At which depth will the glass, which is it not a right cylinder, be volumetrically half-filled? Assume that the thickness of the base is negligible and the thickness of the walls is constant.

I think I would first have to find the derivative of perhaps the circular area or something like that... Out of the box answers are welcome:))
The glass.The glass.

1 Answer
Jan 28, 2018

The split should occur at a height of approx 5.27 \ cm

Explanation:

No Calculus required:

The mathematical name for the given "bucket" , or "glass" shape is a frustum, we can readily calculate the volume using the formula:

V = pi/3(R^2 + rR + r^2)h \ \ \ (See Notes)

We are required to split the volume of the given frustum of lower radius 3.5 \ cm, upper radius 3.9 \ cm and height 10 \ cm into two further frustums, each having equal volume.

Steve MSteve M

Suppose the volumeatic split occurs at a height h \ cm so that the lower frustum is of height h, with lower radius 3.5 \ cm upper radius R. Then the upper frustum will be of height 10-h with lower radius r and upper radius of 3.9 \ cm.

Using this information, and utilising the frustum formula, we can write the identical volume of each frustum as:

pi/3(3.5^2 + 3.5R + R^2)h = pi/3(R^2 + 3.9R + 3.9^2)(10-h)

And by similar triangle's:

h : (R-3.5) = 10: (3.9-3.5)

:. h/(R-3.5) = 10/(0.4)
:. 0.4h = 10(R-3.5)
:. 0.4h = 10R-35
:. 10R = 0.4h +35

Then we solve the two equations simultaneously, which results in the solution:

h ~~ 5.26948 R~~3.71078 (5dp)

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Notes:
We can readily derive the frustum volume by considering the difference in the volumes of standard cones#

http://jwilson.coe.uga.edu/http://jwilson.coe.uga.edu/

Using the standard volume of a cone, V=1/3pir^2h; we have:

Volume of Smaller cone:

V_1 = 1/3pir^2H

Volume of Larger cone:

V_2 = 1/3piR^2(H+h)

So the volume of the frustum is:

V = V_2-V_1
\ \ \ = 1/3piR^2(H+h) - 1/3pir^2H
\ \ \ = 1/3pi(R^2H+R^2h-r^2H)
\ \ \ = 1/3pi(H(R^2-r^2)+R^2h)
\ \ \ = 1/3pi(H(R+r)(R-r)+R^2h)

And by similar triangle's:

R : H+h = r : H

:. R/(H+h) =r/H
:. RH = r(H+h)
:. RH = rH+rh
:. H(R-r) = rh
:. H = (rh)/(R-r)

Substituting into out volume formula we get:

V = 1/3pi((rh)/(R-r)(R+r)(R-r)+R^2h)
\ \ \ = 1/3pih((r)(R+r)+R^2)
\ \ \ = 1/3pi(rR+r^2+R^2) QED