Question

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

The area of the inner loop of the graph of the polar equation \(r=1+2 \cos \theta\) is calculated by evaluating the integral \[ \frac{1}{2} \int_{{-\frac{2\pi}{3}}}^{{\frac{2\pi}{3}}} (1+2 \cos \theta)^2 \, d\theta\]

Step-by-step solution

Graphing the Polar Equation

Using a graphing utility or by manually plotting points, graph the given polar equation \(r=1+2 \cos \theta\). The graph will reveal two loops: an outer loop and an inner loop.

Determining the Limits of Integration

The inner loop begins and ends where \(r=0\). Solve for \(\theta\) in the equation \(r= 1+2 \cos \theta =0\) to find these angles. Upon solving, \(\theta = \pm \frac{2\pi}{3}\) would be obtained.

Calculating the Area of the Inner Loop

Now, the area of a polar curve can be calculated using the formula: \[ A= \frac{1}{2} \int_{{\theta1}}^{{\theta2}} r(\theta)^2 \, d\theta \] Here, \(\theta1=-\frac{2\pi}{3}\) and \(\theta2= \frac{2\pi}{3}\). Substituting \(r(\theta)=1+2 \cos \theta\) into the formula would yield: \[ A= \frac{1}{2} \int_{{-\frac{2\pi}{3}}}^{{\frac{2\pi}{3}}} (1+2 \cos \theta)^2 \, d\theta\] Solving this integral will give the area of the inner loop.