Find the area of the region.interior of r = 5 + 3 sin(𝜃) (below the polar axis

To find the area of the region interior of r = 5 + 3 sin(𝜃) below the polar axis, we need to integrate over the appropriate range of 𝜃.

The polar axis corresponds to 𝜃 = 0. The curve dips below the polar axis when 𝜃 is between π and 2π.

The formula for the area A in polar coordinates is given by:

A = 1/2 ∫ (from α to β) [r(𝜃)]² d𝜃

Here, r(𝜃) = 5 + 3 sin(𝜃), α = π, and β = 2π.

So, we need to compute the following integral:

A = 1/2 ∫ (from π to 2π) [5 + 3 sin(𝜃)]² d𝜃

This integral can be computed using standard techniques of integration.

First, square the function inside the integral:

[5 + 3 sin(𝜃)]² = 25 + 30 sin(𝜃) + 9 sin²(𝜃)

Then, compute the integral term by term. The first two terms are straightforward. The third term can be computed using the power-reduction identity sin²(𝜃) = 1/2 - 1/2 cos(2𝜃).

After computing the integral, don't forget to multiply by 1/2 to get the area.