The sides of a square are initially $4m$ in length and increases at a constant rate of $3m$ per second. One corner of the square moves along a line $L$ at a speed of $2m/s$ for $5$ seconds. During this time the square makes one complete revolution about $L$, revolving at a constant rate and remaining always perpendicular to $L$. Calculate the volume of the solid generated.
I'm a bit confused by this problem.
Now what I have understood is that, if the line $L$ coincides with a side of a square, then the rate of change of side length does not tally with the rate of change of the corner. Also the same problem is there if we consider the $L$ as in the diagonal. So probably it must be a one perpendicular to the plane.
As given in here
But in this case the length of the side will also change.
Also I don't exactly see why the volume of this solid should be equal to (area of a cross section)$^2$$h$.
Appreciate your help.
Best Answer
First, you are correct that the square is in a plane perpendicular to the line $L$ However, you’re misunderstanding the question you referenced, they’re not saying that the volume is equal to the area of the cross section squared times the height, like they said, the cross section has area equal to the side length squared, so in $s^2h$ the square is already part of the cross section. This formula was valid in that problem because the cross sections had equal area, so if you twist the solid, you can make it into a rectangular prism without changing the volume. In this problem, the area of the cross section isn’t constant, so when we untwist the solid, we don’t recover a prism, instead we recover a truncated cone with square base. We just need to integrate the cross section area with respect to height.