inner loop of r = 3 + 6 cos(𝜃)Find the area of the given region.

The area of a polar curve is given by the formula:

A = 1/2 ∫[a,b] (r(θ))^2 dθ

where r(θ) is the polar equation of the curve, and [a,b] is the interval of θ over which the curve is defined.

In this case, r(θ) = 3 + 6cos(θ). The curve completes a full loop when θ goes from 0 to 2π. So, we need to compute the integral:

A = 1/2 ∫[0,2π] (3 + 6cos(θ))^2 dθ

This integral can be computed using standard techniques of integration. First, expand the square:

A = 1/2 ∫[0,2π] (9 + 36cos(θ) + 36cos^2(θ)) dθ

The integral of cos(θ) over [0,2π] is zero, so we only need to compute the integrals of the constant term and the cos^2(θ) term.

The integral of a constant over [0,2π] is just the constant times 2π. So, the contribution of the 9 term to the area is:

1/2 * 9 * 2π = 9π

The integral of cos^2(θ) over [0,2π] can be computed using the power-reduction identity:

∫[0,2π] cos^2(θ) dθ = π

So, the contribution of the 36cos^2(θ) term to the area is:

1/2 * 36 * π = 18π

Adding these up, the total area of the region is:

A = 9π + 18π = 27π