Abstract

Combining techniques from our CAAM 451 class, Introduction to Numerical Algebra (http://www.caam.rice.edu/~caam451/index.html), I find the volume swept out by revolving a banana around an axis. First, I construct a projector to obtain a cross-section of the volume, then I interpolate 55 data points for both the function describing the top half of the banana’s projection and the bottom half. I use a Vandermonde matrix to fit a polynomial of degree 2 to each function. Then I carry out standard techniques of elementary calculus to find the volume swept out by this area enclosed by the two functions.

Introduction

One of the greatest problems plaguing mathematicians today is the inability to express the integral of a banana in closed form. I do not provide a solution to that problem, but I try to make headway in a related problem. From time immemorial, mathematicians have been mystified by those dancers who wear fruit on their heads. More specifically, mathematicians have been bewildered by the fruit on their head. Why is she wearing fruit? Well, no one knows, so we ask another question: what is the volume swept out by that banana as she twirls around? That is the question I try to answer.

Analysis

After devouring a few test subjects, I finally found a candidate for my analysis. It was kind of mushy, but hey, it’d do. Okay, well I sure as heck don’t know how to (easily) find an equation for the banana in 3 dimensions, but I guess I don’t really need it if I just stick to revolving the nanner around the z-axis. Well I guess I’ll just take a cross-section. So how do I do that? Projectors!

Finding a cross-section using projectors:

Okay, so I placed the banana on a sheet of graph paper and traced its outline with a pen. Simple enough. Well, this is a serious mathematical paper, so I’ll make it more complicated.

What I am doing is taking a projector P and multiplying each point x in the banana by P. A matrix P is a projector if P=P^2.

Well, I’m taking my pen, holding it vertically and tracing the heck out of that banana. So pretend the banana is sitting on the x-y plane (in the positive quadrant). What I’m doing is capturing the x-y coordinates of each point on the boundary. Mathematically,

You can show for yourself that

is the projector we’re using. Whoopie doo.

Okay, so I’ve got the cross-section, so I then take data points from the graph paper. Which sucks, by the way.

Well now I got 55 points for both the bottom and top halves. Now I need to complicate this more. So I say, what the heck, let’s interpolate each with a polynomial. Because you know the shapes of bananas are governed polynomially.

How do we do this? Well it’s your lucky day, so I’ll tell you.

Interpolating the cross-section of the banana by polynomials

We use the Vandermonde-matrix method. That is, we solve this equation for the coefficients of a polynomial of degree n-1:

where are the data points. The big matrix on the left is “the” Vandermonde matrix, if you haven’t seen it before.

Funny thing about interpolating with polynomials is that higher degree polynomials don’t necessarily describe the behavior better than lower degree polynomials. But that’s cool because this banana looked like an kind of banana, anyway.

So I need to solve that thing. Well I ain’t about to solve it by hand. Time for Matlab! Well the code’s below, if you’re interested (for the love of God, don’t be that interested).

Matlab Code: banana2.m

You can get a good idea by looking at this graph of how well (how poorly?) the polynomial interpolates this banana. The green lines are the cross-section of the banana (seriously! – okay, I had to approximate at the endpoints since I wouldn’t get simple functions). The black lines are the polynomials approximating the banana.

Anyway, the coefficients I got for the quadratic polynomials defining the top and bottom are as follows:

and

Now all that’s left to do is compute

where those p’s are the polynomials with the respective coefficients above.

So, crunching that out gives you That’s cubic centimeters, btw.

Tada!

Ok, that's off by a factor of two. Just pretend you made a banana split across the y-axis and rotated that as well.

Of course, that’s a Texan banana, so you can consider that volume as an upper bound on any other bananas you find.