math

find the area enclosed betwen cycloid X= a(t-sin(t))? How can i solve this problem ? send me the steps

Note by Er Kundan Sharma
8 years, 3 months ago

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Comments

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

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

Recognize that y=a(1cost)y=a(1-\cos t) and dx=a(1cost)dt\mathrm{d}x=a(1-\cos t)\,\mathrm{d}t, so our integral is: A=02πa2(1cost)2dtA=\int_0^{2\pi}a^2(1-\cos t)^2\,\mathrm{d}t

Can you finish this off?

o b - 8 years, 3 months ago

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A=a202π(12cost+cos2t)dt=a202π[12cost+12+12cos2t]dt=a2[t2sint+12t+14sin2t]02π=a2[2π+π]=3πa2\begin{aligned}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{aligned}

o b - 8 years, 3 months ago
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