What is the intersecting volume of two pipes?

Question

Okay let's try a hard question for a change.

What is the volume of intersection of two pipes whose center axes intersect each other at a perfect right angle, and whose inner diameters are 1 meter? That is, what is the total volume of the set of points which are enclosed by both pipes at the same time?

Hint. Consider the pipes as the equations z^2+x^2=0.25 and z^2+y^2=0.25 .

I am not totally convinced; I suggest people show their work more. For example, is the volume of intersection straight lines like a diamond, or curved, what? If it's a diamond it's two pyramids of volume 1/3 since sides of each of 2 pyramids are 1 m and height is 0.5 so 1/3*1*1*.5 * 2 pyramids. But show your work!

Answer 1

The answer is 2/3 cubic meter.The use of symmetries, as in the answer # 1 (Daniel F.) makes things more complicated. Here is the simplest approach.

If we simply solve for x and y, we get x = +/- sqrt( r^2 - z^2 ) and y = +/- sqrt( r^2 - z^2 ), where r = 0,5 . So, for every z between -r and r, we have a square of side-length 2 sqrt( r^2 - z^2 ). The volume is thus given by integrating the area S(z) = [ 2 sqrt( r^2 - z^2 ) ]^2 = 4 ( r^2 - z^2 ) from -r to r, which is an elementary integral to do. We obtain V(r) = (16/3) r^3. In our case r = 1/2. So, we get (16/3) (1/8) = 2/3.

Or you can solve this without calculus as in the answer #2 (Northstar): The key observation is that, if we replace the square by a circle of same diameter (as the side-length of the square), we get a sphere and the ratio of the area of the square to the area of the circle is a constant : 4 r^2 / (π r^2) = 4/π. Recall that the volume of a sphere is (4/3) π r^3. So, the volume of the intersection of the pipes is V(r) = (4/π) x (4/3) π r^3 = (16/3) r^3. Again, V(1/2) = 2/3.

Answer 2

So the pipes intersect at a perfect right angle, which means that we've got a lot of symmetry. If we use the equations you suggested, then we've got symmetry across xy, xz, yz, and x=y planes. Which means that we need to find only 1/16 of the volume of the intersection right? (and then multiply by 16, of course).

We can set up a triple integral to find the volume of 1/16 of the intersection. We will be using the order of integration dzdydx. The limits on x will be 0 to 1/2, the limits on y will be 0 to x, and the limits on z will be 0 to sqrt(1/4-x^2). It would be much easier to explain why these are the limits using a picture, but note that we are restricted to the first octant because of the symmetries that I just explained.

Evaluating this triple integral yields the answer 1/24. Multiplying this by 16 yields the volume of intersection. The answer is 2/3.

Or You can solve this without calculus.

If you look at the intersection of the two pipes and slice a horizontal cross section you will get a square. Inscribe a circle in the square. As the horizontal cross section moves from the bottom to the top of the intersection of the two pipes, the layers of cross sections of the inscribed circle form a sphere with radius 1/2. In every cross section the ratio of the square intersection to the area of the circle is a constant.

For a square with side s, the circle has us s/2. The ratio is:

Area Circle / Area Square = πp(s/2)² / s² = πp/4

The volume of the sphere is 4πpr³/3 = 4πp(1/2)³/3 = πp/6

The volume of the intersection of the pipes is:

(πp/6) / (πp/4) = 2/3 m³

Answer 3

So the pipes intersect at a perfect right angle, which means that we've got a lot of symmetry. If we use the equations you suggested, then we've got symmetry across xy, xz, yz, and x=y planes. Which means that we need to find only 1/16 of the volume of the intersection right? (and then multiply by 16, of course).

We can set up a triple integral to find the volume of 1/16 of the intersection. We will be using the order of integration dzdydx. The limits on x will be 0 to 1/2, the limits on y will be 0 to x, and the limits on z will be 0 to sqrt(1/4-x^2). It would be much easier to explain why these are the limits using a picture, but note that we are restricted to the first octant because of the symmetries that I just explained.

Evaluating this triple integral yields the answer 1/24. Multiplying this by 16 yields the volume of intersection. The answer is 2/3.

Answer 4

You can solve this without calculus.

If you look at the intersection of the two pipes and slice a horizontal cross section you will get a square. Inscribe a circle in the square. As the horizontal cross section moves from the bottom to the top of the intersection of the two pipes, the layers of cross sections of the inscribed circle form a sphere with radius 1/2. In every cross section the ratio of the square intersection to the area of the circle is a constant.

For a square with side s, the circle has radius s/2. The ratio is:

Area Circle / Area Square = π(s/2)² / s² = π/4

The volume of the sphere is 4πr³/3 = 4π(1/2)³/3 = π/6

The volume of the intersection of the pipes is:

(π/6) / (π/4) = 2/3 m³

Answer 5

If they are concentric the intersection will be equal to the volume of the smaller pipe.

Answer 6

Please stay away from pipes...they lead to more and more use and ruin lives. Hope this helps!