# The volume of the intersection of two similar pipes

##### - Calculus

Thanksgiving break is here and after the feasting, when the food weighs heavy and the mind drifts into that hazy place between sleep and awake thanks to loads of carbs and tryptophan, we might find ourselves wondering what the volume is of the space created by two intersecting pipes. At least, my calculus professor seems to think that would be a good quandary to mull over. In this article we will use Thanksgiving leftovers to help visualize the space created by two intersecting pipes and then try to determine the volume of that space; first via approximating and then by getting serious with the tools we’ve learned in first year calculus.

## Visualizing the space created by intersecting pipes

If I try really hard, I can kind of imagine what the shape of the space created by two intersecting pipes might look like, cube-like, with tapered sides due to the roundness of the pipes. But, stuffed with food as I am, holding that image in my mind long enough to do any real analysis of it is out of the question; I need a visual. Sure, there’s 3D modeling software that could help with this process but what fun would that be? Besides, it’s not like Archimedes had CAD software to help him figure this type of problem out 2000 years ago. Instead, lets create a solid ourselves using nothing more than the leftover cranberry jelly from Thanksgiving that nobody will ever eat anyway!

To summarize the video’s findings, the solid created by two intersecting pipes is known as a Steinmetz solid, a 2-solid, to be specific, as it represents the intersection of two pipes. The most interesting finding though, is that the cross section of this solid is nothing more than a square. Very interesting indeed. Using this information then, what can we do about determining the volume of this Steinmetz solid?

## An approximating start

The first approach I like to take when solving a problem I’m unfamiliar with, is to do some approximating. By approximating, we can generally establish some boundaries on what to expect from the exact answer. If we end up with a final answer that falls outside a previously established upper or lower limit, then we know its either time to check our work or return to the drawing board.

In this case, we can see that the solid produced is roughly a cube with sides that taper inwards. If we treat it as simply a cube, then we can use the simple formula:

\[V = l^3\]In the case of our jellied cylinders, the diameter of the can was 3 inches which gives the following estimate:

\[V = 3^3 = 27\text{in}^3\]Because the actual solid tapers inwards, this estimate is definitely an over-estimate. That is, the exact answer should be less than $ 27\text{in}^3 $. Armed with this sanity check, how might we get a more exact answer?

## Didn’t we just leave this party?

Hey! Wait a sec, that whole approximating by using simpler shapes sounds a lot like something we’ve recently been studying in class: finding the areas beneath a curve. In that problem, the primary idea is to take a complex 2D shape and divide it up into smaller, simpler shapes. The shape of choice in the 2D problem is rectangles.

Let’s look at a graphical representation of this idea. In finding the area of the curve below $ f(x) $, bounded by the x-axis, the left endpoint $ x=0 $, and the right endpoint $ x=4 $, we can approximate by using a rectangle. In this case we use a left endpoint to determine where the box should touch the curve.

The area under the curve then, or at least an approximation of it, is the area of the rectangle shown in figure 1. To get a better approximation of this area, we simply add more rectangles. The length of each rectangle along the $ x $-axis is $ \Delta x $ and its height is $ f(x_i) $, where the $ x_i $ is the left-most $ x $ value of the $ i $th rectangle under the curve.

The basic idea is, in order to find the surface area of a complex shape, we use simpler shapes to approximate it. To get more accurate answers, we take a sum of multiple such rectangular divisions. The sum of these simpler areas is known as a Reimann sum.

\[\sum_{i=1}^{n}f(x_i^\ast)\ \Delta x\]In fact, if we let the number of rectangles approach infinity, we will get an exact answer:

\[\lim_{n \to \infty}\sum_{i=1}^{n}f(x_i^\ast)\ \Delta x\]We’ve also learned a special name for this limit, its a definite integral!

\[\int_{a}^{b}f(x)\ dx = \lim_{n \to \infty}\sum_{i=1}^{n}f(x_i^\ast)\ \Delta x\]## I smell a relationship

After understanding how Reimann sums help solve the problem of finding the surface area under curves, it isn’t too much of a stretch then to take this concept and apply it to the problem of finding the volume of a Steinmetz solid. Instead of using rectangles as the fundamental divider, we will instead use boxes to divide up the Steinmetz solid. Because the cross section of the Steinmetz solid is a square, we can use a simple circle to diagram this idea, keeping in mind that we must square the length to get the volume.

Here we can see the volume is made up of 4 boxes with a height of 0.5in. One box has a length and width of $ 2\sqrt{\frac{5}{4}} $ and the other $ 2 \sqrt{2} $. So our new, better, guess for the volume can be found by adding the volumes of these 4 boxes:

\[\begin{aligned} A& \approx2\left[0.5\left(2\left(\sqrt{\frac{5}{4}}\right)^2\right)\right] + 2[0.5(2\sqrt{2})^2]\\ & =2\left[\frac{10}{4}\right] + 2[4]\\ & = 5 + 8 = 13\text{in}^3 \end{aligned} \]By adding more boxes, we get a better approximation of $ 13\text{in}^3 $. To get to an exact answer, we can divide our solid into an infinite number of boxes and find their sum. This would be an application of the Reimann sum in a 3D realm.

Two totally different problems, one for finding surface areas and another for finding volumes, using the same strategy to arrive at an answer, that can’t be just a coincidence. In fact, it implies there might be a relationship between surface areas and volumes. Earlier in the semester, when we were first learning to take derivatives, we were asked to find the derivative of the general equation for a volume of a sphere.

\[\begin{aligned} V& =\frac{4}{3}\pi r^3\\ \frac{d}{dr}(V)& =\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)\\ \frac{dV}{dr}& =4\pi r^2 \end{aligned} \]That’s right, the surface area is a derivative of the volume! We’ve already established a relationship between these two concepts, one that can be expressed unequivocally in the language of mathmatics. Looking at our problem, the end goal is finding the volume, so we must have in our data somewhere the ability to construct a general equation for the surface area at any height of the solid.

Let’s start with what we are given, and with what we know. From the video, we know our surface area is a square. This we can represent as the length of any side, squared:

\[A = l^2\]So now, how do we determine the side $ l $? Lets look at the solid from the side, which is effictively same as a cross section of a cylinder.

In this case we already know that the radius of the can is half of its 3in diameter, $ r = \frac{3}{2}\text{in} $. Therefore, we can find the length $ l $ through the following,

\[l = 2b\] \[\begin{aligned} a^2 + b^2& =r^2\\ a^2 + b^2& =\frac{3}{2}^2\\ b^2& =\frac{9}{4} - a^2\\ b& =\sqrt{\frac{9}{4} - a^2} \end{aligned} \] \[\begin{aligned} l = 2b = 2\sqrt{\frac{9}{4} - a^2}\]Putting that back into the area function gives the following general function:

\[\begin{aligned} A& =l^2\\ & =\left(2\sqrt{\frac{9}{4} - a^2}\right)^2\\ & =4\left(\frac{9}{4}-a^2\right)\\ & =9-4a^2 \end{aligned} \]## Let’s get clever

Lets look at what we have here. The equation for the surface area, $ A = 9-4a^2 $, is in terms of $ a $. Referring back to the graph, $ a $ represents a height from the center of a solid positioned at $ (0,0,0) $. Therefore, the equation gives us a surface area of a slice of the solid at any given height. In this case, the heights can range from $ -1.5\text{in} $, the bottom of the solid, to $ 1.5\text{in} $, the top.

In addition, we’ve identified a relationship between surface area and volume, that the surface area is the derivative of the volume. Since we have an equation for surface area, we need to “go back” to a volume equation. The tool we’ve just learned for doing this in class, undoing the derivative process, is the integration process. Here, we’re looking for a specific answer, namely the volume of our Steinmetz solid, so we want to set up a definite integral. We know the equation we’re integrating, as well as its bounds, so lets solve this thing!

\[\begin{aligned} \int_{-\frac{3}{2}}^{\frac{3}{2}} (9-4a^2)\ da& =\int_{-\frac{3}{2}}^{\frac{3}{2}} (9-4a^2)\ da\\ & =\int_{-\frac{3}{2}}^{\frac{3}{2}} 9\ da - \int_{-\frac{3}{2}}^{\frac{3}{2}} (4a^2)\ da\\ & =9a\Big|_{-\frac{3}{2}}^{\frac{3}{2}} - \frac{4a^3}{3}\Big|_{-\frac{3}{2}}^{\frac{3}{2}}\\ & =27-9\\ & =18\text{in}^3 \end{aligned} \]So there we have it, the area of our holiday jellied intersection is $ 18\text{in}^3 $. Pretty neat!

## Conclusion

By leveraging the skills aquired throughout the semester, we’re able to solve a problem more complex than the ones we are normally use to. But, after breaking the problem down into its individual components, was it really all that hard to solve? The difficulty of our task is largely defined by our perspective. Tasked with this same problem before this semester, the best we could likely do from that vantage point is estimate. But, given our new tools of differentiation and integration, we can approach problems from a new perspective.