# liquid in a cylinder

Solution

The solution to this problem can be achieved through purely geometric means; see the diagram below.

We can see that exactly half of the upper part of the cylinder is filled. This region of the cylinder has a volume of $V_U = \pi r^2h = \pi(2^2)(4) = 16\pi$. Of this part of the cylinder, exactly half is occupied.

The total volume of liquid in the cylinder is therefore $V_{Liquid} = V_L + \frac{1}{2}V_U = \pi r^2h + 8\pi = \pi(2)^2(6) + 8\pi = 32\pi$. The entire cylinder has a volume of $40\pi$, so $\frac{4}{5}$ of the liquid remains.