Find the area of the region enclosed by the inner loop of the curve.
r = 4 + 8 sin(θ)
r=0 at θ=7π/6, 11π/6
So, using symmetry, the area is
a = 2∫[7π/6,3π/2] 1/2 (4+8sinθ)^2 dθ = 8(2π-3√3) ≈ 8.6963
To find the area of the region enclosed by the inner loop of the curve, we can use the formula for the area bounded by a polar curve. The formula is given by:
A = (1/2) ∫[θ1,θ2] (r(θ))^2 dθ
In this case, the equation of the curve is r = 4 + 8 sin(θ).
To find θ1 and θ2, we need to determine where the curve intersects itself and where the inner loop starts and ends.
Since the sine function has a period of 2π, the inner loop of the curve is completed in the interval [0, 2π].
We can now calculate the area:
A = (1/2) ∫[0, 2π] (4 + 8 sin(θ))^2 dθ
= (1/2) ∫[0, 2π] (16 + 64 sin(θ) + 64 sin^2(θ)) dθ
= (1/2) ∫[0, 2π] (16 + 64 sin(θ) + 32 (1 - cos(2θ))) dθ
= (1/2) ∫[0, 2π] (16 + 64 sin(θ) + 32 - 32 cos(2θ)) dθ
= (1/2) ∫[0, 2π] (48 + 64 sin(θ) - 32 cos(2θ)) dθ
The integral can now be evaluated to find the area of the inner loop of the curve.
To find the area of the region enclosed by the inner loop of the curve, we can use the formula for the area of a polar region which is given by:
A = (1/2) ∫[θ₁, θ₂] (r(θ))² dθ
First, let's find the values of θ at which the inner loop is formed. The inner loop is formed when r(θ) = 0. In this case, r = 4 + 8sin(θ), so we can set r equal to zero and solve for θ:
4 + 8sin(θ) = 0
Solving this equation, we get:
sin(θ) = -1/2
The values of θ at which sin(θ) is equal to -1/2 are θ = 7π/6 and θ = 11π/6.
Now, we can calculate the area using the formula stated above. In this case, θ₁ = 7π/6 and θ₂ = 11π/6.
A = (1/2) ∫[7π/6, 11π/6] (4 + 8sin(θ))² dθ
To evaluate this integral, expand the expression inside the integral, square it, and then integrate term by term:
A = (1/2) ∫[7π/6, 11π/6] (16 + 64sin(θ) + 64sin²(θ)) dθ
Simplify the integrand:
A = (1/2) ∫[7π/6, 11π/6] (16 + 64sin(θ) + 32(1 - cos(2θ))) dθ
Now, integrate term by term:
A = (1/2) [16θ + 64(-cos(θ)) + 16sin(2θ)] evaluated from θ = 7π/6 to θ = 11π/6
Finally, compute the values and subtract to get the area of the region enclosed by the inner loop.