help with area of polar graphs please...?

Question

Find the area of a common interior.

r= 5-3sin(θ))
r= 5-3cos(θ))

Answer 1

½∫ r² dθ = ½∫{0→2π} min²(5 - 3sinθ, 5 - 3cos θ) dθ

On the interval of (0, 2π),
• cosθ ≥ sinθ when θ ≤ π/4 or θ ≥ 5π/4
• sinθ ≥ cosθ when π/4 ≤ θ ≤ 5π/4

Thus we have:

½∫{0→2π} min²(5 - 3sinθ, 5 - 3cos θ) dθ
= ½[∫{0→π/4} (5 - 3cosθ)² dθ + ∫{π/4→5π/4} (5 - 3sinθ)² dθ + ∫{5π/4→2π} (5 - 3cosθ)² dθ]
= ½[[59θ/2 - 30sinθ + (9/4)sin(2θ)]{0→π/4} + [59θ/2 + 30cosθ - (9/4)sin(2θ)]{π/4→5π/4} + [59θ/2 - 30sinθ + (9/4)sin(2θ)]{5π/4→2π}]
= ½[[59(π/4)/2 - 30sin(π/4) + (9/4)sin(2(π/4))] - [59(0)/2 - 30sin0 + (9/4)sin(2(0))] + [59(5π/4)/2 + 30cos(5π/4) - (9/4)sin(2(5π/4))] - [59(π/4)/2 + 30cos(π/4) - (9/4)sin(2(π/4))] + [59(2π)/2 - 30sin(2π) + (9/4)sin(2(2π))] - [59(5π/4)/2 - 30sin(5π/4) + (9/4)sin(2(5π/4))]]
= ½[[59π/8 - 30(√2/2) + (9/4)sin(π/2)] - [0 - 30(0) + (9/4)sin(0)] + [295π/8 + 30(-√2/2) - (9/4)sin(5π/2)] - [59π/8 + 30(√2/2) - (9/4)sin(π/2)] + [59π - 30(0) + (9/4)sin(4π)] - [295π/8 - 30(-√2/2) + (9/4)sin(5π/2)]]
= ½[[59π/8 - 15√2 + (9/4)(1)] - [-0 + (9/4)(0)] + [295π/8 - 15√2 - (9/4)(1)] - [59π/8 + 15√2 - (9/4)(1)] + [59π - 0 + (9/4)(0)] - [295π/8 + 15√2 + (9/4)(1)]]
= ½[[59π/8 - 15√2 + 9/4] - 0 + [295π/8 - 15√2 - 9/4] - [59π/8 + 15√2 - 9/4] + [59π + 0] - [295π/8 + 15√2 + 9/4]]
= ½[[59π/8 - 15√2 + 9/4] + [295π/8 - 15√2 - 9/4] - [59π/8 + 15√2 - 9/4] + 59π - [295π/8 + 15√2 + 9/4]]
= ½[59π/8 - 15√2 + 9/4 + 295π/8 - 15√2 - 9/4 - 59π/8 - 15√2 + 9/4 + 59π - 295π/8 - 15√2 - 9/4]
= ½[59π - 60√2]
= 59π/2 - 60√2/2
= 59π/2 - 30√2

Answer 2

The integral of an area using polar coordinates is of the form:

Area = ∫(1/2)r²dθ

The curve r = 5 - 3sinθ [From θ = π/4 to 5π/4] has the smaller radius and therefore the area in common. The same could be said of the other curve for the remainder of possible values for θ on [0, 2π]. So take the area of the once curve over the interval and double it.

Area = 2∫(1/2)r²dθ = 2∫(1/2)(5 - 3sinθ)² dθ

= ∫(25 - 30sinθ + 9sin²θ)dθ

= ∫[25 - 30sinθ + (9/2)(1 - cos 2θ)]dθ

= ∫[59/2 - 30sinθ - (9/2)(cos 2θ)]dθ

= (59/2)θ + 30cosθ - (9/4)(sin 2θ) | [Eval from θ = π/4 to 5π/4]

= [(59/2)(5π/4) - 30/√2 - (9/4)(1)]
- [(59/2)(π/4) + 30/√2 - (9/4)(1)]

= [295π/8 - 30/√2 - 9/4] - [59π/8 + 30/√2 - 9/4]

= 236π/8 - 30√2

= 59π/2 - 30√2

Answer 3

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)