Chapter 9: Vector Calculus
Section 9.6: Surface Integrals
Example 9.6.11
Working in spherical coordinates, obtain the flux of the field F=x i+y j+z k through the surface of the unit sphere centered at the origin. Use a parametric representation of the surface.
Solution
Mathematical Solution
Figure 9.6.11(a) shows the sphere, a representative normal on its surface, and field arrows.
The arrows suggest the field is radially outward, so in spherical coordinates it would be ρ eρ, and on the sphere it is just eρ. The outward normal on the sphere is also just eρ, so F·N=1. Thus, the integral of F·N dσ over the surface of the sphere is
∫02 π∫0πsinφ ⅆφ ⅆθ=4 π
use Student:-VectorCalculus in module() local F,p1; F:=VectorField(<x,y,z>); p1:=Flux(F,Sphere(<0,0,0>,1),output=plot,caption="",tickmarks=[3,3,3],axes=frame,orientation=[-55,65,0],fieldoptions=[grid=[5,5,5]]); print(p1); end module: end use:
Figure 9.6.11(a) Sphere, field, normal
Maple Solution - Interactive
Table 9.6.11(a) provides a solution via task template. The vector field F is given in Cartesian coordinates, but could also be expressed in spherical coordinates. Table 9.6.11(b) shows how the same task template can be used for the alternate calculation.
Tools≻Tasks≻Browse: Calculus - Vector≻Integration≻Flux≻3-D≻Through a Sphere
Flux through a Sphere
For the Vector Field:
Select Coordinate SystemCartesian [x,y,z]Cartesian - othercylindricalsphericalbipolarcylindricalbisphericalcardioidalcardioidcylindricalcasscylindricalconicalellcylindricalhypercylindricalinvcasscylindricallogcylindricallogcoshcylindricaloblatespheroidalparaboloidalparacylindricalprolatespheroidalrosecylindricalsixspheretangentcylindricaltangentspheretoroidal
Table 9.6.11(a) Solution by task template
Table 9.6.11(b) uses the same task template, but expresses the vector field F in spherical coordinates.
Table 9.6.11(b) Task-template solution for field in spherical coordinates
Maple Solution - Coded
Table 9.6.11(c) provides a solution in which the vector field is given in Cartesian coordinates, but in which the Flux command in the Student VectorCalculus package writes the integral in spherical coordinates.
Initialize
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
Use the VectorField command to define F.
F≔VectorFieldx,y,z:
Use the Flux command with the Sphere option to compute the flux
FluxF,Sphere0,0,0,1,output=integral
∫02π∫0πsinφⅆφⅆθ
FluxF,Sphere0,0,0,1 = 4π
Table 9.6.11(c) Solution via the Flux command
Figure 9.6.11(a) can be obtained by changing the option "output = integral" to "output = plot" in Table 9.6.11(c). The version of this command used for this figure is hidden behind the cell containing the graph.
Table 9.6.11(d) provides a solution from first principles. The surface is represented by a position vector that parametrizes the surface via spherical coordinates. However, the position vector is essentially a Cartesian vector, so the vector field F is also given in Cartesian coordinates. In the expression F·N, the Cartesian variables x,y,z are simply replaced with their equivalents on the surface of the sphere.
Set the display of vectors with the BasisFormat command.
BasisFormatfalse:
Define the sphere as the position vector R.
R≔sinφcosθ,sinφsinθ,cosφ:
Obtain Rφ×Rθ
In the Student VectorCalculus package, diff maps onto the components of vectors.
Apply the CrossProduct, Norm, and simplify commands.
simplifyNormCrossProductdiffR,φ,diffR,θ assuming φ∷RealRange0,π = sinφ
Obtain a unit outward normal on the surface of the sphere
Apply the CrossProduct, Normalize, and simplify commands.
N≔simplifyNormalizeCrossProductdiffR,φ,diffR,θ assuming φ∷RealRange0,π
Evaluate F·N on the surface of the sphere
Use the DotProduct command to obtain F·N.
Use the eval command to make the substitutions for x,y,z as returned by the Equate command.
Apply the simplify command.
simplifyevalDotProductF,N,Equatex,y,z,R = 1
Form and evaluate the flux integral
Use the Int and int commands, accessing the top-level versions via the prefix operator :-
:-Int1⋅sinφ,φ=0..π,θ=0..2 π= :-int1⋅sinφ,φ=0..π,θ=0..2 π
∫02π∫0πsinφⅆφⅆθ=4π
Table 9.6.11(d) Solution from first principles with F given in Cartesian coordinates
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