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{Question}
{\text{Compute $\iint_S1dA$, where $S$ is the region between the x-axis and the curve }\\
\text{$g(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}1-\sin(t)\\1-\cos(t)\end{pmatrix}$ for $0\le t\le 2\pi$. (This curve is called a cycloid.)}}$
Let $Q=x$ and $P=0$, apply Green's Theorem we have
\begin{align}
\int_S 1-0~dA=\int_S\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dA=&\int_{\partial S}Pdx+Qdy\\
=&\int_{\partial S}xdy\\
=&\int_0^{2\pi-\sin(2\pi)}0~dt+\int_{2\pi}^{0}(t-\sin(t))\sin(t)dt\\
=&3\pi
\end{align}
Is my solution correct ?
Best Answer
I don't understand how you derived
\begin{align} &\int_{\partial S}xdy\\ =&\int_0^{2\pi-\sin(2\pi)}0~dt+\int_{2\pi}^{0}(t-\sin(t))\sin(t)dt\\ \end{align}
I think it should be
\begin{align} &\int_{\partial S}xdy\\ =&\int_{2\pi}^{0}(1-\sin t)\sin tdt\\ =&\pi \end{align}
Also, note that the equation for the area of one arch of a cycloid equals $3\pi a^2$ only applys to $$x = a(\boldsymbol{\theta}-\sin \theta)$$ $$y = a(1-\cos \theta)$$