Cylinders

Priscilla Crespo { pcrespo88@yahoo.com }
Math 213 -- Mathematics with Mathematica, Fall 2009
Decemeber 3, 2009

Abstract

This project will focus on the different aspects of cylnders.  It is hard to just imagine what 3D objects really look like without having them drawn properly or directly in front of you; for this reason we are lucky to have Mathematica.  I will examine how the volume and surface area are affected when different factors of the cylinder are changed.  Next, I will try to find how many cylinders fit inside of the largest cylinder, which has a radius and height both of 25.  After that, I will see what is the total surface area that can be consumed in a cylinder of height and radius both of 25.  The main commands used in this project was Manipulate and Table.  These commands allowed me to see the different cylinders made depending on the variables that changed and the information obtained.

Section 1 : Introduction

I set the volume and surface area as functions that depend on the height and radius of the cylinder.

v[r_,h_]:=N[Pi*h*(r^2)]

sa[r_,h_]:=N[2*Pi*(r^2)+2*Pi*r*h]

I wanted to calculate what the largest possible volume and surface area would be when the radius and height are both 25.  I did this just to make sure that mathematica was doing what I wanted it to do (and that I was putting in the correct functions).

One problem I had with adding text was when it was directly next to a compuatation.  What Mathematica did was it output the computation followed by the text.  To fix this problem I used <> ToString which creates the computation within the text.

"The largest volume will be: "<>ToString[N[Pi*25*(25^2)]]

"The largest surface area will be: "<>ToString[N[2500*Pi]]

This is the basic manipulation of a cylinder where the radius and height range from 1 to 25.  Each cylinder will be centerd at the origin.  If you look at the cylinder on a 2D plane, it is just a circle that grows out of the paper (this being the z-axis, or h in this case).  In this manipulation the relation can fool the eye. It is a frame of reference.  It may seem that the variables are dependent on each other but it actually the view of the box that is changing.

Manipulate[Graphics3D[Cylinder[{{0,0,0},{0,0,h}},r]],{h,1,25},{r,1,25}]

Section 2 : Changing Volume and Surface Area

This is the revised  manipulation of the cylinders.  Notice that the Cuboid command has been added; I did this so that I could set the limitations of how large the cylinder can grow.

Manipulate[{"The Volume is " <>ToString[v[r,h]]<>" and the surface area is "  <>ToString[sa[r,h]], Graphics3D[{Red,Cylinder[{{0,0,0},{0,0,h}},r],Opacity[.1],Cuboid[{-25,-25,0},{25,25,25}]}]},{h,1,25},{r,1,25}]

Let' s take a look at the table related to the radius, height, volume, and surface area.  Using the Table command gives all the values asked for in this table.  But to make it easier to read, I included the TableForm command.

TableForm[Table[{r,h,v[r,h],sa[r,h]},{r,1,25,5},{h,1,25,5}]]

Section 3: Surface Area Relative to the Volume ... (Here there is a constant height of 25 units)

As we know from calculus if we take the sum of surface areas of cylinders with a constant height and the tops and bottoms removed we will approach the volume.  We must take into account the thickness of each cylinder, so this manipulate command shows the sum of surface areas of cylinders with a thickness going from 1 to 1/1000.  Remember the smaller the thickness gets the closer we are to the true volume.

Manipulate[Total[Table[sa2[r,25] 10^d,{r,0,25-(10^d),10^d}]],{d,0,-3,-1}]

The way we can think of this is when we use integrals to find the area under curves.  Recall that the smaller the width became for the rectangle, the closer we got to the true area.  As you can see this volume is approaching the true volume of 49087.4

Section 4 : Is there a difference between the number of cylinders which fit and the amount of surface area that is enclosed? (Now height and radius change together)

With the largest cylinder having a radius and height both of 25, my hypothesis is that 24 cylinders which differ by one unit fit inside of this "original" cylinder.

This is the diagram of cylinders inside each other.  Each of these inserted cylinders are one unit less in height and radius from the previous.

Graphics3D[{Opacity[.2],EdgeForm[Opacity[1]],Cylinder[{{0,0,0},{0,0,25}},25],Cylinder[{{0,0,0},{0,0,24}},24],Cylinder[{{0,0,0},{0,0,23}},23],Cylinder[{{0,0,0},{0,0,22}},22],Cylinder[{{0,0,0},{0,0,21}},21],Cylinder[{{0,0,0},{0,0,20}},20],Cylinder[{{0,0,0},{0,0,19}},19],Cylinder[{{0,0,0},{0,0,18}},18],Cylinder[{{0,0,0},{0,0,17}},17],Cylinder[{{0,0,0},{0,0,16}},16],Cylinder[{{0,0,0},{0,0,15}},15],Cylinder[{{0,0,0},{0,0,14}},14],Cylinder[{{0,0,0},{0,0,13}},13],Cylinder[{{0,0,0},{0,0,12}},12],Cylinder[{{0,0,0},{0,0,11}},11],Cylinder[{{0,0,0},{0,0,10}},10],Cylinder[{{0,0,0},{0,0,9}},9],Cylinder[{{0,0,0},{0,0,8}},8],Cylinder[{{0,0,0},{0,0,7}},7],Cylinder[{{0,0,0},{0,0,6}},6],Cylinder[{{0,0,0},{0,0,5}},5],Cylinder[{{0,0,0},{0,0,4}},4],Cylinder[{{0,0,0},{0,0,3}},3],Cylinder[{{0,0,0},{0,0,2}},2],Cylinder[{{0,0,0},{0,0,1}},1]}]

Now I will make a table of the cylinders is this diagram so that I can add up how many there are in total : Notice that the step size is 1, this is because the radius and height differ by 1 unit each time.

t1=Table[Cylinder[{{0,0,0},{0,0,i}},r],{i,1,25,1},{r,1,25,1}];

(*I used the semicolon so that the long table would not appear*)

Length[t1]

My hypothesis was correct!!

This would be the added surface areas including the tops and bottoms of the cylinder: Notice that I am excluding the last cylinder so that I only get the volume inside of the outer-most cylinder.

s1=Table[sa[r,r],{r,1,24,1}]

Now I want to add up the total surface area enclosed :

Plus@@s1

Now what would happen if they differ by .5 units, my hypothesis is that twice as many cylinders would fit.  Let's see if I'm correct.....

This is the insertion of cylunder with the difference of .5 units (in the height and radius) rather than a whole unit.

Graphics3D[{Opacity[.1],EdgeForm[Opacity[1]],Cylinder[{{0,0,0},{0,0,25}},25],Cylinder[{{0,0,0},{0,0,24.5}},24.5],Cylinder[{{0,0,0},{0,0,24}},24],Cylinder[{{0,0,0},{0,0,23.5}},23.5],Cylinder[{{0,0,0},{0,0,23}},23],Cylinder[{{0,0,0},{0,0,22.5}},22.5],Cylinder[{{0,0,0},{0,0,22}},22],Cylinder[{{0,0,0},{0,0,21.5}},21.5],Cylinder[{{0,0,0},{0,0,21}},21],Cylinder[{{0,0,0},{0,0,20.5}},20.5],Cylinder[{{0,0,0},{0,0,20}},20],Cylinder[{{0,0,0},{0,0,19.5}},19.5],Cylinder[{{0,0,0},{0,0,19}},19],Cylinder[{{0,0,0},{0,0,18.5}},18.5],Cylinder[{{0,0,0},{0,0,18}},18],Cylinder[{{0,0,0},{0,0,17.5}},17.5],Cylinder[{{0,0,0},{0,0,17}},17],Cylinder[{{0,0,0},{0,0,16.5}},16.5],Cylinder[{{0,0,0},{0,0,16}},16],Cylinder[{{0,0,0},{0,0,15.5}},15.5],Cylinder[{{0,0,0},{0,0,15}},15],Cylinder[{{0,0,0},{0,0,14.5}},14.5],Cylinder[{{0,0,0},{0,0,14}},14],Cylinder[{{0,0,0},{0,0,13.5}},13.5],Cylinder[{{0,0,0},{0,0,13}},13],Cylinder[{{0,0,0},{0,0,12.5}},12.5],Cylinder[{{0,0,0},{0,0,12}},12],Cylinder[{{0,0,0},{0,0,11.5}},11.5],Cylinder[{{0,0,0},{0,0,11}},11],Cylinder[{{0,0,0},{0,0,10.5}},10.5],Cylinder[{{0,0,0},{0,0,10}},10],Cylinder[{{0,0,0},{0,0,9.5}},9.5],Cylinder[{{0,0,0},{0,0,9}},9],Cylinder[{{0,0,0},{0,0,8.5}},8.5],Cylinder[{{0,0,0},{0,0,8}},8],Cylinder[{{0,0,0},{0,0,7.5}},7.5],Cylinder[{{0,0,0},{0,0,7}},7],Cylinder[{{0,0,0},{0,0,6.5}},6.5],Cylinder[{{0,0,0},{0,0,6}},6],Cylinder[{{0,0,0},{0,0,5.5}},5.5],Cylinder[{{0,0,0},{0,0,5}},5],Cylinder[{{0,0,0},{0,0,4.5}},4.5],Cylinder[{{0,0,0},{0,0,4}},4],Cylinder[{{0,0,0},{0,0,3.5}},3.5],Cylinder[{{0,0,0},{0,0,3}},3],Cylinder[{{0,0,0},{0,0,2.5}},2.5],Cylinder[{{0,0,0},{0,0,2}},2],Cylinder[{{0,0,0},{0,0,1.5}},1.5],Cylinder[{{0,0,0},{0,0,1}},1],Cylinder[{{0,0,0},{0,0,.5}},.5]}]

Now I will make a table of the cylinders is this diagram so that I can add up how many there are in total : Notice that the step size is .5, this is because the radius and height differ by .5 units each time.

t2=Table[Cylinder[{{0,0,0},{0,0,i}},r],{i,1,25,.5},{r,1,25,.5}];

(*I used the semicolon so that the long table would not appear*)

Now let' s see how many time cylinders fit if the radius and height differ by one unit.

Length[t2]

Notice that my hypothesis here was incorrect.  This is because the cylinders radii can be 1, 1.5, 2, 2,5, ... , 24, 24.5, 25  --- but there is no 25.5 , this is why there is one less.

This is the surface area of these cylinders :

s2=Table[sa[r,r]*.5,{r,1,24.5,.5}]

Plus@@s2

... ... Just for curiosity let' s see what happens when the height and radius differ by .1 units :

t3=Table[Cylinder[{{0,0,0},{0,0,i}},r],{i,1,25,.1},{r,1,25,.1}];

Length[t3]

s3=Table[sa[r,r] .1,{r,1,24.9,.1}];

Plus@@s3

s4=Table[sa[r,r],{r,1,25,.01}];

Plus@@s4

As the difference in height and radius decreases, I noticed that the ' volume' is increasing when adding up the surface areas.  Something has to be wrong.  I think the problem is that the tops and bottoms are adding extra volume. 


I decided to take the approach where I did not include the tops and bottoms of the cylinders: In doing this I realized that I was getting a smaller volume because this was creating the image of an hour glass.  Since the heights are changing at a constant rate, there are two cones missing (the top and the bottom).

sa2[r_,h_]:=N[2*Pi*r*h]

(*This is the equation of the surface area excluding the tops and bottoms*)

Table[sa2[r,r],{r,1,24,1}]

Plus@@Table[sa2[r,r],{r,1,24,1}]

(*As you can see this volume is much smaller than the actual volume of the cylinder with height and radius of 25*)

Table[sa2[r,r] .5,{r,1,24.5,.5}]

Plus@@Table[sa2[r,r] .5,{r,1,24.5,.5}]

Section 5 : Discussion, Conclusion and Future Work

Summary

Well my project did not go as planned.  What I got from this project is that "supposedly" there is an infinite amount of volume.  But in my previous math courses I have learned that there is a finite amount of volume in a given containter.  But what this project has shown me is that the smaller the variables get the more "space" there is.  
	At the beginning of this proect I was just going to investigate the volume and surface area of different cylinders.  As you would assume, the volume and surface area go hand in hand.  As the volume increase so does the surface area, and vice versa; the negative is also true.  So I decided to take this to the next level and investigate how many cylinders would be able to fit perfectly inside of a larger cylinder with a set volume.  It turns out that this question is based on calculus.  We know that there cannot be an infinite number of cylinders which fit into a given space ... ... so how many is that going to be?  This is actually the limit.  Does this limit ever stop?  The smaller the radius and height become, the more cylinders will be able to fit.  But technically I can make these variables as small as I'd like to.  So the answer to this question is that the amount of cylinders which fit will approach infinity.  I came to the conculsion that I can fit infinetly smaller cylinders into a larger one with a set volume.
	The volume in a given cylinder is actually the surface areas of all the smaller cylinders added up.  Theoretically, the sum of the surface areas of a bunch of smaller cylinders inside of a given larger one is finite.  When I added up the sums of the surface areas I got sums that were different from the largest given volume.  I do not know why this is.  If I had more time to investigate this, I would try to find out why is it that the surface areas do not add up to this maximum volume.

What I learned

But I did learn how to use several useful commands in Mathematica.  One problem I had was with adding text when it was directly next to a compuatation.  What Mathematica did was it output the computation followed by the text.  To fix this problem I used <> ToString which creates the computation within the text.  I learned how to use the Plus @@ command that adds up all the elements in a given list.  I finally learned how to work with Manipulate properly.  This is good because this command is very usefull when investigating how this change visually.  I also learned how to use Graphics3D which allowed me to see different graphics in a 3 D format rather than in 2 D.  Inside of my Manipulate command, I was able to insert the  Opacity [ ] command which allows you to make certain graphics more and less transparent.  Using the different opacity levels is helpful so that you can see through specified graphics. Another command that I added to my Manipulate was EdgeForm which allows there to be specified fillings to edges of polygons and other graphics.  This was helpful in my project because it allowed the edges of each cylinder to stand out from one another.

For the Future

I would like to investigate where my project went wrong.  Apparently, the formulas I was putting in when adding up the different surface areas with the heights changing were incorrect in some way.  Where did it go wrong.  Something needs to be changed in this formula, whether it be the height, radius or thickness.  My guess is that the height might be the biggest issue here.  With the heights changing as well, the tops and bottoms of the cylinders where overlapping in a way that mad the sums exceed the true volume.  Maybe the height needs to be in halves or reduced by 2d.  I'm not sure exactly what needs to be done.  But if you find out before I do, please let me know.  Good luck!

*******From these two diagrams, you could see that the amount of cylinders which fit depend on the size of their radius and height.  Some would think there is and infinite number that fit, but in reality there is only a finite amount of space inside of the cylinder. ******** The same goes for the surface area. *********