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Volume of Intersecting Pipes


Date: 10/27/2000 at 06:10:35
From: Sam
Subject: Volume of Intersecting Pipes

Can you please explain how to calculate the volume of the intersection 
of two perpendicular pipes of the same radius? I have heard that the 
result is 16/3 r^3, but I can't fathom how to get this answer.

Could you explain it to me in simple terms?


Date: 10/27/2000 at 14:55:21
From: Doctor Rob
Subject: Re: Volume of Intersecting Pipes

Thanks for writing to Ask Dr. Math, Sam.

There are two ways to proceed, depending on whether or not you know 
integral calculus. First I'll assume you don't, then that you do.

Set up a three-dimensional Cartesian coordinate system with the origin 
at the intersection of the axes of the pipes, the axis of one pipe 
along the x-axis, and the axis of the other pipe along the y-axis. 
Then the equations of the surfaces of the two pipes are:

     y^2 + z^2 = r^2
     x^2 + z^2 = r^2

Notice that the object is symmetrical with respect to the xy-plane, 
the xz-plane, the yz-plane, and the plane x = y. These four symmetries 
mean that we only have to compute the volume of 1/16 of the object, 
then multiply by 16. Thus we can add the restrictions

     x >= y >= 0
     z >= 0

Then the curved surface of this piece lies on the surface of the 
second pipe, x^2 + z^2 = r^2. Observe that a horizontal cross-section 
of this piece by the plane z = z0 will be an isosceles right triangle 
defined by

     sqrt(r^2-z0^2) >= x >= y >= 0.

Its base and altitude are both sqrt(r^2-z0^2), so its area is half the 
square of that number, or (r^2-z0^2)/2.

The volume V of the piece of the solid is bounded by

     n-1                                n
     SUM (r/n)*(r^2-[i*r/n]^2)/2 < V < SUM (r/n)*(r^2-[i*r/n]^2)/2.
     i=0                               i=1

This is obtained by enclosing within the solid a stack of n prismoidal 
slices of height r/n and upper base a section of the piece at height 
z = i*r/n, and by enclosing the solid within a stack of n slices of 
height r/n and lower base a section of the piece at height z = i*r/n. 
Manipulating and simplifying the two sums, 

                               n-1
     n*(r/n)*r^2/2 - (r/n)^3/2*SUM i^2 < V <
                               i=0
                                                               n      
      
                                    n*(r/n)*r^2/2 - (r/n)^3/2*SUM i^2
                                                              i=0

     r^3/2 - (r/n)^3/2*(n-1)*n*(2*n-1)/6 < V <
                                  r^3/2 - (r/n)^3/2*n*(n+1)*(2*n+1)/6
 
     r^3*[1/2-(1-1/n)*(2-1/n)/12] < V < r^3*[1/2-(1+1/n)*(2+1/n)/12]

Now, taking the limit as n increases without bound of both sides, we 
see that

     r^3*[1/2 - 2/12] <= V <= r^3*[1/2 - 2/12]
     r^3/3 <= V <= r^3/3
     V = r^3/3

Thus the total volume is (16/3)*r^3.

If you know integral calculus, you can express the volume of the 1/16 
piece of the solid by

        r   x  sqrt(r^2-x^2)
   V = INT INT     INT       1 dz dy dx
        0   0       0

        r    x
     = INT  INT  sqrt(r^2-x^2) dy dx
        0    0

        r
     = INT  x*sqrt(r^2-x^2) dx
        0

                           r
     = [-(r^2-x^2)^(3/2)/3]   
                           x=0

     = r^3/3

Thus the total volume is 16 times this, or 16*r^3/3.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
College Higher-Dimensional Geometry
High School Calculus
High School Higher-Dimensional Geometry

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