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 enclosed by the spiral $r=\theta$and the $x$-axis on

the interval $0\le \theta \le \mathrm{\pi }$

Short answer

The area of the spiral$r=\theta$is$\frac{{\pi }^3}{6}$

Step-by-step solution

Given information

The spiral$r=\theta$on the interval$0\le \theta \le \pi$

The objective is to find the area of the spiral.

The region's corresponding limits are $0to\mathrm{\pi }$

The interval is $[0,\pi ]$

The formula of the area is$A={\int }_{\alpha }^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$or$A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$

The area of the function is calculated as below

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{\pi }{\theta }^{2}d\theta [\text{since}r=\theta ] \\ A=\frac{1}{2}{\left[\frac{{\theta }^3}{3}\right]}_0^{\pi } \end{gathered}$$

Limits are established by applying them

$A=\frac{1}{2}\left[\frac{{\pi }^3}{3}-0\right]$

$$\begin{gathered} A=\frac{1}{2}·\frac{{\mathrm{\pi }}^3}{3} \\ \Rightarrow A=\frac{{\mathrm{\pi }}^3}{6} \end{gathered}$$

The spiral's enclosing area$r=\theta$is$A=\frac{{\pi }^3}{6}$