Area swept by a moving line Consider a line segment on an $\Bbb R^2$ plane with one end (call it $p_1$) fixed along x-axis and another end (call it $p_2$) fixed along y-axis. The point $p_1$ starts at $(0,0)$ and the point $p_2$ starts at $(0,1)$. I allow the line segment to move with one of the following constraints:
$1$. The point $p_1$ moves to the right with the constraint that the length of the line segment is fixed. The line segment stops moving when $p_1$ reaches $(1,0)$.
$2$. The point $p_1$ moves to the right and the point $p_2$ moves downward with the constraint that the distances travelled by each of the points are equal at any moment, i.e. when $p_1$ moves by $(t,0)$, $p_2$ shall move by $(0,-t)$. The line segment stops moving when $p_1$ reaches $(1,0)$.
In each of both cases, there is a region swept by the line segment. What are their areas respectively?
It may look like a simple integration problem, but I have absolutely no idea how to find the function for the curve above the region. Do I miss out something obvious, or is it really a non-trivial problem?
 A: It's not trivial, but it's not that hard either.  Each family of line segments crosses each vertical $x=x_0$ (for $x_0\in[0,1]$) at a range of heights $\{ y(x_0) \}$; you want the function $f(x)$ such that $f(x_0)=\sup\{y(x_0)\}$.
In the first case, you have line segments from $(0,\cos\theta)$ to $(\sin\theta,0)$, crossing $x_0$ at $\cos\theta - x_0 \cot\theta$.  This is maximized when
$$
-\sin\theta + x_0\csc^2\theta=0,
$$
or $\sin\theta=x_0^{1/3}$; so $f(x)=\cos\theta - \sin^3\theta\cot\theta=\cos^3\theta=(1-x^{2/3})^{3/2}$.
In the second case, you have line segments from $(0,t)$ to $(1-t,0)$, crossing $x_0$ at $t - x_0 t/(1-t)$.  This is maximized when
$$
1 - \frac{x_0}{1-t} - \frac{x_0 t}{(1-t)^2}=0,
$$
or $t=1-\sqrt{x_0}$; so $f(x)=(1-\sqrt{x})- x(1-\sqrt{x})/\sqrt{x}=(1-\sqrt{x})^2$ in this case.
A: Isn't this $1 - \frac{\pi}{4}$ ?
The line describes a quarter-circle of radius 1 centred at $(1,1)$, which sits in inside a unit square.
So the area underneath the 'slipping ladder' is a unit square minus a quarter of a unit circle. 
