Question

In Exercises 33-36, find the length of the curve over the given interval. $$ \begin{array}{ll} \text { Polar Equation } & \text { Interval } \\ \hline r=a & 0 \leq \theta \leq 2 \pi \end{array} $$

Short answer

The length of the curve over the given interval is \(2a\pi\)

Step-by-step solution

Recall the formula for the arc length in polar coordinates.

The arc length (L) in polar coordinates is calculated using the formula \( L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \), where r is the polar function, and \( \alpha \) and \( \beta \) are the limits of the interval. Here, \( r = a \) (a constant) so the derivative \( \frac{dr}{d\theta} \) will be zero.

Substitute the given values into the formula.

When you substitute the given values into the formula, you will have: \( L = \int_{0}^{2\pi} \sqrt{a^2 + \left(0\right)^2} d\theta = \int_{0}^{2\pi} a d\theta \). This turns out to be an integral of a constant, which is a straightforward problem.

Compute the integral.

The integral of a constant 'a' over the interval from 0 to \( 2\pi \) is \( a * (\theta \mid_{0}^{2\pi}) = a * (2\pi - 0) = 2a\pi \). Therefore, the length of the curve over the given interval is \( 2a\pi \)