Question

In Exercises 11 and 12, use a graphing utility to graph the polar equation and find the area of the given region. Inner loop of \(r=1+2 \cos \theta\)

Short answer

After calculating the integral in step 3, the final answer for the area within the inner loop of \(r = 1+2cos(\theta)\) is found to be \(\pi\) square units.

Step-by-step solution

Graphing the Polar Curve

Using a graphing utility or tool like Desmos or Geogebra, plot the polar curve represented by \(r=1+2 \cos \theta\). This polar equation reveals a limaçon, which is a kind of spiralled curve, with an inner loop due to the nature of the cosine function and the coefficients.

Identifying the Inner Loop

From the graph, observe that the inner loop of the curve lies within the range \(-\pi/2 \leq \theta \leq \pi/2\). The loop starts form the pole (origin), spirals out and then back in to the pole again over this interval of \(\theta\).

Calculating the Area

To find the area (A) inside the inner loop of our curve, we integrate half the square of our radius from \(-\pi/2\) to \(\pi/2\). Mathematically, this is represented as: \(A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} (1+2cos(\theta))^2 d\theta\). Evaluate this definite integral to get the area.