# Find the area bounded by one arc of the cycloid x = a(0-sin(0)), y = a(1-cos(0)),where a > 0, and 0 <= 0 <= 2pi, and the axis(use Green's theorem).

Question
Find the area bounded by one arc of the cycloid $$\displaystyle{x}={a}{\left({0}-{\sin{{\left({0}\right)}}}\right)},{y}={a}{\left({1}-{\cos{{\left({0}\right)}}}\right)}$$,where a > 0, and $$\displaystyle{0}\le{0}\le{2}\pi$$, and the axis(use Green's theorem).

2021-02-01
Step 1
Note that the atea of the region D bounded by C del D is $$\displaystyle{A}{\left({D}\right)}=\frac{{1}}{{2}}\int_{{C}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}$$.
The given cycloid and the x-axis intersect each other at the points (0,0) and (2 a pi,0).
Note that, $$\displaystyle\frac{{1}}{{2}}\int_{{C}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}=\frac{{1}}{{2}}\int_{{{{y}_{{1}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}+\frac{{1}}{{2}}\int_{{{{y}_{{2}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}$$.
The parametrization of the segment $$\displaystyle{y}_{{1}}{i}{s}{\left\lbrace\begin{array}{c} {x}{\left({t}\right)}={t}\\{y}{\left({t}\right)}={0}\end{array}\right.}{o}{v}{e}{r}{0}\le{t}\le{2}{a}\pi$$.
Step 2
The derivative of the above function is computed as follows,
$$\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}={1}$$
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}={0}$$
Now,
$$\displaystyle\frac{{1}}{{2}}\int_{{{{y}_{{1}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\int_{{0}}^{{{2}{a}\pi}}}{\left({x}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}-{y}\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=\frac{{1}}{{2}}{\int_{{0}}^{{{2}{a}\pi}}}{\left({1}{\left({0}\right)}-{\left({0}\right)}{\left({1}\right)}\right)}{\left.{d}{t}\right.}$$
=0
Step 3
The parametrization of the segment $$\displaystyle{y}_{{1}}{i}{s}{\left\lbrace\begin{array}{c} {x}{\left({0}\right)}={a}{\left({0}-{\sin{{0}}}\right)}\\{y}{\left({0}\right)}={a}{\left({1}-{\cos{{0}}}\right)}\end{array}\right.}{o}{v}{e}{r}{0}\le{0}\le{2}\pi$$.
The derivative of the above function is computed as follows,
$$\displaystyle\frac{{{\left.{d}{x}\right.}}}{{{d}{0}}}={a}{\left({1}-{\cos{{0}}}\right)}$$
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{d}{0}}}={a}{\sin{{0}}}$$
Now,
$$\displaystyle\frac{{1}}{{2}}\int_{{{{y}_{{2}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\int_{{0}}^{{{2}\pi}}}{\left({x}\frac{{{\left.{d}{y}\right.}}}{{{d}{0}}}-{y}\frac{{{\left.{d}{x}\right.}}}{{{d}{0}}}\right)}{d}{0}$$
$$\displaystyle\frac{{1}}{{2}}{\int_{{0}}^{{{2}\pi}}}{\left({\left({a}{\left({0}-{\sin{{0}}}\right)}\cdot{a}{\sin{{0}}}\right)}-{\left({a}{\left({1}-{\cos{{0}}}\right)}\right)}{\left({a}{\left({1}-{\cos{{0}}}\right)}\right)}\right)}{d}{0}$$
$$\displaystyle=\frac{{1}}{{2}}{\int_{{0}}^{{{2}\pi}}}{\left({a}^{{2}}{\left({0}{\sin{{0}}}-{{\sin}^{{2}}{0}}\right)}-{\left({a}^{{2}}\right)}{\left({1}-{\cos{{0}}}\right)}^{{2}}\right)}{)}{d}{0}$$
$$\displaystyle=\frac{{a}^{{2}}}{{2}}{\left[{\int_{{0}}^{{2}}}\pi{0}{\sin{{0}}}{d}{0}-{\int_{{0}}^{{{2}\pi}}}{{\sin}^{{2}}{0}}{d}{0}-{\int_{{0}}^{{{2}\pi}}}{\left({1}-{\cos{{0}}}\right)}^{{2}}{d}{0}\right]}$$
$$\displaystyle=\frac{{a}^{{2}}}{{2}}{\left[{\left(-{2}\pi\right)}-{\left(\pi\right)}-{\left({3}\pi\right)}\right]}$$
$$\displaystyle\frac{{a}^{{2}}}{{2}}{\left[{\left(-{6}\pi\right)}\right]}$$
$$\displaystyle=-{3}{a}^{{2}}\pi$$
Step 4
Obtain the area of region as follows:
$$\displaystyle{A}{\left({D}\right)}=\frac{{1}}{{2}}\int_{{C}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}$$
$$\displaystyle=\frac{{1}}{{2}}\int_{{{{y}_{{1}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}+\frac{{1}}{{2}}\int_{{{{y}_{{2}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}$$
$$\displaystyle={0}-{\left(-{3}{a}^{{2}}\pi\right)}$$
$$\displaystyle={3}{a}^{{2}}\pi$$

### Relevant Questions

Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{5}{\cos{{\left({y}\right)}}},{8}{\sin{{\left({y}\right)}}}\right\rangle}$$.Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across the boundary of the rectangle $$\displaystyle{0}\le{x}\le{5},{0}\le{y}\le\frac{\pi}{{2}}$$ using the vector form of Green's Theorem.
(Use symbolic notation and fractions where needed.)
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$ ?
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. F = (x - y)i + ( y - x)j
C: The square bounded by x = 0, x = 1, y = 0, and y = 1
Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}$$ where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}$$.
Use Green's Theorem to evaluate F * dr. (Check the orientation of the curve before applying the theorem.)
$$\displaystyle{F}{\left({x},{y}\right)}=\sqrt{{x}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}$$
C consists of the arc of the curve $$\displaystyle{y}={\sin{{\left({x}\right)}}}$$ from (0, 0) to $$\displaystyle{\left(\pi,{0}\right)}$$ and the line segment from $$\displaystyle{\left(\pi,{0}\right)}$$ to (0, 0)
Use Green’s Theorem to evaluate around the boundary curve C of the region R, where R is the triangle formed by the point (0, 0), (1, 1) and (1, 3).
Find the work done by the force field F(x,y)=4yi+2xj in moving a particle along a circle $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ from(0,1)to(1,0).
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. $$\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}$$
C: The triangle bounded by y = 0, x = 3, and y = x
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{4}{\cos{{\left({y}\right)}}},{2}{\sin{{\left({y}\right)}}}\right\rangle}$$. Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across the boundary of the rectangle $$\displaystyle{0}\le{x}\le{5},{0}\le{y}\le\frac{\pi}{{2}}$$ using the vector form of Green's Theorem.
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$?
$$\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}$$