91影视

Given equation : $r=sec胃鈭�2\cos 胃$

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 $r=sec胃-2\cos 胃$.

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

localid="1651471166287" $$\begin{gathered} sec胃-2\cos 胃=0 \\ 鈬�{\cos }^2胃=\frac{1}{2} \\ 鈬�\cos 胃=卤\frac{1}{\sqrt{2}} \end{gathered}$$

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

Strophoid loops are symmetric around the horizontal axis.

The arc of the strophoid loop can be represented as

$A=2{鈭�}_{3蟺/4}^{蟺}\frac{1}{2}{r}^{2}d胃$.

Put$r=sec胃鈥�2\cos 胃$

$$\begin{gathered} A={鈭�}_{3蟺/4}^{蟺}{(sec胃-2\cos 胃)}^{2}d胃 \\ ={鈭�}_{3蟺/4}^{蟺}(se{c}^2胃-4sec胃路\cos 胃+{\cos }^2胃)d胃 \\ ={鈭�}_{3蟺/4}^{蟺}(se{c}^2胃-4+2(1+\cos 2胃\left)\right) \end{gathered}$$

Integrate in relation to $胃$:

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

Set the boundaries

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

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