If you're going to parametrize the cycloid at least give us both \(x,y\):
\[x=a(t-\sin t)\\
y=a(1-\cos t)\]

You want to determine the area from \(t=0\) to \(t=2\pi\) (i.e. one whole arch), so we write:
\[A=\int_{t=0}^{t=2\pi}y\,\mathrm{d}x\]

Recognize that \(y=a(1-\cos t)\) and \(\mathrm{d}x=a(1-\cos t)\,\mathrm{d}t\), so our integral is:
\[A=\int_0^{2\pi}a^2(1-\cos t)^2\,\mathrm{d}t\]

Can you finish this off?
–
O B
·
3 years, 10 months ago

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@O B
–
\[\begin{align*}A&=a^2\int_0^{2\pi}(1-2\cos t+\cos^2 t)\,\mathrm{d}t\\&=a^2\int_0^{2\pi}\left[1-2\cos t+\frac12+\frac12\cos 2t\right]\,\mathrm{d}t\\&=a^2\left[t-2\sin t+\frac12t+\frac14\sin 2t\right]_0^{2\pi}\\&=a^2\left[2\pi+\pi\right]\\&=3\pi a^2\end{align*}\]
–
O B
·
3 years, 10 months ago

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TopNewestIf you're going to parametrize the cycloid at least give us both \(x,y\): \[x=a(t-\sin t)\\ y=a(1-\cos t)\]

You want to determine the area from \(t=0\) to \(t=2\pi\) (i.e. one whole arch), so we write: \[A=\int_{t=0}^{t=2\pi}y\,\mathrm{d}x\]

Recognize that \(y=a(1-\cos t)\) and \(\mathrm{d}x=a(1-\cos t)\,\mathrm{d}t\), so our integral is: \[A=\int_0^{2\pi}a^2(1-\cos t)^2\,\mathrm{d}t\]

Can you finish this off? – O B · 3 years, 10 months ago

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– O B · 3 years, 10 months ago

\[\begin{align*}A&=a^2\int_0^{2\pi}(1-2\cos t+\cos^2 t)\,\mathrm{d}t\\&=a^2\int_0^{2\pi}\left[1-2\cos t+\frac12+\frac12\cos 2t\right]\,\mathrm{d}t\\&=a^2\left[t-2\sin t+\frac12t+\frac14\sin 2t\right]_0^{2\pi}\\&=a^2\left[2\pi+\pi\right]\\&=3\pi a^2\end{align*}\]Log in to reply