6.2 Volumes

Mathematica script by Chris Parrish,
   cparrish@sewanee.edu

Sources and references for some of these problems include
    James Stewart, "Calculus: Concepts and Contexts," Second Edition, Brooks/Cole, 2001
    Deborah Hughes-Hallett, Andrew M. Gleason, et. al., "Calculus," Second Edition, John Wiley & Sons, 1998
    Robert Fraga, ed., "Calculus Problems for a New Century," The Mathematical Association of America, 1993
    Selwyn Hollis, "CalcLabs with
Mathematica" for Stewart's "Single Variable Calculus, Concepts and Contexts, Second Edition," Brooks/Cole, 2001

Volume of a Solid of Revolution: The Trigonometric Football

Reference: In this example we follow the programming style used by Selwyn Hollis in "CalcLabs with Mathematica," Section 5.2, "Volume."

Let's calculate the volume of the solid of revolution obtained by revolving one loop of the Sin function around the x-axis.

In[73]:=

<<Graphics`Colors`

In[74]:=

Clear[f, g, x]  f[x_] := Sin[x]  Plot[f[x], {x, 0, π}, PlotStyle PermanentGreen, Background->Mint] ;

[Graphics:HTMLFiles/6.2_volumes_3.gif]

Mathematica has a "FilledPlot" function for displaying such regions.

In[77]:=

<<Graphics`FilledPlot`

In[78]:=

FilledPlot[f[x], {x, 0, π}, FillsCinnabarGreen, Background->Mint] ;

[Graphics:HTMLFiles/6.2_volumes_6.gif]

A vertical cross-section which cuts through the solid of revolution at the point x is a disk with radius f[x], hence has area π f[x]^2, for 0 ≤ x ≤ π.

In[79]:=

a[x_] = π f[x]^2

Out[79]=

π Sin[x]^2

Therefore the volume of the solid of revolution is given by

In[80]:=

∫_0^πa[x] x %//N

Out[80]=

π^2/2

Out[81]=

4.9348

Here's a view of the Trigonometric Football.

In[82]:=

<<Graphics`SurfaceOfRevolution`

In[83]:=

SurfaceOfRevolution[Sin[x], {x, 0, Pi}, RevolutionAxis {1, 0, 0}] ;

[Graphics:HTMLFiles/6.2_volumes_15.gif]

Volume of a Solid of Revolution: The Trigonometric Donut

Reference: In this example we follow the programming style used by Selwyn Hollis in "CalcLabs with Mathematica," Section 5.2, "Volume."

Let's first create a region in the xy-plane bounded above by one loop of the Sin function and bounded below by a corresponding loop of -Sin.

In[84]:=

Clear[f, x]  f[x_] := Sin[x] g[x_] := -Sin[x]  Plot[{f[x], g[x]}, {x, 0, π}, PlotStyle Peru, Background->Mint] ;

[Graphics:HTMLFiles/6.2_volumes_17.gif]

Mathematica has a "FilledPlot" function for displaying such regions.

In[88]:=

FilledPlot[{f[x], g[x]}, {x, 0, π}, Background->Mint] ;

[Graphics:HTMLFiles/6.2_volumes_19.gif]

Now let's rotate this region about the y-axis, creating a sort of "Trigonometric Donut."
A vertical cross-section which cuts through the template region at the point x will sweep out a ring with volume dv = 2πx (f[x] - g[x]) dx, for 0 ≤ x ≤ π.

In[89]:=

v[x_] = 2π x (f[x] - g[x])

Out[89]=

4 π x Sin[x]

Therefore the volume of the solid of revolution is given by

In[90]:=

∫_0^πv[x] x %//N

Out[90]=

4 π^2

Out[91]=

39.4784

Here is a view of the "Trigonometric Donut."

In[92]:=

RowBox[{RowBox[{top, =, RowBox[{SurfaceOfRevolution, [, RowBox[{Sin[x], ,, {x, 0, Pi}, ,, > ... ty}], ]}]}], ;}], }] Show[top, bottom, DisplayFunction$DisplayFunction] ;

[Graphics:HTMLFiles/6.2_volumes_26.gif]


Created by Mathematica  (April 29, 2004)