Find the area.
-Find the area under one arch of the cycloid x = 2(t - sin t), y = 2(1 - cos t).
A) $12 \pi$
B) $6 \pi$
C) $12 \pi ^ { 2 }$
D) $8 \pi$

Question

Quizplus · Accepted Answer

The area under one arch of the cycloid can be calculated using the formula for the area under a parametric curve, $A = \int y \, dx$. Given $x = 2(t - \sin t)$ and $y = 2(1 - \cos t)$, we find $dx = 2(1 - \cos t)dt$. The limits of integration for one complete arch of the cycloid are from $t = 0$ to $t = 2\pi$, as these values correspond to one complete cycle of the trigonometric functions involved. Substituting $y$ and $dx$ into the integral and evaluating it from $0$ to $2\pi$, we get:$$A = \int_{0}^{2\pi} 2(1 - \cos t) \cdot 2(1 - \cos t)dt = 4\int_{0}^{2\pi} (1 - \cos t)^2 dt$$Expanding and simplifying the integrand, and then integrating, yields an area of $12\pi$.

Find the Area 12π12 \pi12π B) 6π6 \pi6π C) 12π212 \pi ^ { 2 }12π2

Find the Area $12 \pi$ B) $6 \pi$ C) $12 \pi ^ { 2 }$