How to find interval $t$ in calculating volume of parametric equations rotated. $$x=2(t-\sin t),~y=2(1-\cos t)$$ Find the volume as curves are rotated around $x$-axis. The interval of $t$ is not given, Is there any way to calculate $t$?
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Indeed, the answer very much depends on the interval of $t$ that should be used. By looking at the graph, my guess is that $0 \leq t \leq 2\pi$ is one possible option. Or maybe you can express the answer as a function of the maximum value of $t$! – Matti P. Nov 05 '19 at 12:23
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Since the curve repeats itself at each period $2\pi$ we can take the first period interval for $y(t)$ that is $t\in [0,2\pi]$.
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The curve $$x=2(t-\sin t),~y=2(1-\cos t)$$
has its y-intercepts at $$y=2(1-\cos t)=0$$ which includes $t=0$ and $t=2\pi$
Thus if you are interested in the first region rotated, your limits are $t=0$ and $t=2\pi$
Mohammad Riazi-Kermani
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