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Volume of Liquid in the Bottom of a Tank

Date: 08/11/2003 at 17:02:12
From: George
Subject: Volume

I have a cylindrical vertical tank. the bottom of the tank is a 
parabolic bowl. I need to be able to calculate the volume of stored 
liquid in the bowl of the tank based on the liquid level. The tank 
has a parabolic equation of y=kx^2 until it reaches a Level H; then 
the tank walls are straight and vertical.

As the liquid level rises, the cross-sectional area of the liquid is 
continually changing.

Date: 08/11/2003 at 18:31:01
From: Doctor Jeremiah
Subject: Re: Volume

Hi George,

The easiest way I know of to solve this is to use calculus. However, 
it is most definitely NOT the easiest way to explain it.

The vertical direction is z so the graph of the parabola is really 
z = kx^2. The cross section of the 3D parabola is a circle where the 
radius is the x value of z = kx^2 at any particular z.  So r^2 = x^2 = 
z/k because the radius is the x value of the z = kx^2 graph.  The area 
of the circular cross section is A = Pi r^2, but because r^2 = z/k
the area is also A = Pi z/k.

To find the volume, basically imagine filling the tank with infinitely 
thin circles of paper.  If we add up all the volumes of all those 
individual pieces of paper we end up with the volume of the tank. The 
volume of one of these sheets is dV = A dz, where A is the cross-
sectional area, dz is the thickness of the infinitely thin piece of 
paper and dV is the infinitely small volume of that piece.

If the thickness were not infinitely thin we could use a regular sum.  
But since the thickness is infinitely thin we need to add up the 
volumes of an infinite number of pieces. A regular sum of an infinite 
number of terms would take too long so we use a calculus integral 
instead. It does the same thing as a sum but adds up an infinite 
number of infinitely thin objects.

An integral sign looks like an elongated capital S.  I can't really 
draw that here so I will do the best I can. The volume of the parabola 

     z=h    z=h      z=h                z=h
      /      /        /                  /
  V = } dV = | A dz = | Pi z/k dz = Pi/k | z dz
      /      /        /                  /
     z=0    z=0      z=0                z=0

Where we are integrating from z=0 up to a height of z=h

Using the rules of solving integrals this integral becomes

  V = Pi/k z^2/2 at z=h minus Pi/k z^2/2 at z=0
  V = Pi/k h^2/2 - Pi/k 0^2/2
  V = Pi/k h^2/2

And when the parabola is full h=N and V = Pi/k N^2/2 but when z is 
some arbitrary value of h then V = Pi/k h^2/2

It turns out that there is an easier way to look at it. The volume of 
a paraboloid is 1/2 the area of the circular base times the height, or 
written mathematically:

  V = Pi x^2 z/2

Since your equation was z = k x^2, then x^2 = z/k and:

  V = Pi (z/k) z/2
  V = Pi/k z^2/2

Which is the same thing.

- Doctor Jeremiah, The Math Forum 

Date: 08/12/2003 at 14:18:57
From: George
Subject: Thank you (Volume)

Thank you for your quick response to my question. It is very helpful, 
and I now have the information I need. Thanks again.
Associated Topics:
College Calculus
High School Calculus

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