The question asks to find the area under one arch of the cycloid:
$$x=a(t-\sin t), y=a(1-\cos t)$$
The solution says that $A=\int_{0}^{2\pi}ydx$
I'm just confused about how they were able to get this integral. Can someone please explain how to set up this integral? Why is it not $\int y dy$?
On
$$ \begin{aligned} A &=\int_{0}^{2 a\pi} y d x \\ &=\int_{0}^{2 \pi} a(1-\cos t) a(1-\cos t) d t \\ &=a^{2} \int_{0}^{2 \pi}(1-\cos t)^{2} d t \\ &=a^{2} \int_{0}^{2 \pi}\left(1-2 \cos t+\cos ^{2} t\right) d t \\ &=a^{2}\left([t-2 \sin t]_{0}^{2 \pi}+ \int_{0}^{2 \pi} \frac{1+\cos 2 t}{2} d t\right) \\ &=a^{2}\left(2 \pi+\frac{1}{2}\left[t+\frac{\sin 2 t}{2}\right]_{0}^{2 \pi}\right)\\&=a^{2}(2 \pi+\pi) \\ &=3 \pi a^{2} \end{aligned} $$
This is the same as your standard area integral. The integral $\int_a^b f(x) dx$ represents area above the $x$-axis and below the curve $y=f(x)$, so we can replace the $f(x)$ in the integrand with a $y$.
You can also see that this should indeed be a $dx$ integral because the limits of integration are the $x$-values $0$ and $2\pi$. The cycloid never gets near the height $y=2\pi$.
That said, how do we interpret the expression $\int y dx$ in this context? We can replace the "$y$" with our expression for $y$: $a(1-\cos t)$. The "$dx$" can be replaced with $\frac{dx}{dt}dt$, taking the derivative of the $x(t)$ expression and then calculating at $dt$ integral, which also runs from $0$ to $2\pi$.