Curve fitting the cross sectional area of a cake. For my final Calculus project I have to find the area of a Bundt cake through the use of cross sectional areas. (Cakeulus)
While most seniors in High School who run into this popular calculus project may do the minimal amount possible; I wish to take this a step forward. 
I cut a cross section of the cake and remade it in Photoshop

Both axis are in inches. Each interval is 1 inch. The origin represents the center bottom of the plate.
I used some software to find the points of about 180 points on the curve.
Here it is on a google sheets
To find the volume of the entire cake by rotating the curve around the y axis, I need to find a function that can most accurately fit to the curve I have at hand. 
I understand that I could use something like an ellipse, but is there any other method by which I can find a curve that "perfectly" fits my data? The size of the equation would not be an issue I just want a high level of accuracy.
Many other students are just using a quadratic to keep things simple but that ignores the curved nature of the base of the cake.
Any help would be appreciated in finding the best function that I can then rotate along the y axis. 
 A: Actually, you could get an excellent approximation of the area without using a "curve" to fit your data. You say you have about $180$ points on the curve. That means you have a polygon that closely approximates the cross-section of the bundt cake. That's all you need.
Pappus's Centroid Theorem (the second one) says the volume of the cake is the product of the area of the slice and the length of the circle that the centroid makes when it revolves around the $y$-axis. You can use the shoelace formula to find the area from your polygon's points, and another, similar formula to find the location of the centroid. (Note that the centroid reference gives another formula for the area.) Finding the arc length made by the centroid is trivial, so the volume will be easy.
From the links I gave you should be able to use a spreadsheet (such as google sheets) to find the area and centroid, and the rest is easy. Ask if you need more help. This should easily give you high credit for your project. If your teacher asks how calculus was used, point to Pappus's Centroid Theorem (which was discovered by geometry but these days is used in calculus class and justified by calculus).

By the way, I exported your Data to MS Excel and got these values:
Area=$7.170317894$ (actually, the negative of that, since your points are in "backwards," clockwise order)
Centroid= $(3.072863031, 1.504220763)$
(These figures are corrected from a previous version of this answer.) I'll leave the final volume calculation to you.

If you are serious about finding "a curve that 'perfectly' fits my data" and it is true that "the size of the equation would not be an issue I just want a high level of accuracy", the polygon leads you to a perfect answer. This particular answer works since no two consecutive $x$-values in your data are equal, so none of the bordering line segments is vertical. Here is a relation, using a point-point form of the equation of a line:
$$\left(2.7263985\le x<2.8254118 \quad\text{and}\quad
 \frac{y-3.1299872}{x-2.7263985}=\frac{3.1291170-3.1299872}{2.8254118-2.7263985}\right)$$
$$\text{or}$$
$$\left(2.8254118\le x<2.9258595 \quad\text{and}\quad
 \frac{y-3.1291170}{x-2.8254118}=\frac{3.0881827-3.1291170}{2.9258595-2.8254118}\right)$$
$$\text{or}$$
$$\vdots$$
If you want to avoid the logical operators "and" and "or" there are numeric ways of simulating them, but that would make the relation even longer. Note that I did not use the usual "definition by cases" of a function since the relation is not a function: there are two values of $y$ for all values of $x$ between $1.5996108$ and $4.5847140$. You could use "definition by cases" of two functions if you use a parametric form of the relation rather than a Cartesian one, or if you define the upper and the lower curves separately.
A: You could use a Fourier series to represent the shape of the cross section of the cake 
$$
r(\theta) = \sum_{k=0}^{\infty}(a_k \sin(k\theta) + b_k \cos(k\theta))
$$
One would have to set an origin inside the cross section of the cake. With some luck, the function $r$ will be one-valued and the expansion could be used.
