Finding the Surface Area of a Guitar with Area Under the Curve
My Yamaha F600 guitar is something that I play frequently. Upon jamming out with it the past few days, I began to wonder how the intricately crafted curves of the guitar could be applied to a graph, and from there, how the surface area of the entire guitar could be estimated using “area under the curve.” This then became the subject of a project for my Honors Precalculus class at Concordia International School Shanghai. After completing various calculations, these are my findings.
Graphing the Guitar
In order to apply “area under the curve” to a guitar, we must understand one simple fact: the guitar is symmetrical. Therefore, if we split the guitar directly down the middle horizontally, we would only need to find the area of half the guitar, and then double it.
I added the picture of my guitar into Desmos and then resized it so that one unit on Desmos would be equivalent to 4cm in real life. This would then mean that one square unit in Desmos is equivalent to 16 square centimeters, and this laid the foundation upon which I could convert graphs on screen into real life measurements. I shifted and rotated the image so that the x-axis sliced the guitar in half and the end of the guitar’s body was right at the point (0,0).
Next, to graph the shape of the body, I decided to break it down into smaller parabolas and linear lines. To find the parabolic equations, I added three points into Desmos that I thought would fit a section of the guitar, and then I noted the coordinates of these points. I then derived a parabolic equation from these three points by using a systems-of-equations-solving calculator. This is possible because parabolas are graphed with the formula y=ax²+bx+c, in which there are three constants. Thus, if you have 3 inputs for x and y, a system of equations will find the a, b, and c values that fit this curve.To find the linear equations, I added two points that fit the body of the guitar and determined the slope using (y1-y2)/(x1-x2). I then plugged in the slope value and the coordinates of one the points to solve for b, the y-intercept of the linear equation.
After finding these parabolic and linear equations, the next crucial step was determining a domain in which this graph would appear. Because the overall graph of the guitar would consist of small chunks of different equations, it was imperative that I only selected the needed portions. For the lower bound, the equation would begin at the x-value at which the previous equation ended. For the upper bound, the equation would end at the x-value at which it began veering off the desired curve of the guitar.
I repeated this process several times until I had graphed the entire upper half of the guitar’s body, reaching a total of 9 equations: 7 parabolic equations and 2 linear equations.
The process for the neck and head of the guitar was similar. Since this part of the guitar was composed of straight lines, I only needed to find linear equations. On the other hand, to graph the shape of the sound hole, I used the equation for a circle and centered it at (0,3.5) by referencing the center of the sound hole in the image. I then adjusted the r, the radius, until it fit almost perfectly. Upon adjusting domain values with the aforementioned method, I had a complete graph of the guitar, totaling 14 equations.
The Top and Bottom of the Body
I decided to start with tackling the most complicated curve of the guitar: the body. The area of the entire curve would be equal to the sum of the areas under each of the 9 smaller equations.
To begin, I decided to estimate each area under the 9 equations using 5 rectangles, 10 rectangles and 20 rectangles. This was a process that I had learned about in my Honors Precalculus class.
To do so, I used the following formula:
This formula takes the curve for the function f(x) within the interval bounded by a and b and divides that into n rectangles. Then, it takes the sums of all these rectangles to approximate the area under the curve. Recalling that the area of a rectangle is its length times its width, the equation makes sense as it can still be broken down in this manner. The width of each rectangle is equal to (b-a)/n as you are taking the distance and breaking it into these equally wide chunks. The lengths, or heights, of the rectangles is equal to the summation of f(a+(b-a)i/n) since the sum is determining the y-values at each increment of (b-a)/n and adding them together to form a collective height.
I inputted 5, 10 and 20 for n into each of the nine equations and inputted their domains for a and b. For reference, here are my handwritten calculations:
By adding up all these estimated areas, I found that when each section was broken down into 5 rectangles, 10 rectangles and then 20 rectangles, the total area was 60.267 units², 60.217 units² and 60.157 units² respectively. Multiplying each of these values by 16 to convert them into square cm, we get 964.272 cm2, 963.472 cm2 and 962.512 cm² respectively.
Although using n rectangles can provide a rough estimation of the area under the curve, it will always be slightly overestimating or slightly underestimating the true area. I will go into further detail on this after explaining how to find the true area using infinite rectangles.
The process of using infinite rectangles actually seems very similar to using n rectangles, but unlike the latter, using infinite rectangles requires applying limits. The new formula looks like this:
The reason that the limit is applied is because there is no practical way in doing the algebraic and arithmetic calculations with the value of infinity itself. Rather, the best course of action is to evaluate the summation in terms of n; then, the limit is applied to cancel out terms that become extremely small and therefore negligible as n approaches infinity. To do so, I had to apply the summation formulas:
Upon evaluating the summation, there will usually be one term in the numerator and one term in the denominator with n raised to the highest power. Let’s call this n^h. Dividing the numerator and denominator by n^h, the coefficient of n^h should now remain on its own while the other terms become divided by n^h. When the limit is applied, the n^h becomes infinity, so all of these terms become infinitely small and negligible. Now, only the coefficients of n^h in the original equation remain. Adding up the terms that are not canceled out by the application of the limit yields the area under the curve for infinite rectangles. For reference, here are my handwritten calculations:
Adding these values together, I got a total of 60.084 units², and multiplying this by 16 to convert it to square cm yields 961.344 cm². Doubling this to get to the other half of the body, I found that one surface of the body was equal to 1922.688 square cm².
I added all the values that I calculated between 5, 10, 20 and infinite rectangles to see if I could determine a general trend as the number of rectangles increased. Here is the table that I created:
In order to determine the percent accuracy of how much an area calculation overestimated or underestimated, I divided the area estimation for each equation and n value by the area for infinite rectangles and then subtracted that by 1. I then converted this decimal into a percentage. Predictably, the percent of overestimation or underestimation declines substantially when the n value increases, proving that getting closer and closer to infinite rectangles will yield more and more accurate results.
Some equations overestimated the area while others underestimated because it depended on whether the function itself had an upward or downward slope. All the calculations formed rectangles based on points to the right. Thus, when the equation had an upward slope, the area was an overestimate, and when the equation had a downward slope, the area was an underestimate. Overall, it appears that the n rectangle calculations yielded an overestimate.
Moreover, the percent of overestimation or underestimation seems to form an inverse function with the number of rectangles, although the resemblance is closer in some instances than others.
I realized after conducting the calculations that I had applied the same number of rectangles to different domain sizes as, within the 9 equations, the intervals varied from 0.1 to 5.5. Clearly, the accuracy would depend on how widely you stretch the rectangles over this range, so that caused some big discrepancies in accuracy percentages between equations. The overall curvature of the equation also played a major role in determining how accurate the estimations were.
The Sound Hole
Looking at the guitar, there is clearly a big sound hole in the center of the body that we must take into account. I attempted to integrate the area under the semicircle I graphed, but the answer came out to be undefined. Instead, I decided to apply the classic area of a circle formula:
The two endpoints of the graph were (0, 2.158) and (0, 4.842). The distance between these two points is 2.684, which is the diameter. Dividing that in half, I got a radius of 1.342 units. Plugging that into the equation, I arrived at an area of 5.658 units² for the sound hole, which is equal to 90.528 cm².
Now, putting together all the values I had calculated thus far, I could find the surface area of the top and bottom surfaces of the body. The bottom surface, which is pure wood, is equal to 1922.688 cm². The top surface, which has the sound hole in it, is equal to
1922.688 square cm²–90.528 cm², totaling 1832.16 cm². The top and bottom surfaces therefore have a combined area of 3754.848 cm².
The Sides of the Body
I recognized that the sides of the body was essentially one long rectangle that was wrapped around the curves of the body. If I imagined myself wrapping a string around this part of the guitar and then I straightened out the string, that would be equivalent to the width of the rectangle. The length, which is constant around the guitar, was measured to be 11 centimeters.
Applying the string idea, I realized that I needed to find the arc length of the curve I had previously graphed. This was the equation for determining the arc length of a curve:
To carry this out, I needed calculus knowledge beyond the scope of what my Honors Precalculus class had taught me. So I opted to use Desmos instead, which gave me answers that looked like this:
Summing up the arc lengths that I calculated with Desmos, I found that the total arc length of the curve was 21.038 units, giving me 84.152 cm. I then got 168.304 cm by doubling this to get the length around the entire body. Multiplying this length by the 11 cm width, the total surface area of the sides of the body came out to be 1851.34 cm².
Adding this to the top and bottom surfaces, the total surface area of the body is 5606.188 cm².
The Neck and Head
Because the neck and head are composed of purely linear lines, the area calculation could be done by simply breaking down the area into a triangle and a rectangle. From there, the area calculation could be completed using A=1/2(base)(height) and Area=length*width. Here are my handwritten calculations:
This yielded a total of 65.024 cm² and 198.08 cm² for the head and neck respectively. Adding them together and doubling it to get the top and bottom surfaces of the head and neck, I arrived at 526.208 cm².
For the sides of the head and neck, I once again used the arc length of the curve as they were mostly rectangular shapes. This gave me a total of 159.132 cm². Adding this with the previous calculations, the surface area of the head and neck totaled 685.34 cm².
The Final Answer
For simplicity, I omitted the surface area of the strings, frets, pegs and guitar ornamentations. If I were to use these calculations to determine the amount of paint I needed for a guitar, I probably would not decide to paint these parts of the guitar anyway.
Combining the 5606.188 cm² of the body and the 685.34 cm² of the neck and head, the total surface area of this guitar is 6291.528 cm² or 0.629 m².
