Falling rain drop problem EDIT: I've read that a ball moving in a rectilinear motion with a non-constant radio, $r$ satisfies that
$$\frac{dV_c}{dy} = \pi r^2,$$
where $V_c$ is the volume swept by the ball and $y$ is the position of the ball along the moving direction. I don't get why it should be like that, any idea?
Previous post: I've been trying to solve this physics problem today "Spherical raindrop falls through a cloud with uniform density. After falling 1km it has a radius of 5mm. What volume fraction of the cloud is made of condensed water. Assuming that the raindrop starts with negligible size and picks up any condensed water it passes through"
The problem I'm having is understanding the solution, the first step of the model answer is: "Let the raindrop have a radius r, after falling a distance y through the cloud. The swept volume of the cloud is $$\ V_c \,$$ so that $$\frac{dV_c}{dy}=\pi r^2\,$$
But I really don't understand how they came to that conclusion, could somebody point out what I'm missing? Thanks!
 A: Ain't nothing quite like a picture to explain this:
 
The drop falls a distance $v$ in one second, so the volume of the cylinder is $\pi r^2 v$. Of course, the drop grows as it picks up all the moisture in the gray region - so $r$ will be a function of distance fallen, and the volume swept will increase accordingly (right hand diagram).
In essence the drop will "suck up" all the moisture in a cone shaped volume - although it's not intuitively clear whether the cone has straight edges, or is convex or concave. This will depend on whether the velocity changes with size (presumably it does - but since it's picking up stationary water as it is falling it's not immediately clear to me that you reach terminal velocity).
Very interesting problem.
A: If after falling some distance $y$ the rain drop has radius $r(y)$ and cross-sectional area $A(y)=\pi (r(y))^2$when it falls an additional small distance $\delta y$ it sweeps out an additional volume $\delta V_v\approx A(y)\delta y= \pi (r(y))^2\delta y$. Then taking limits as $\delta y \to 0$ gives:
$$
\frac{dV_c}{dy}=\pi (r(y))^2
$$
