Respuesta :
(Green's Theorem" The area is bound by the x-axis on the bottom part from x = 0 to x = 2Ď€, and
by the cycloid on the top.
C = the bounding curve
Csub1 = the x-axis part of C
Csub2 =the cycloid part.
You will take an integeral
2 x the Area will end up being the integral from 2pi to 0 of cos(t)dt with is 6pi
So 2 x Area = 6pi so the area = 3pi.
The area under one arch of the cycloid x = t − sin(t), y = 1 − cos(t). is [tex]3\pi[/tex]
Explanation:
Green's theorem is a relationship between the two-dimensional vector field line integral over the closed path in the plane and the double integral over the region as it encloses.
A cycloid is the curve that traced by point on the rim of a circular wheel as the wheel rolls along a straight line without slipping. Cycloid is when a circle rolls along a straight line the path.
What is the area under one arch of the cycloid x = t − sin(t), y = 1 − cos(t) ?
The area is bounded by the x-axis on the bottom from [tex]x = 0[/tex] and [tex]x = 2 \pi[/tex] and by the cycloid on the top. Whereas the C is the bounding curve that consist of the x-axis traversed from [tex]x = 0[/tex] to [tex]x = 2 \pi[/tex]followed by the cycloid going from [tex]t = 2\pi[/tex] to [tex]t = 0[/tex]. [tex]C_1[/tex] is the x-axis part of C and [tex]C_2[/tex] is the cycloid part
[tex]2 * area(D) = \int\limits^._C {-y dx+x dy} \, = \int\limits^._{C1} {-y dx+x dy +} \, \int\limits^._{C2} {-y dx+x dy \, \\[/tex]
[tex]2 * area(D) = \int\limits^._{C2} {-y dx+x dy} \,[/tex] (because [tex]y=0[/tex] and [tex]dy=0[/tex] on [tex]C_1[/tex]}
Because of x = t − sin(t), y = 1 − cos(t) then
[tex]2 * area(D) = \int\limits^{t=0}_{t=2\pi} {(-(1-cost)(1-cost)+(t-sint)sint)dt} \,[/tex]
[tex]2 * area(D) = \int\limits^{t=0}_{t=2\pi} {(-2+2cost+tsint)dt=4\pi-tcost|_{t=2\pi}^{t=0} + \int\limits^{t=0}_{t=2\pi} {cos t} \, dt = 6\pi } \,[/tex]
Therefore the area under one arch of the cycloid is [tex]area (D) = 3\pi[/tex]
Learn more about one arch of the cycloid : https://brainly.com/question/10582117 and https://brainly.com/question/14168141
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