An industrial cylindrical water tank has a radius of r meters and a height of h meters, where h is twice the radius. When the tank is full, its water level decreases at a rate of 0.25 meters per hour due to usage. After 4 hours of usage, what is the volume of water remaining in the tank in terms of r?

Sure, let's solve the problem step by step.

1. Determine the initial volume of the tank: The volume $V$ of a cylinder is given by the formula: $V = \pi r^2 h$ Given that the height $h$ is twice the radius $r$, we can write: $h = 2r$ Substituting $h$ into the volume formula, we get: $V = \pi r^2 (2r) = 2\pi r^3$ So, the initial volume of the tank is $2\pi r^3$.

2. Calculate the decrease in water level after 4 hours: The water level decreases at a rate of 0.25 meters per hour. After 4 hours, the total decrease in water level is: $\text{Decrease in water level} = 0.25 \times 4 = 1 \text{ meter}$

3. Determine the new height of the water level: The initial height of the water level is $h = 2r$. After a decrease of 1 meter, the new height $h_{\text{new}}$ is: $h_{\text{new}} = 2r - 1$

4. Calculate the remaining volume of water: Using the new height $h_{\text{new}}$, the remaining volume $V_{\text{remaining}}$ is: $V_{\text{remaining}} = \pi r^2 h_{\text{new}}$ Substituting $h_{\text{new}}$: $V_{\text{remaining}} = \pi r^2 (2r - 1)$ Simplifying this expression, we get: $V_{\text{remaining}} = \pi r^2 (2r - 1) = 2\pi r^3 - \pi r^2$

So, the volume of water remaining in the tank after 4 hours of usage is $2\pi r^3 - \pi r^2$.