(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am trying to solve for the partial volume of a horizonal cylinder that has two frustums attached to the sides. The cylinder radius is equal to the large radius of each frustum. The frustum bases and caps are open. The container is full when liquid begins to run out of the sides of the container at the bottom of the caps.

2. Relevant equations

Constants -

L=cylinder length

R1=cylinder radius = frustum base radius

R2- frustum cap radius

d=frustum distance from base to cap

h=height of liquid

Equations -

1) Cylinder volume is: V = Pi * R1^2 * L

2) Frustum full volume is: V=Pi * (d/3) * (R1^2 +R2^2 + R1*R2)

3) Partial volume of Cylinder is: V= L*(R1^2 * acos((R1-h)/R1) - (R1-h)*(2*R1*h-h^2)^.5

3. The attempt at a solution

I attempted to subsitute (R1^2 +R2^2 + R1*R2) for the cylinder's R1 into#3 for one frustum's partial volume but it doesn't seem to work out. Can you help? I'm interested in the liquid volume that would begin to overflow out of the small frustum cap and smaller h's also.

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# Homework Help: Partial Vol of horizonal cylinder with 2 frustums

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