Napkin Ring Problem Analysis

Introduction

How many of you have heard of a Napkin Ring? Well for those of you who have no idea I would like for you to imagine a sphere. A circular hole is then drilled through this sphere and whatever remains is called a Napkin Ring. Napkin Rings are fascinating to every mathematician. Do you know that all Napkin Rings that have the same height have the same volume?

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This is a fascinating paradox. Despite the radius of any sphere, as long as it is reduced to a napkin ring of a given height, it would have a volume equal to an entirely different napkin ring of the same height from a different sphere. Because of this fantastic discovery, we are going to look at two different ways of proving that napkin rings from different spheres but have the same height have similar volume. We are going to analyze and validate the cross-section area method and method of cylindrical shells to calculate the size of different spheres.

For you to fully understand these methods, you must have gone through the basics of calculus. By that, I mean you know integration methods. You should be able to integrate any differential equation given to you. It does not matter if you are not fluent in all ways, but as long as you understand the principles of integration, then this would be very easy for you.

Now let us get to business. Imagine you have a sphere of radius R with a cylindrical hole of height h drilled through it. The shell you have is called a napkin ring. You are expected to calculate the volume of the napkin ring.

First, we are going to use the cylindrical shells method. Most of you should find this easy to use. From the above diagram, we can be able to see the differential equation for the volume of a part of the napkin ring above the x-axis. Using Pythagoras theory, we can be able to see the value of the y-axis. From the equation x2+y2=R2, finding y gives us: y2=R2-x2 resulting in y=R2-x2.

Finding a representative height of the rectangle h(x) gives us:hx=R2-x2(i). We then have to find a radius r(x) from the cylindrical wall to the sphere's surface which is:rx=x(ii). Within the range of the cylinder's radius and the sphere's radius. Therefore, finding the cylinder's radius through Pythagoras theory gives us: r2+y2=R2 where y=h2 and r=rx=x. This gives us r2=R2-y2 which is similar to x2=R2-(h2)2 giving: x=R2-(h2)2. Therefore, the radius r(x) is within the range R2-(h2)2xR(iii).

From the above equations (i), (ii) and (iii)we can be able to find the differential equation to find the volume of the napkin ring. This gives us: v=2pR2-(h2)2Rx(R2-x2)dx(iv). We then have to integrate the above equation. Using the integral principles, we shall use substitution to integrate. Taking u=R2-x2, we differentiate to get: dudx=-2x. Substituting for dx, dx=-du2x we get: v=2pR2-(h2)2R-x(u12)du2x(v). From equation(v) we can simplify it into: v=-pR2-(h2)2R(u12)du(vi). Equation(vi) can now be integrated into: v=-23p(u)32(vii) within the range R2-(h2)2xR.

By substituting for u in equation (vii), you can obtain v=-23p(R2-x2)32(viii) within the range R2-(h2)2xR. We can now simplify the equation(viii) by replacing the value for x. This results in: v=-23pR2-R232--23pR2-R2-h22232(ix). Equation(ix) is further simplified to: v=--23ph2232(x) which when further simplified results to: v=ph312(xi).

The gotten volume is for one half of the napkin ring. Thus, we have to multiply by 2 to get the final volume of the ring. Hence the capacity of the napkin ring obtained from any sphere of radius R is gotten by v=ph36(xii). The equation can be used to get the volume of any napkin ring and you can be able to see that the volume is not dependent on the radius but the height of the ring. Thus, two rings obtained from different spheres but have similar height have the same volume from equation(xii).

As you have seen the method of calculating the volume of a napkin ring is that simple when using the method of cylindrical spheres. For those of you who find the process too complicated for you or too easy, you can be able to compare with the next one; the way of cross-section area. Before we go through this method, you have to learn about a principle that is essential to the method's success. Has any of you heard of Cavalieri's principle?

Well, this guy Cavalieri was fascinated by geometrical shapes so much that he found out that two solids were of equal volume as long as the height of both rings is similar. As such, the Cavalieri's principle was proposed. This principle states that: If in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

To understand the method of cross-section area, we will have to first grasp and go through Cavalieri's principle and derive a differential equation that we will have to use to find the volume of a napkin ring.

Given a napkin ring of height h obtained from a sphere of radius R, we will have to see the cross-section area of the ring at a height dh from the x-axis. Using Pythagoras theory, we can be able to find the radius of the cylindrical ring r. This will result in: r2+h22=R2 giving the value of r as: r2=R2-h22(xiii). We can also fond the radius of the sphere r0at a height dh from the y-axis. This gives the equations: r02+dh2=R2 which gives the radius of the sphere at a height dh is: r02=R2-(dh)2(xiv).

Now that we have the radius of the sphere at a height dh and the range of the cylindrical hole, we can be able to calculate the cross-sectional area of the napkin at the height of dh. dA=pR2-(dh)2-pR2-h22(xv). By simplification, we obtain the following equation: dA=ph22-p(dh)2(xvi). With the cross-sectional area at the height dh, you can be able to obtain the volume dv by multiplying with the height dh. dv=ph22-p(dh)2dh(xvii).

Equation(xvii) when simplified forms the equation: dv=ph22dh-p(dh)3(xviii). The value -p(dh)30 therefore is neglected when it comes to integrating the equation(xviii). The volume of the napkin ring is obtained from the integration of the equation within a range of 0hh. Thus, integrating the equation gives: v=p0hh22dh. The resultant of the integral is: v=ph36. From this, you can be able to see that the volume of the napkin ring is similar regardless of the method used to calculate.

Conclusion

Thus, this far you can be able to choose for yourself the most suitable method to use in any situation. You should, however, be able to use both ways. In the case of succeeding in your examinations, I recommend that all processes be well within your grasp of mental conviction.