Please help me with these area under polar curve problems!!?

Area = (1/2) ∫ r^2 dθ [ integrate from 0 to 2pi]

= (1/2) ∫ 9 cos^2(theta) dtheta

= (9/2) ∫ (1+cos(2theta))/2 dtheta

= (9/4) theta + (9/8) sin(2theta)

F(x) = (9/4) x +(9/8) sin (2x)

F(2pi) = 9pi/2

F(0) = 0

Area = 9pi/2

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r = 4 + 2sin(theta)

Area = (1/2) ∫ r^2 dθ [ integrate from 0 to 2pi]

= (1/2) ∫ (16+ 16 sin(θ) + 4 sin^2(θ) ) dθ

= (1/2) ∫ (16+16 sin x + 4 sin^2x ) dx

= 8 ∫ dx + 8 ∫ sin x dx + 2 ∫ (1-cos (2x)) / 2 dx

= 8x -8 cos x + x - sin(2x) / 2

= 9x - 8 cos x - (1/2) sin 2x

Let F(x) = 9x - 8 cos x - (1/2) sin 2x

substitute the upper limit 2pi

F(2pi) = 18pi - 8

substitute the lower limit 0

F(0) = -8

Area = F(2pi)-F(0) = 18pi

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1+cos(theta) = 1

cos(theta) = 0

theta = pi/2 , 3pi/2

theta = -pi/2 to pi/2

Area = (1/2) ∫ (1+cos(theta) )^2 dtheta - (1/2) ∫ 1^2 dtheta , each integral from theta=-pi/2 to pi/2

(1+cos(theta))^2 = 1+ 2 cos(theta) + cos^2(theta)

Use the identity cos^2(theta) = (1+cos(2theta) )/2 and integrate

Area = (8+pi)/4

Cardidoid