Q. 25 - Calculus

Question

In Exercises 17-25 find a definite integral expression that represents the area of the given region in the polar plane, and then find the exact value of the expression.

The graph of the polar equation $r=\sec \theta -2\cos \theta$ for $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ is called a strophoid. Graph the strophoid and find the area bounded by the loop of the graph.

Step-by-Step Solution

Answer

Ans: The area of the equation is  $A=\left(2-\frac{\pi }{2}\right)$

Step 1. Given information:

The polar equation $r=\sec \theta -2\cos \theta$

Step 2. Implying formula:

Formula to find the area is $A={\int }_{\alpha }^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$ or $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$.

By equating, $\sec \theta -2\cos \theta =0$

Thus,

$$\begin{gathered} \sec \theta =2\cos \theta \\ \cos \theta =\frac{1}{\sqrt{2}} \\ \cos \theta =\cos \frac{\pi }{4} \\ \theta =\frac{\pi }{4} \end{gathered}$$

The limits are from 0 to $\frac{\pi }{4}$.

Step 3. Continue:

Then,

$$\begin{gathered} A=2·\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(\sec \theta -2\cos \theta {)}^{2}d\theta \\ A={\int }_0^{\frac{\pi }{4}}\left({\sec }^2\theta +4{\cos }^2\theta -2·2\sec \theta ·\cos \theta \right)d\theta \\ A={\int }_0^{\frac{\pi }{4}}\left({\sec }^2\theta +4{\cos }^2\theta -4\right)d\theta \end{gathered}$$

Step 4. Simplification:

$$\begin{gathered} A={\int }_0^{\frac{\pi }{4}}\left({\sec }^2\theta +4\frac{1+\cos 2\theta }{2}-4\right)d \left[since{\cos }^2\theta =\frac{1+\cos 2\theta }{2}\right] \\ A={\int }_0^{\frac{\pi }{4}}\left({\sec }^2\theta +2+2\cos 2\theta -4\right)d\theta \\ A=(\tan \theta -2\theta +\sin 2\theta {)}_0^{\frac{\pi }{4}} \end{gathered}$$

Step 5. Continue:

Thus,

$$\begin{gathered} A=\left(\tan \frac{\pi }{4}-2·\frac{\pi }{4}+\sin 2·\frac{\pi }{4}\right) \\ A=\left(1-\frac{\pi }{2}+1\right) \\ A=\left(2-\frac{\pi }{2}\right) \end{gathered}$$

Therefore, the area of the equation is $A=\left(2-\frac{\pi }{2}\right)$.