A cylindrical tank with axis horizontal has a diameter d and a length L. It holds oil to depth d ( d<D ) when a leak starts to drain off the oil at a constant rate k. How fast is the level falling

https://en.wikipedia.org/wiki/Circular_segment

So you want the area not shaded, PI*r^2-areasegment.
I suspect you will have to work this in two cases:
a) D/2 <d<D and b) 0<d<D/2, and the levels rates will be falling at diffent rates depending on which case.

To find the rate at which the level of oil is falling, we need to use the concept of related rates. We can set up the following variables:

V = Volume of oil in the tank
h = Height of oil in the tank
r = Radius of the tank

Given:
d = Diameter of the tank
L = Length of the tank
k = Constant rate at which the oil is draining off

We know that the volume of a cylinder is given by V = πr^2h.
Since the tank is cylindrical, we can find the radius using the formula r = d/2.

To find the rate at which the level h is falling, we need to differentiate the volume equation with respect to time (t). This will give us an equation relating the rate of change of volume (dV/dt) with the rate of change of height (dh/dt).

Differentiating V = π(d/2)^2h with respect to t, we get:
dV/dt = π(d/2)^2 * dh/dt

Since the rate at which the oil is draining off (k) is a constant, we know that dV/dt = -k (negative because the volume is decreasing).

Therefore, we have:
-k = π(d/2)^2 * dh/dt

Now, we can rearrange the above equation to solve for dh/dt:
dh/dt = -k / (π(d/2)^2)

Simplifying further, we have:
dh/dt = -4k / (πd^2)

So, the rate at which the level of oil is falling is given by dh/dt = -4k / (πd^2).

Note: The negative sign indicates that the level of oil is decreasing with time.