Math 166
Computer Lab #5
Applications of Integration I

Goal:  To investigate applications of volumes of solids of revolution.

Solids of equal Volume

A striped watermelon has an ellipsoidal shape that is formed by revolving the curve y = Sqrt[a^2 - x^2]/2 for x between -a and a about the x-axis.  The length 2a feet is the length of the watermelon along the long axis.  A cross section of such a melon with a = 1 (that is with long diameter 2 feet) is  given below.

Clear[a]
a=1;
Plot[{Sqrt[a^2 - x^2]/2, -Sqrt[a^2 - x^2]/2},
{x, -a, a}, AspectRatio->Automatic,
Ticks->{{-a, a}, {-.5a, .5a}}]

Note that the number "a" simply scales the melon the same in all directions.  To see this, plot the  cross section of a melon whose long diameter is 3 feet (that is take a = 3/2) and a giant melon whose long diameter is 5 feet.  To check your results double click on the arrow below.

Clear[a]
a=3/2;
Plot[{Sqrt[a^2 - x^2]/2, -Sqrt[a^2 - x^2]/2},
{x, -a, a}, AspectRatio->Automatic,
Ticks->{{-a, a}, {-.5a, .5a}}]

Clear[a]
a=5/2;
Plot[{Sqrt[a^2 - x^2]/2, -Sqrt[a^2 - x^2]/2},
{x, -a, a}, AspectRatio->Automatic,
Ticks->{{-a, a}, {-.5a, .5a}}]

The volume of the melon can be found by the method of slices where we take slices perpendicular to the x-axis.  It is given by

Clear[a]

volmelon = Pi*Integrate[(Sqrt[a^2 - x^2]/2)^2, 
{x, -a, a}]

We want to slice the melon perpendicular to its long axis so that we get three pieces of equal volume.  (Don't worry about the fact that some of the slices have more rind than others.)  A cross-section of the melon is shown below.  We want to choose p such that the volume of the slice generated by rotating the cross sectional region from x = p to x = a about the x-axis is one-third the total volume of the melon.  (Alternately, we could consider revolving the section from x = -p to x = p about the x-axis.) That is, the volume of the slice should be Pi a^3/9.

The volume of the slice formed by rotating the section from x = p to x = a can easily be found by the disk method.  We have

vol3 = Pi*Integrate[(Sqrt[a^2 - x^2]/2)^2, {x, p, a}]

We note that this is a cubic equation for our unknown p.  Recall that a is one-half the length of the long diameter of the melon and we are looking for the percent of distance from the center.  As we noted earlier the parameter a simply scales the melon by the same factor in all directions.  Thus, without loss of generality we can set a = 1.  In order to determine p, all we have to do is set the volume above equal to one-third the total volume of the melon and solve for p.  We have

a=1;
N[Solve[vol3==volmelon/3, p]]

Note that we solved a cubic equation so we have 3 roots, all of which are real.  (We take 10-19 â 0.)  We must determine which of the roots is physically reasonable.  Since only one of the roots is positive and less than 1 and the half-diameter is 1, we should cut it at ± .226074 feet from the center.  For a melon whose long diameter is 2a feet, we should cut it approximately 22.6% away from the center on each side.  Below we plot a cross-section of the cut melon that shows the cuts.

Plot[{Sqrt[1-x^2]/2,-Sqrt[1-x^2]/2},{x,-1,1},
AspectRatio->Automatic,
  PlotStyle->RGBColor[1.000, 0.382, 0.130],
  PlotLabel->"Cross Section of cut melon; 3 equal pieces",
  Ticks->{ {{.25,"p"},-1,1}, {-.5, .5} },
  Epilog->{{Line[{{.226,-.5},{.226,.5}}],
           Line[{{-.226,-.5},{-.226,.5}}]}}]

In the exercises you will determine where to cut the watermelon so that we have 4 or 5 pieces of equal volume.

Problems (to be handed in)

1.  Consider the striped watermelon we had before whose long diameter is 2 feet (that is a = 1).  Determine where we should cut the melon (perpendicular to the long axis) in order to have 4 pieces of equal volume.  Draw a cross section of your melon with the cut lines drawn.  (To see how to generate the cross section plot with the cut lines, open the closed group before the plot of the cross section of our cut melon.)

2. Consider the same striped melon with long diameter 2 feet.  Determine where we should cut the melon (perpendicular to the long axis) in order to 	have 5 pieces of equal volume.  Draw a cross section of your melon with the cut lines drawn.

3. Suppose you want to design a pan, such as a wok, for stir frying food.  Assume that  we can view the pan as being formed by taking a sphere of radius 5 inches and slicing off a portion as shown in the graph.  Determine the radius and height of the wok which has a volume of 2.5 quarts. (1 quart = 57.75 cubic inches)

4.  Suppose that we want to determine the volume of a three-dimensional object which is 5 feet long and has a cylindrical hole of radius 1 foot down its middle.  If we choose the x-axis to lie along the center of the hole running through the object, then the object is symmetric about the x-axis and we can determine its volume by rotating a cross-sectional area about the x-axis.  However, instead of having an explicit function which determines the shape of the cross section, we have a set of data.  The data for the region which generates the outer boundary of the object is given in the lists below.  The hole is generated by rotating the line y = 1 about the x-axis.

(a)  Plot the cross section.  

(b)  Use this data to approximate a lower and upper bound for the volume of the solid.  In particular, use a right-hand and left-hand Riemann sum to approximate the cross-sectional area and rotate each rectangle to form a washer.

(c)  If you later determine that the outer boundary of the cross section can be explicitly determined by the function f(x) = 1.5 + Sqrt[x], calculate the exact volume of the object and compare it to your approximations.

datax={0, .5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5,
5.0};
datay={1.5, 2.1, 2.5, 2.8, 3.0, 3.1, 3.3, 3.4, 3.45,
3.55, 3.7};
data=Table[{datax[[i]], datay[[i]]}, {i, 1, 11}];