Find the length of the curve over the given interval. $$r = 9 + 9 \sin \theta , 0 \leq \theta \leq 2 \pi$$
A) 72
B) 54
C) 9
D) 36
E) 18

Question

Quizplus · Accepted Answer

The length $L$ of a curve given in polar coordinates $r = f(\theta)$ for $\theta$ in $[a, b]$ can be found using the formula $$L = \int_{a}^{b} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta.$$ For $r = 9 + 9\sin\theta$, we have $\frac{dr}{d\theta} = 9\cos\theta$. Substituting $r$ and $\frac{dr}{d\theta}$ into the formula gives $$L = \int_{0}^{2\pi} \sqrt{(9 + 9\sin\theta)^2 + (9\cos\theta)^2} d\theta = \int_{0}^{2\pi} \sqrt{81 + 162\sin\theta + 81\sin^2\theta + 81\cos^2\theta} d\theta.$$ Simplifying inside the square root using $\sin^2\theta + \cos^2\theta = 1$, we get $$L = \int_{0}^{2\pi} \sqrt{162 + 162\sin\theta} d\theta.$$ This integral evaluates to 72, making the correct answer A.

Find the Length of the Curve Over the Given Interval r=9+9sin⁡θ,0≤θ≤2πr = 9 + 9 \sin \theta , 0 \leq \theta \leq 2 \pir=9+9sinθ,0≤θ≤2π

Find the Length of the Curve Over the Given Interval $r = 9 + 9 \sin \theta , 0 \leq \theta \leq 2 \pi$