6.3 Arc Length and Surface Area

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

Arc Length of a Sin Wave

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

The distance "as the crow flies" from one endpoint to the other of a Sin wave is 2π. The crow simply flies straight down the x-axis.
But how long is a Sin wave from one end to the other if the crow were to actually hike along the curve itself?

In[255]:=

<<Graphics`Colors` <<Graphics`FilledPlot`

In[257]:=

Clear[f, x]  f[x_] := Sin[x]  Plot[f[x], {x, 0, 2π}, PlotStyle CinnabarGreen, Background->Moccasin] ;

[Graphics:HTMLFiles/6.3_arcLength_3.gif]

We are asking for the arclength of a Sin wave, f[x] = Sin[x].
The element of arclength is

In[260]:=

(* dl = (1 + f '[x]^2)^(1/2) dx *)

so the total length of the curve is

In[261]:=

(* l = ∫ l = ∫_0^(2π) (1 + f '[x]^2)^(1/2) x . *)

Thus,

In[262]:=

arcLength = ∫_0^(2π) (1 + f '[x]^2)^(1/2) x %//N

Out[262]=

4 2^(1/2) EllipticE[1/2]

Out[263]=

7.6404

The integrals which arise from considerations of arclength often cannot be evaluated in terms of elementary functions, and that is the case here.
Nevertheless, we can apply numerical procedures to approximate the value of the integral.
How much longer than 2π is the hike along the curve?

In[264]:=

arcLength/(2π)//N

Out[264]=

1.21601

About 21.6% longer.

Surface Area of a Trigonometric Football

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

Spin one arc of a Sin wave around the x-axis, creating a surface which looks rather like a football

In[265]:=

Clear[f, g, x]  f[x_] := Sin[x] g[x_] := -Sin[x]  FilledPlot[{f[x], g[x]}, {x, ... 1;BackgroundMoccasin, PlotStyle CinnabarGreen, Fills->Melon] ;

[Graphics:HTMLFiles/6.3_arcLength_12.gif]

Here it is in 3D.

In[269]:=

<<Graphics`SurfaceOfRevolution`

In[270]:=

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

[Graphics:HTMLFiles/6.3_arcLength_15.gif]

We are asking for the surface area of a surface of revolution generated by sweeping one arc of a Sin wave, f[x] = Sin[x], about the x-axis.
The element of surface area is

In[271]:=

(* dS = 2π f[x] (1 + f '[x]^2)^(1/2) dx *)

so the total surface area is

In[272]:=

(* S =  ∫ S =  ∫_0^π 2π f[x] (1 + f '[x]^2)^(1/2) x . *)

Thus,

In[273]:=

surfaceArea = ∫_0^π2π f[x] (1 + f '[x]^2)^(1/2) x %//N

Out[273]=

2 π (2^(1/2) + ArcSinh[1])

Out[274]=

14.4236

Surface Area of a Trigonometric Donut

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

Consider the football-shaped region between the curves f[x] = Sin[x] and g[x] = -Sin[x] to be a sort of template, and spin the template around the y-axis to generate a "Trigonometric Donut." What is the surface area of this donut?

In[275]:=

Clear[f, g, x]  f[x_] := Sin[x] g[x_] := -Sin[x]  FilledPlot[{f[x], g[x]}, {x, ... 1;BackgroundMoccasin, PlotStyle CinnabarGreen, Fills->Melon] ;

[Graphics:HTMLFiles/6.3_arcLength_22.gif]

Here is the Trigonometric Donut in 3D.

In[279]:=

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

[Graphics:HTMLFiles/6.3_arcLength_24.gif]

We are asking for the surface area of a surface of revolution generated by sweeping the region between the curves f[x] and g[x] about the y-axis.
The element of surface area in this case is

In[282]:=

(* dS = 2π x (1 + f '[x]^2)^(1/2) dx  + 2π x (1 + g '[x]^2)^(1/2) dx = 4π x (1 + f '[x]^2)^(1/2) dx *)

so that the total surface area is

In[283]:=

(* S = ∫ S =  ∫_0^π 4π x (1 + f '[x]^2)^(1/2) x . *)

In[284]:=

surfaceArea = ∫_0^π4π x (1 + f '[x]^2)^(1/2) x %//N

Out[284]=

∫_0^π4 π x (1 + Cos[x]^2)^(1/2) x

Out[285]=

75.4077

Well, it seems that this integral might be even more difficult to deal with than the previous one.
Mathematica exhausted all of its tricks, and just returned the original integral. That's equivalent to saying "uncle."
Nevertheless, numerical procedures have no difficulty at all in estimationg the value of the definite integral.

Surface Area of a "Real" Donut

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

Just out of curiosity, what would be the surface area of a "real" donut -- one obtained by spinning a circle about the x-axis.
We will have to specify two radii in order to describe the donut, the radius of the generating circle and the radius from the origin out to the center of the circle.

In[286]:=

Clear[f, g, r1, r2, x]  r1 = 1 ; r2 = 2 ;  f[x_] := ( r1^2 - (x - r2)^2)^(1/2) ... 62371;Fills->Melon, AspectRatioAutomatic, AxesOrigin {0, 0}] ;

[Graphics:HTMLFiles/6.3_arcLength_31.gif]

Here is a "real" donut in 3D.

In[292]:=

RowBox[{RowBox[{top, =, RowBox[{SurfaceOfRevolution, [, RowBox[{f[x], ,, {x, r1, r1 + r2}, ,,  ... ty}], ]}]}], ;}], }] Show[top, bottom, DisplayFunction$DisplayFunction] ;

[Graphics:HTMLFiles/6.3_arcLength_33.gif]

We are asking for the surface area of a surface of revolution generated by sweeping the region between the curves f[x] and g[x] about the y-axis.
The element of surface area in this case is

In[295]:=

(* dS = 2π x (1 + f '[x]^2)^(1/2) dx + 2π x (1 + g '[x]^2)^(1/2) dx  = 4π x (1 + f '[x]^2)^(1/2) dx *)

so the total surface area is

(* S =  ∫ S =  ∫_ (r2 - r1)^(r2 + r1) 4π x (1 + f '[x]^2)^(1/2) x . *)

In[297]:=

surfaceArea = ∫_ (r2 - r1)^(r2 + r1) 4π x (1 + f '[x]^2)^(1/2) x %//N

Out[297]=

8 π^2

Out[298]=

78.9568

Well, the "real" donut has slightly more surface area than the trigonometric donut.
Does that seem reasonable?


Created by Mathematica  (April 22, 2004)