Chapter 9: Vector Calculus
Section 9.6: Surface Integrals
Example 9.6.2
Integrate the scalar fx,y,z=x y z on the surface of the sphere whose center is at the Cartesian point A,B,C and whose radius is a.
Solution
Mathematical Solution
Let R=xyz=A+a sin(φ)cos(θ)B+a sin(φ)sin(θ)C+a cos(φ) be a position-vector parametrization of the sphere.
Then, except for the differentials dφ and dθ, the surface-area element is
∂R∂φ×∂R∂θ = a cos(φ)cos(θ)a cos(φ)sin(θ)−a sin(φ)×−a sin(φ)sin(θ) a sin(φ)cos(θ)0 = a2 sinφ
Hence, the requisite surface integral is
∫02 π∫0πA+a sinφcosθB+a sinφsinθC+a cosφ a2sinφ ⅆφ ⅆθ=4 π a2 A B C
Maple Solution - Interactive
Table 9.6.2(a) provides a solution via task template.
Tools≻Tasks≻Browse Calculus - Vector≻Integration≻Surface Integration≻Over a Sphere
Surface Integral on a Sphere
Integrand
fx,y,z=
Sphere Radius:
Sphere Center:
x=
y=
z=
Table 9.6.2(a) Solution via task template
If the sphere is parametrized by the equations
x=A+a sinφcosθ
y=B+a sinφsinθ
z=C+a cosφ
then the task-template solution in Table 9.6.2(b) is available. Note that φ→u and θ→v.
Tools≻Tasks≻Browse Calculus - Vector≻Integration≻Surface Integration≻Over a Parametrically Defined Surface
Surface Integral on a Surface Defined Parametrically
xu,v=
yu,v=
zu,v=
Fu,v≡fxu,v,yu,v,zu,v
N≡∂y,z∂u,v2+∂z,x∂u,v2+∂x,y∂u,v2
∫∫Sf ⅆσ =∫u=au=b∫v=guv=huFu,vN ⅆv ⅆu
=
b=
∫∫Sf ⅆσ=∫v=av=b∫u=Gvu=HvFu,vN ⅆu ⅆv
a=
Table 9.6.2(b) Alternate solution by task template
A final simplification sets csgna2, the complex sign of a2, to 1 since a>0.
Maple Solution - Coded
Table 9.6.2(c) contains a solution based on the SurfaceInt command.
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
SurfaceIntx y z,x,y,z=SphereA,B,C,a,output=integral=SurfaceIntx y z,x,y,z=SphereA,B,C,a
∫02⁢π∫0πa2⁢sin⁡φ⁢−cos⁡θ⁢sin⁡θ⁢cos⁡φ3⁢a3−C⁢cos⁡θ⁢sin⁡θ⁢cos⁡φ2⁢a2+A⁢sin⁡φ⁢sin⁡θ⁢cos⁡φ⁢a2+B⁢sin⁡φ⁢cos⁡θ⁢cos⁡φ⁢a2+cos⁡θ⁢sin⁡θ⁢cos⁡φ⁢a3+A⁢C⁢sin⁡φ⁢sin⁡θ⁢a+B⁢C⁢sin⁡φ⁢cos⁡θ⁢a+C⁢cos⁡θ⁢sin⁡θ⁢a2+A⁢B⁢cos⁡φ⁢a+A⁢B⁢Cⅆφⅆθ=4⁢A⁢B⁢C⁢π⁢a2
Table 9.6.2(c) Solution via the Sphere option in the SurfaceInt command
Table 9.6.2(d) contains a parametric solution obtained from first principles.
Initialize
Set the display of vectors with the BasisFormat command.
BasisFormatfalse:
Write parametric equations in which u→φ and v→θ
Use spherical coordinates to define the parametrization.
X≔A+a sinucosv:Y≔B+a sinusinv:Z≔C+a cosu:
Express the surface of the sphere via the position vector R
Define R as a free vector.
R≔X,Y,Z:
Obtain the vectors Ru=∂uR and Rv=∂vR tangent to coordinate curves
Apply the diff command, which automatically maps onto the components of vectors.
Set the names Ru and Rv as Atomic Variables, something best done in typeset math.
R__u≔diffR,u
R__v≔diffR,v
Except for the differentials, dσ=Ru×Rv
Use the CrossProduct and Norm commands, and then simplify with appropriate conditions.
dsig≔simplifyNormCrossProductR__u,R__v assuming a>0,u∷RealRange0,π
sin⁡u⁢a2
Form and evaluate the appropriate surface integral
Use the top-level Int and int commands. To access the top-level int command, not the modified int command in the Student VectorCalculus package, prefix with :- (that is, with "colon dash).
IntX Y Z dsig,u=0..π,v=0..2 π= :-intX Y Z dsig,u=0..π,v=0..2 π
∫02⁢π∫0πA+a⁢sin⁡u⁢cos⁡v⁢B+a⁢sin⁡u⁢sin⁡v⁢C+a⁢cos⁡u⁢sin⁡u⁢a2ⅆuⅆv=4⁢A⁢B⁢C⁢π⁢a2
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