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So, the cycloid is given with parametric equations: $$x=r(t-\sin{t})$$ $$y=r(1-\cos{t})$$ The teacher solved it like this: $$P=\int_a^by(x)dx$$ $x=x(t)$; $\alpha<t<\beta$ $$P=\int_\alpha^{\beta}y(t)x'(t)dt$$ $x'(t)=r(1-\cos{t})$ $$P=\int_0^{2\pi}r(1-\cos{t})r(1-\cos{t})dt=r^2\int_0^{2\pi}(1-2\cos{t}+\cos^2{t})dt=$$ $$=r^2(t|_0^{2\pi}-2\sin{t}|_o^{2\pi}+\frac{1}{2}t|_0^{2\pi}-\frac{1}{4}\sin{2t}|_0^{2\pi}=r^2(2\pi+\pi)=3r^2\pi$$ So, we get that the area below one arch of a cycloid equals three areas of a circle which forms that cycloid. My question is: I don't understand anything about this problem :) How did the teacher integrate this parametric equation, why did he write the integral of $y(t)x'(t)$, why did he need a derivative of x(t) and what does it represent. Can you please explain this to me geometrically?

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  • $\begingroup$ The Wikipedia page on Cavalieri's principle has noncalculus way to do this, by the way. $\endgroup$ Commented Mar 29, 2015 at 13:17
  • $\begingroup$ This is actually a special case of Green's Theorem. $\endgroup$
    – john
    Commented Jun 4, 2017 at 19:13

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This problem is a nice illustration of Green's theorem.

Let $R$ be the region enclosed by the piecewise defined curve $C= C_1 \cup C_2$, where $C_1$ and $C_2$ are defined as$$ \begin{align} C_1:x&=r(t-\sin{t}), ~~~y=r(1-\cos{t}), &0 \le t \le 2 \pi \\ C_2: x &= 2\pi (1-t) , ~~~~~y = 0, &0 \le t \le 1 \end{align}$$ Note that our curve $C$ is negatively (clockwise) oriented, and Green's Theorem requires positively oriented (counterclockwise) curves. We can denote the positively oriented 'reverse curve' by $ C^{*}$, where $C^{*} =-C$.

Green's Theorem states that for a simple closed positively oriented curve $C$, we have: $$\int_{C} (P\, dx + Q\, dy) = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, dA $$ We can exploit Green's theorem to evaluate the area enclosed by the curve $C$ using the vector field $ \vec{\mathbf { F} } = \langle -y, 0 \rangle $. We get $$\int_C (-y ~dx + 0 ~dy) = \iint_R (0 - (-1)) dA =\iint_R dA $$ The latter integral is the area we want. Now using the fact that for any line integral and curve $\Gamma$ $$ \int_{-\Gamma} P dx + Q dy = - \int_{\Gamma} P dx + Q dy$$ we obtain $$ \iint_R dA = \int_{C^{*}} -y~ dx + 0~ dy =-\int_{C} -y ~dx + 0~ dy =\int_{C} y~ dx $$ Now we have a tangible result we can use. Evaluating the line integral we have $$ \begin{align} \int_{C} y~ dx &=\int_{C_1 \cup C_2} y~ dx \\ &= \int_{C_1} y~ dx + \int_{C_2} y~ dx \\ &= \int_{0}^{2\pi} (r (1-\cos t) ) ( r( 1- \cos t) ~dt) + \int_{0}^{1} 0 (-2 \pi dt ) \\ &= 3 \pi r^2 \end{align}$$

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Here is a picture of cycloid:

http://en.wikipedia.org/wiki/Cycloid

Do you know that the area under a curve $y(x)$ between an interval $[a,b]$ is

$$\int^b_a y(x) dx$$

By using change of variable $x=x(t)$, you can change the upper and lower limits to $0$ and $2\pi$, and $dx=x'(t)dt$. $y(x)$ then of course should be changed to $y(x(t))$ which is also $y(t)$.

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    $\begingroup$ I don't understand why $y(x(t))=r(1-\cos(x(t)))=r(1-\cos(r(t-\sin(t)))=y(t)$? $\endgroup$
    – zoli
    Commented Mar 29, 2015 at 13:16
  • $\begingroup$ $y(x(t))=y(t)=r(1-\cos{t})$. Since $y(t)$ is already defined, you don't need to define it using $x(t)$. Otherwise, you have to represent $t$ by $x$, and $x$ by $t$ again, which gives you the original $y(t)$. $\endgroup$
    – KittyL
    Commented Mar 29, 2015 at 16:06
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    $\begingroup$ I have the same issue as the OP. It is not clear how $ y(x(t))$ becomes $y(t)$. $\endgroup$
    – john
    Commented Jun 4, 2017 at 22:13
  • $\begingroup$ @john: $y$ in terms of $x$ is defined through the parametric equations $x(t)$ and $y(t)$. So when you go back to find $y(x(t))$, it is implicitly just $y(t)$. $\endgroup$
    – KittyL
    Commented Jun 5, 2017 at 9:32
  • $\begingroup$ This seemed very weird to me as well, but this site finally helped me to understand: tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx Essentially (if I understand it correctly) the thing is that the $y(x)$ written in the beginning is not the same function (at least from a programming standpoint / perspective) as the $y(t)$ at the end. $y(x)$ would be the value of $y$ as it depends on $x$, but we do not know the formula for that, only when it depends on $t$. Still I find it a bit counter intuitive, but I think that I can accept it now. $\endgroup$
    – Isti115
    Commented Oct 25, 2018 at 22:36
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For the integral of the parametric equation, he used one of the duplication formulae in trigonometry with an error (but he was lucky enough to get a correct final answer): $$1-2\cos t+\cos^2t=1-2\cos t+\frac{1+\cos 2t}2$$ hence the value of the integral is $$t-2\sin t+\frac t2 +\frac14\sin 2t\Biggr\lvert_0^{2\pi}= 3\pi.$$

As for the change of variable induced by the parametrisation of the cycloid, this comes from the very definition of the differential: $$\mathrm d\mkern1.5mu x=x'(t)\,\mathrm d\mkern1.5mu t.$$

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