Chapter 9: Vector Calculus
Section 9.9: Stokes' Theorem
Example 9.9.4
Apply Stokes' theorem to the vector field F=x y i+y z j+x z k; the curve C, the triangle with vertices 30,0,0,0,24,0,0,0,20; and the first-octant portion of the plane 4 x+5 y+6 z=120 as the capping surface S.
Solution
Mathematical Solution
Begin by calculating
∇×F=ijk∂x∂y∂zx yy zx z = −y−z−x
A unit (upward) normal on S is N=177456 where z=20−2 x/3−5 y/6. Hence,
dσ=1+zx2+zy2 dA=1+2/32+5/62 dA=77/6 dA
and
∇×F·N dσ
=−1774 y+5 z+6 x 776 dA
=−4 y+5 z+6 x/6 dA
which, on S becomes
=−4 y+520−2 x/3−5 y/6+6 x dA
=y/36−4 x/9−50/3 dA
The projection of S onto the plane z=0 is a right triangle whose hypotenuse is y=24−4 x/5, from which it follows that
∫∫S∇×F·N dσ=∫030∫024−4 x/5y/36−4 x/9−50/3 dy dx = −7520
The line integral of F around C requires parametrizing each of the three sides of a triangle. To this end, label the vertices 30,0,0,0,24,0,0,0,20 as a,b,c, respectively, and let a, b, c be their respective position-vector representations. If R is the position vector for the point x,y,z, then Table 9.9.4(a) provides the necessary parametrizations of the sides of the triangle and the line integral of F·dr along each such side.
Lab
R=a+tb−a = 30(1−t)24 t0
∫01301−t⋅−720 t ⅆt= −3600
Lbc
R=b+tc−b = 024(1−t)20 t
∫01241−t⋅−480 t ⅆt= −1920
Lca
R=c+ta−c = 30 t020(1−t)
∫01201−t⋅−600 t ⅆt= −2000
Table 9.9.4(a) Parametrization of the sides of C
Consequently, the line integral of F around C is the sum −3600−1920−2000 = −7520.
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
Tools≻Tasks≻Browse: Calculus - Vector≻ Vector Algebra and Settings≻ Display Format for Vectors
Press the Access Settings button and select "Display as Column Vector"
Display Format for Vectors
Obtain z=zx,y
Control-drag the equation of the plane.
Context Panel: Solve≻Obtain Solutions for≻z
Context Panel: Assign to a Name≻Z
4 x+5 y+6 z=120→solutions for z20−23x−56y→assign to a nameZ
Define the vector field F
Enter a free vector whose components are those of F. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Conversions≻To Vector Field
Context Panel: Assign to a Name≻F
x y,y z,x z = →to Vector Field →assign to a nameF
Obtain ∇×F
Common Symbols palette: Del and cross-product operators
Context Panel: Evaluate and Display Inline
∇×F =
To evaluate ∫∫S∇×F·N dσ, use the task template in Table 9.9.4(b). Should the "Clear All and Reset" button in the Task Template be pressed, all the data that has been input to the template will be lost. In that event, the reader should simply re-launch the example to recover the appropriate inputs to the template.
Tools≻Tasks≻Browse: Calculus - Vector≻Integration≻Flux≻3-D≻Through a Surface Defined over a Triangle
Flux through a Surface Defined over a Triangle
For the Vector Field:
Select Coordinate SystemCartesian [x,y,z]Cartesian - othercylindricalsphericalbipolarcylindricalbisphericalcardioidalcardioidcylindricalcasscylindricalconicalellcylindricalhypercylindricalinvcasscylindricallogcylindricallogcoshcylindricaloblatespheroidalparaboloidalparacylindricalprolatespheroidalrosecylindricalsixspheretangentcylindricaltangentspheretoroidal
Table 9.9.4(b) Task template used to evaluate ∫∫S∇×F·N dσ
Table 9.9.4(c) accesses the LineInt command through the Context Panel.
Form and evaluate the line integral of F around C
Write the name F and press the Enter key.
Context Panel: Student Vector Calculus≻Line Integral (Complete the dialog as per Figure 9.9.4(a).)
Context Panel: Evaluate Integral
Figure 9.9.4(a) Line Integral Domain dialog
F
→line integral
∫01−600t20−20tⅆt+∫01−72030−30ttⅆt+∫01−48024−24ttⅆt
=
−7520
Table 9.9.3(c) Evaluation of the line integral of F around C
Note that in order for Maple to trace the edges of the triangle C, the starting vertex must be repeated as the last node to be reached.
Maple Solution - Coded
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
Set display of vectors with BasisFormat command.
BasisFormatfalse:
Define F with the VectorField command.
F≔VectorFieldx y,y z,x z:
Use the Curl and Flux commands to obtain the flux of ∇×F through S1
FluxCurlF,Surfacex,y,20−23 x−56 y,x,y=Triangle0,0,0,24,30,0,output=integral
∫030∫0−45x+24136y−503−49xⅆyⅆx
FluxCurlF,Surfacex,y,20−23 x−56 y,x,y=Triangle0,0,0,24,30,0
Table 9.9.4(d) uses the LineInt command to obtain the value of ∳CF·dr, where C is the triangle whose vertices are 0,0,20,30,0,0,0,24,0.
LineIntF,LineSegments0,0,20,30,0,0,0,24,0,0,0,20,output=integral
LineIntF,LineSegments0,0,20,30,0,0,0,24,0,0,0,20 = −7520
Table 9.9.4(d) Line integral of the tangential component of F around C
Table 9.9.4(e) implements the line integral of F around C from first principles.
Define the position vectors a, b, c.
a,b,c≔30,0,0,0,24,0,0,0,20:
Define the general position vector R.
R≔x,y,z:
Obtain parametric representations of each side of C
Lab≔a+t b−a
Lbc≔b+t c−b
Lca≔c+t a−c
Obtain F·dr on each side of C
Obtain dr with the diff command.
Obtain F·dr with the DotProduct command.
Obtain parametric equations x=xt,y=yt,z=zt with the Equate command.
Evaluate F·dr on each edge with the eval command.
λab≔evalDotProductF,diffLab,t,EquateR,Lab:
λbc≔evalDotProductF,diffLbc,t,EquateR,Lbc:
λca≔evalDotProductF,diffLca,t,EquateR,Lca:
Use the top-level Int and int commands to integrate along each side of C, and sum
Intλab,t=0..1+Intλbc,t=0..1+Intλca,t=0..1= :-intλab,t=0..1+:-intλbc,t=0..1+:-intλca,t=0..1
∫01−600t20−20tⅆt+∫01−72030−30ttⅆt+∫01−48024−24ttⅆt=−7520
Table 9.9.4(e) Evaluation of ∳CF·dr from first principles
The simple syntax of the top-level int command is accessed by prefixing the int command in the Student VectorCalculus package with "colon dash".
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