Polar Coordinates Integration?

Question

Set up the integral and use a calculator to find the area inside the polar graph of r=2+cos(x) (x being theta).

I don't need help on the calculating part, it's setting up the integral that's getting to me.

Is the limits of integration from -pi/2 < x < pi/2? If I'm wrong, could you tell me how you would find the limits of integration?

Answer 1

Integrate to find area of curve r = 2 + cosθ.

Area = ∫(1/2)r²dθ

By symmetry we can integrate from 0 to π and multiply by 2.

Area = 2∫(1/2)(2 + cosθ)²dθ = ∫(4 + 4cosθ + cos²θ)dθ

= ∫[4 + 4cosθ + (1 + cos 2θ)/2]dθ

= ∫[9/2 + 4cosθ + (cos 2θ)/2]dθ

= 9θ/2 + 4sinθ + (sin 2θ)/4 | [Evaluated from 0 to π]

= [9π/2 + 0 + 0] - [0 + 0 + 0] = 9π/2

Answer 2

i think the limits of cosine with polar integration is from 0 to pi