Question

Find the length of the curve over the given interval. $$ \begin{array}{ll} \text { Polar Equation } & \text { Interval } \\ \hline r=1+\sin \theta a \cos \theta & 0 \leq \theta \leq 2 \pi \end{array} $$

Short answer

The length of the curve over the given interval is \( L = \int_0^{2\pi} \sqrt{1 + \sin^2\theta} d\theta \)

Step-by-step solution

Find the derivative of r with respect to \( \theta \)

The derivative of \( r = 1 + \sin \theta \cos \theta \) with respect to \( \theta \) is obtained by applying the product rule. Thus, \( \frac{{dr}}{{d\theta}} = \sin^2\theta - \cos^2\theta \).

Substitute r and \( \frac{{dr}}{{d\theta}} \) into the formula

Substitute \( r = 1 + \sin \theta \cos \theta \) and \( \frac{{dr}}{{d\theta}} = \sin^2\theta - \cos^2\theta \) into the formula for the length of a polar curve, which gives \( L = \int_0^{2\pi} \sqrt{(1 + 2\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta) + (\sin^2\theta - \cos^2\theta)^2} d\theta \).

Simplify the integral

Simplify the integral to get a more manageable expression. After simplification, the integral to be evaluated becomes \( L = \int_0^{2\pi} \sqrt{1 + \sin^2\theta} d\theta \).

Evaluate the integral

Evaluate the integral from \( 0 \) to \( 2\pi \) of \( \sqrt{1 + \sin^2\theta} \). Unfortunately, this integral cannot be expressed in terms of elementary functions, so the answer will be presented in its integral form.