Find the length of the curve.
-The spiral $r = 6 \theta , 0 \leq \theta \leq \pi$
A) 36
B) $3 \left( \pi \sqrt { \pi ^ { 2 } + 1 } + \ln \left( \pi + \sqrt { \pi ^ { 2 } + 1 } \right) \right)$
C) $6 \left( \sqrt { \pi ^ { 2 } + 1 } + \ln \left( \frac { 1 + \sqrt { \pi ^ { 2 } + 1 } } { \pi } \right) \right)$
D) $16 \left( \left( \pi ^ { 2 } + 1 \right) ^ { 3 / 2 } - 4 \right)$

Question

Quizplus · Accepted Answer

The length of a curve given in polar coordinates $r = f(\theta)$ from $\theta = a$ to $\theta = b$ is calculated using the formula $L = \int_{a}^{b} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta$. For $r = 6\theta$, $\frac{dr}{d\theta} = 6$, so the integral becomes $\int_{0}^{\pi} \sqrt{(6\theta)^2 + 6^2} d\theta$, which simplifies to $3 \left( \pi \sqrt { \pi ^ { 2 } + 1 } + \ln \left( \pi + \sqrt { \pi ^ { 2 } + 1 } \right) \right)$ after evaluation.

Find the Length of the Curve r=6θ,0≤θ≤πr = 6 \theta , 0 \leq \theta \leq \pir=6θ,0≤θ≤π

Find the Length of the Curve $r = 6 \theta , 0 \leq \theta \leq \pi$