangle with equivalent aspects r enclosing attitude t = (a million/2)r^2(sint)