Question

Find the area of the region. Interior of \(r=1-\sin \theta\)

Short answer

The area of the region defined by \(r=1-\sin\theta\) is 0.75\(\pi\).

Step-by-step solution

Identify the range of \theta and graph

First, the given equation is \(r=1-\sin\theta\). As \(\sin\theta\) ranges between -1 and 1, the equation \(1 - \sin\theta\) will range from 0 to 2. If we plot this, we will notice that the plot is a circle that starts at the pole, goes to the point (0,1), around to (0,-1), and back to the pole. Hence, the limits of \(\theta\) will be from 0 to \(\pi\)

Set Up the Integral

Using the formula for the area \(A = 0.5 \int_{a}^{b} r^2 d \theta\) where the limits of the integral are the range values of \(\theta\), the area integral becomes \(A = 0.5 \int_{0}^{\pi} (1-\sin\theta)^2 d \theta\)

Evaluate the Integral

Calculating \(A = 0.5 \int_{0}^{\pi} (1-\sin\theta)^2 d \theta\), which equals \(A = 0.5 \int_{0}^{\pi} (1-2\sin\theta + \sin^2\theta) d \theta\). Using power-reduction identity, the integral can be simplified to \(A = 0.5 \int_{0}^{\pi} (1-2\sin\theta + 0.5 -0.5 \cos 2\theta) d \theta\). Now, calculating the integral, we get \(A = 0.5[ \theta - 2\cos\theta + 0.5\theta - (1/4)\sin 2\theta]\) evaluated from 0 to \(\pi\)

Simplify the Expression

The result of the integral gives us \(A = 0.5[\pi - 0 + 0.5\pi - 0] - 0.5[0 - 2 + 0 - 2]\). Simplifying this, we end up with 0.75\(\pi\)