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).

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).

Answers (1)

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:
\(\displaystyle{A}{\left({D}\right)}=\frac{{1}}{{2}}\int_{{C}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{1}}{{2}}\int_{{{{y}_{{1}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}+\frac{{1}}{{2}}\int_{{{{y}_{{2}}^{+}}}}{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\)
\(\displaystyle={0}-{\left(-{3}{a}^{{2}}\pi\right)}\)
\(\displaystyle={3}{a}^{{2}}\pi\)
0

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