Question

In exercise 26-30 Find a definite integral that represents the length of the specified polar curve and then find the exact value of integral

$Thespiralr=e^{\alpha \theta }for0\le \theta \le 2\pi$

Short answer

The integral can be given as ${\int }_0^{2\pi }\sqrt{1+{\alpha }^2}{e}^{\alpha \theta }d\theta$and the length of the spiral can be given as $\sqrt{(1+{\alpha }^2)}\left(\frac{e^{2\pi \alpha }-1}{\alpha }\right)$

Step-by-step solution

Given information

We are given a spiral with equation$r=e^{\alpha \theta }for0\le \theta \le 2\pi$

Find the integral and evaluate it

We know that

${\int }_0^{2\pi }\sqrt{(f{\left(\theta \right))}^2+(f\text{'}{\left(\theta \right))}^2}d\theta$

We are given

$$\begin{gathered} r=e^{\alpha \theta } \\ r\text{'}=\alpha e^{\alpha \theta } \end{gathered}$$

Substituting in the formula we get

localid="1650460246282" $$\begin{gathered} {\int }_0^{2\pi }\sqrt{e^{2\alpha \theta }+{\alpha }^2{e}^{2\alpha \theta }}d\theta \\ =\sqrt{1+{\alpha }^2}{\int }_0^{2\pi }\sqrt{e^{2\alpha \theta }}d\theta \\ =\sqrt{1+{\alpha }^2}{\int }_0^{2\pi }{e}^{\alpha \theta }d\theta \\ =\sqrt{(1+{\alpha }^2)}\left(\frac{e^{2\pi \alpha }-1}{\alpha }\right) \end{gathered}$$