



 
Hi Brian, We received a somewhat similar question a few years ago from Richard but the tilt was at an arbitrary angle. I developed an expression for the volume and it appears at the bottom of my response to Richard. Since your tilt is 45 degrees and $\tan(45) = \cot(45) = 1$ the expression for the volume $V$ simplifies to \[V = \frac23 \left( 2rh  h^2\right)^{3/2} + 2(hr) \left(\frac{\pi r^2}{4}  \frac{r^2}{2}\sin^{1}\left(\frac{rh}{r}\right)  \frac{rh}{2}\sqrt{2rhr^2}\right)\mbox{ cubic units.}\] When using this expression make sure your calculator is in radian mode not degree mode. Still a somewhat complicated expression that requires $r$ and $h$ where $r$ is the radius of the drum and $h$ is the slant height of the liquid in the drum. Can you measure $h?$ If not what can you measure to determine the amount of liquid in the tank? Harley  


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