Q. 18. - 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 region inside one loop of the lemniscate $r^2=\sin 2\theta$

Step-by-Step Solution

Answer

The area inside one loop of the lemniscate $r^2=\sin 2\theta$is $\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}2\theta d\theta =\frac{\pi }{8}$

Step 1: Given information

Think about the polar function.$r^2=\sin 2\theta$

Step 2: The objective is to find the area inside by one loop of the function.

The function $r^2=\sin 2\theta$Calculate the shaded region's area.

The darkened region's corresponding bounds are $0 to \frac{\mathrm{\pi }}{2}$

 The interval is $\left[0,\frac{\pi }{2}\right]$

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

Step 3: The area of the function is calculated as below

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

Limits are established by applying them

$$\begin{gathered} A=\frac{1}{4}\left[\frac{\pi }{2}-\frac{\sin 2·\frac{\pi }{2}}{2}-0\right] \\ \Rightarrow A=\frac{1}{4}\left[\frac{\mathrm{\pi }}{2}-0\right] \\ \Rightarrow A=\frac{\mathrm{\pi }}{8} \end{gathered}$$

The area inside one loop is $A=\frac{\pi }{8}$