Question

In Exercises 29–38, find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral.

38.The graph of the polar equation $r=sec\theta -2\cos \theta$is called a strophoid. Graph the strophoid and find the area bounded by the loop of the graph.

Short answer

The iterated integral that represents the area of the given region is $\mathit{A}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}$

We've plotted the required plot of the given strophoid.

Step-by-step solution

Given Information

Given equation : $r=sec\theta -2\cos \theta$

Graphing the strophoid and find the area bounded by the loop of the graph 

The goal of this task is to graph the polar equation and determine the area enclosed by the graph's loop.

The strophoid equation is $r=sec\theta -2\cos \theta$.

Determining the tangent at the pole by putting $r=0$, we have :

localid="1651471166287" $$\begin{gathered} sec\theta -2\cos \theta =0 \\ \Rightarrow {\cos }^2\theta =\frac{1}{2} \\ \Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}} \end{gathered}$$

Thus, localid="1651471171666" $\theta =\boxed{\frac{3\pi }{4}}\mathrm{and}\boxed{\frac{5\pi }{4}}$:

Finding an iterated integral that represents the area of the given region 

Strophoid loops are symmetric around the horizontal axis.

The arc of the strophoid loop can be represented as

$A=2{\int }_{3\pi /4}^{\pi }\frac{1}{2}{r}^{2}d\theta$.

Put$r=sec\theta –2\cos \theta$

$$\begin{gathered} A={\int }_{3\pi /4}^{\pi }{(sec\theta -2\cos \theta )}^{2}d\theta \\ ={\int }_{3\pi /4}^{\pi }(se{c}^2\theta -4sec\theta ·\cos \theta +{\cos }^2\theta )d\theta \\ ={\int }_{3\pi /4}^{\pi }(se{c}^2\theta -4+2(1+\cos 2\theta \left)\right) \end{gathered}$$

Integrate in relation to $\theta$:

localid="1651471257788" $A={\left[\tan \theta -4\theta +2(\theta +\frac{1}{2}\sin 2\theta ))\right]}_{3\pi /4}^{\pi }$

Set the boundaries

localid="1651471264053" $$\begin{gathered} A=\tan \left(\pi \right)-4\left(\pi \right)+2(\pi +\frac{1}{2}\sin 2\pi )-\tan \left(\frac{3\pi }{4}\right)-4\left(\frac{3\pi }{4}\right)+2(\frac{3\pi }{4}+\frac{1}{2}\sin 2(\frac{3\pi }{4}\left)\right) \\ =\left[\{-4\pi +2\pi \}-\{-1-3\pi +(\frac{3\pi }{2}-1\left)\right\}\right] \\ =2-\frac{\pi }{2} \end{gathered}$$

As a result, the strophoid loop's area islocalid="1651471271290" $A=\mathbf{2}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}$