I’ve now been in calculus for about 23 weeks, and through all the stress, frustration and many hours of homework 4 days a week, I loath math. I’m beyond tired of integrals and differentiation. As I was sitting thinking about what interested me most in the whole 23 weeks I actually had a change in heart. The one thing I’ve learned that I found so amazing is how you can find the volume and surface area of objects. But not just any object, lots of objects. Such as Darth Vaders head. So this is why I decided to do my project on finding the volume and surface area of a Cabo Wabo Tequila bottle. It’s not like the other tequila bottles tall, straight and boring. No Cabo Wabo bottles are a teal blue color and have a short stocky shape. I’m personally not a big tequila drinker, but they do have the most interesting bottle.
Using the disk method breaks up each section into Infinite many disks and adds them all together and as long as you’re on the axis the radius is your equation. Using this method I was able to get the total volume of the bottle. Using the surface area equation which uses the formula for arc length formula and the surface area of a cylinder I obtained the total surface area also.
In order to find the volume and surface area of the bottle I needed a model which I could graph. So I placed my bottle on a piece of graph paper and traced. I then measure both ends and found the middle of the drawing and sketched a line. I then used this line as my xaxis. Then using the bottom of the bottle as my yaxis I was ready to make my measurements form the xaxis to the side of the bottle. I made measurements every centimeter. I then went to my handy TI89 under apps data matrix editor. I plugged in my 19 points and then used the calc button, which gave me the QuartReg function and I got a quadratic equation. I had it saved to Y1(X), and graphed it. As I looked and looked I realized it looked nothing like my bottle. So I decided that I needed more points. I then measured ever millimeter. I had a total of onehundred and ninetyone points. I thought well if this is accurate what will be. I then graphed the 191 points and it was closer but definitely not my bottle.
I then asked an expert and he gave me a great idea. He said break the bottle in three parts the base, the neck, and the top. So upon reentering the points in three different sections I had three separate equations. I graphed each of the equations and I now had my bottle. The first section went from {0,12} the second {12,18} and the third {18,19} along the xaxis. This was perfect I would have three integrals and be able to just add them together after integrating. I was so glad I asked an expert. Thanks by the way.
I now have three equations and two formulas time to get to work. I used the three equations in separate integrals using the disk method to find the volume ( [R(X)]^2 dx). I then entered each of the equations into the calculator with the appropriate bounds and got my volumes for each section. Adding them together gave me the total volume for the bottle. Equations follow.
V= [R(x)]^2
V1= ((.001718 x^4) + (.031841 x^3) + (.214158 x^2)+ (.686257 x)+ (4.490651))^2 dx = 304.95 cm ^3
V2 = ((.001182 x^4) + (.077599 x^3) + (1.908211 x^2)+ (20.831546 x) + (86.480731)) ^2 dx = 11.03 cm ^3
V3= ((2.039627 x^4) + (151.340326 x^3) + (4209.035548 x^2) + (52002.049534 x) + (240815.769231)) ^2 dx = 2.70 cm ^3
V= (304.95) + (11.03) + (2.70) Total Volume of Bottle = 318.68 cm ^3
After all that I actually still wanted to know the surface area. So using my trusty TI89, my three equations, and my formula for surface area ( 2 r(x) 1 + (f (x))^2). I was on my way. I first had to find the derivatives of each of my equations. Then square the derivatives to put them in the three separate integrals using the same bounds. I then used the formula and multiply by 2 . Add them all up and I have the total surface area of my bottle. I had a little problem with the last one though. I typed in the whole equation to get a value and for about fortyfive minutes all my calculator said was busy. Finally it gave me an answer, but I was worried I might have broken it. Equations follow.
S = 2 r(x) 1+ (f (x))^2
S1 = 2 ((.001718 x^4) + (.031841 x^3) + (.214158 x^2) + (.686257x) + (4.490651)) ( 1 + ((.006872 x^3) +(.095523 x^2) + (.428316 x) + (.686257))^2) dx = 420.372 cm^2
S2 = 2 ((.001182 x^4)+ (.077599 x^3) + (1.908211 x^2) +(20.831546 x)+(86.480731)) ( 1+((.004728x^3) + (.232797 x^2)+(3.81642x)+ (20.831546)) ^2) dx = 51.411 cm^2
S3 = 2 ((2.039627 x^4) + (151.340326 x^3) + (4209.035548 x^2) + (52002.049534 x) + (240815.769231)) ( 1 + (( 8.15815 x^3) + (454.021 x^2) + (8418.07 x) + (52002.0495))^2 dx = 14.135 cm^2
S = (420.372) + (51.411) + (14.135) Total Surface area of Bottle = 485.918 cm^2
I have created a graph of the three functions color coded to match the color coding on the equations. So this concludes my project. Although I’m tired of the endless homework I have to say this project has shown me that I can actually apply this to areas in real life. I was starting to feel like I was wasting my time and I’d never use this stuff again especially all the trig. So now you know the volume and surface area of a Cabo Wabo Tequila bottle. Just in case you ever think about drinking a whole one yourself, you might want to think a bit more about it.
