Volume of liquid in a hopper / inverted frustum of a pyramid, based on the liquid level. This question is part of a larger investigation into specific machinery process rates.
I know the dimensions of the hopper, and thus the volume of the hopper (inverted frustum of a pyramid). However, I would like to know how to calculate the volume of a liquid in the hopper based on the level of the liquid within the hopper.

e.g. 


*

*When the level of a liquid in the hopper is $75.1$ cm ($50\%$ in terms of distance between the top and the bottom of the hopper), what is the volume of the liquid?

*When the level of a liquid in the hopper is $95\%$ ($142.69$ cm in terms of distance from the bottom of the hopper), what is the volume of the liquid?
I would like to be able to input a formula into an excel spreadsheet to calculate this volume, and ultimately further integrate this data into the control system.
FYI By my calculations the volume of the hopper is $2.04198802$ $\text{m}^3$.

For the side 208.1cm we have the side increased to at a height of h:
$$35.2 + \frac{208.1 - 35.2}{150.2}h = 35.2 + 1.1511h$$
Similarly for the other side :
$$34.2+ \frac{159.8 - 34.2}{150.2}h = 34.2 + 0.8362h$$
Hence at a height of H the volume is
$$\int_0^H(35.2 + 1.1511h)(34.2 + 0.8362h)dh$$
$$= 1203.84H + 34.4H^2 + 0.32H^3$$
To verify if H is 150.2 we have volume 2.041 $m^3$
