What's the area of one arch of a cycloid? 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?

Here is a picture of cycloid:

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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 $y(t)$.