Chapter 9: Vector Calculus
Section 9.9: Stokes' Theorem
Example 9.9.5
Apply Stokes' theorem to the vector field F = x y i + y z j + x z k ; the curve C , the triangle with vertices 0 , 0 , 0 , 12 , 0 , 0 , 0 , 8 , 0 ; and the first-octant portion of the plane 2 x + 3 y + 4 z = 24 along with the coordinate planes x = y = 0 as the capping surface S . Hint : S has three sides but is open at the bottom.
Solution
Mathematical Solution
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The surface S is shown in Figure 9.9.5(a), where outward normal vectors are drawn on each of the three faces S k , 1 , 2 , 3 , comprising S , and listed in Table 9.9.5(a).
S 1
first-octant part of the plane 2 x + 3 y + 4 z = 24
S 2
the region in the xz -plane bounded by the triangle whose vertices are 0 , 0 , 0 , 12 , 0 , 0 , and 0 , 0 , 6
S 3
the region in the yz - plane bounded by the triangle whose vertices are 0 , 0 , 0 , 0 , 8 , 0 , and 0 , 0 , 6
Table 9.9.5(a) S = S 1 ⋃ S 2 ⋃ S 3
>
use plots,Student:-VectorCalculus in module() local p1,p2,p3,p4,p5,N1,N2,N3; N1:=RootedVector(root=[3,0,3],<0,-1,0>); N2:=RootedVector(root=[0,2,2],<-1,0,0>); N3:=RootedVector(root=[3,2,3],<2,3,4>/sqrt(29)); p1:=plot3d(6-x/2-3*y/4,x=0..12,y=0..8-2*x/3,color=red); p2:=plot3d([x,0,z],x=0..12,z=0..6-x/2,color=gold); p3:=plot3d([0,y,z],y=0..8,z=0..6-3*y/4,color=green); p4:=PlotVector([N1,N2,N3],color=black,width=.3); p5:=display(p1,p2,p3,p4,axes=frame,tickmarks=[3,3,4],labels=[x,y,z],scaling=constrained,style=surface,orientation=[-20,75,0],lightmodel=none); print(p5); end module: end use:
Figure 9.9.5(a) The surface S
Begin by calculating
∇ × F = i j k ∂ x ∂ y ∂ z x y y z x z = − y − z − x
A unit (upward) normal on S 1 is N = 1 29 2 3 4 where z = 6 − x / 2 − 3 y / 4 . Hence,
dσ = 1 + z x 2 + z y 2 dA = 1 + 1 / 2 2 + 3 / 4 2 dA = 29 / 4 dA
and
∇ × F · N dσ
= − 1 29 2 y + 3 z + 4 x 29 4 dA
= − 2 y + 3 z + 4 x / 4 dA
which, on S 1 becomes
∇ × F · N dσ
= − 2 y + 3 20 − x / 2 − 3 y / 4 + 4 x dA
= 6 + 5 x / 2 − y / 4 dA
The projection of S 1 onto the plane z = 0 is a right triangle whose hypotenuse is y = 8 − 2 x / 3 , from which it follows that
∫ ∫ S 1 ∇ × F · N dσ = ∫ 0 12 ∫ 0 8 − 2 x / 3 6 + 5 x / 2 − y / 4 dy dx = − 328
An outward normal on S 2 where y = 0 is the unit vector − j . Hence, ∇ × F · N = z on S 2 , a right triangle with hypotenuse z = 6 − x / 2 in the xz -plane. This gives the integral
∫ ∫ S 2 ∇ × F · N dσ = ∫ 0 12 ∫ 0 6 − x / 2 z dz dx = 72
An outward normal on S 3 where x = 0 is the unit vector − i . Hence, ∇ × F · N = z on S 3 , a right triangle with hypotenuse z = 6 − 3 y / 4 in the yz -plane. This gives the integral
∫ ∫ S 3 ∇ × F · N dσ = ∫ 0 8 ∫ 0 6 − 3 y / 4 y dz dy = 64
From these results it follows that ∫ ∫ S ∇ × F · N dσ = − 328 + 72 + 64 = − 192 .
The surface S caps C , the right triangle with vertices 0 , 0 , 0 , 12 , 0 , 0 , 0 , 8 , 0 . Since this triangle lies in the plane z = 0 , F · dr = 0 on each leg, and is x 8 − 2 x / 3 on the hypotenuse. See Table 9.9.5(b) for the details.
y = z = 0
x = z = 0
8 − 2 x / 3 , z = 0
x y y z x z · 1 0 0 = x y → 0
x y y z x z · 0 − 1 0 = − y z → 0
x y y z x z · 1 − 2 / 3 0 = x y − 2 y z / 3 → x 2 x − 3
Table 9.9.5(b) F · dr on C
From Table 9.9.5(b), it follows that ∳ C F · dr = ∫ 12 0 x 8 − 2 x / 3 dx = − 192 .
Maple Solution - Interactive
Initialize
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Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
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Tools≻Tasks≻Browse: Calculus - Vector≻ Vector Algebra and Settings≻ Display Format for Vectors
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Press the Access Settings button and select "Display as Column Vector"
Display Format for Vectors
Obtain z = z x , y
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Control-drag the equation of the plane.
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Context Panel: Solve≻Obtain Solutions for≻ z
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Context Panel: Assign to a Name≻ Z
2 x + 3 y + 4 z = 24 → solutions for z 6 − 1 2 x − 3 4 y → assign to a name Z
Define the vector field F
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Enter a free vector whose components are those of F . Context Panel: Evaluate and Display Inline
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Context Panel: Student Vector Calculus≻Conversions≻To Vector Field
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Context Panel: Assign to a Name≻ F
x y , y z , x z =
→ to Vector Field → assign to a name F
Obtain ∇ × F
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Common Symbols palette: Del and cross-product operators
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Context Panel: Evaluate and Display Inline
∇ × F =
To evaluate ∫ ∫ S 1 ∇ × F · N dσ , use the task template in Table 9.9.5(c). 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 System Cartesian [x,y,z] Cartesian - other cylindrical spherical bipolarcylindrical bispherical cardioidal cardioidcylindrical casscylindrical conical ellcylindrical hypercylindrical invcasscylindrical logcylindrical logcoshcylindrical oblatespheroidal paraboloidal paracylindrical prolatespheroidal rosecylindrical sixsphere tangentcylindrical tangentsphere toroidal
Table 9.9.5(c) Task template used to evaluate ∫ ∫ S 1 ∇ × F · N dσ
To evaluate ∫ ∫ S 2 ∇ × F · N dσ , use the task template in Table 9.9.5(d). 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 Parametric Surface
Flux through a Parametrically Defined Surface
For the Vector Field:
Select Coordinate System Cartesian [x,y,z] Cartesian - other cylindrical spherical bipolarcylindrical bispherical cardioidal cardioidcylindrical casscylindrical conical ellcylindrical hypercylindrical invcasscylindrical logcylindrical logcoshcylindrical oblatespheroidal paraboloidal paracylindrical prolatespheroidal rosecylindrical sixsphere tangentcylindrical tangentsphere toroidal
Select Coordinate System Cartesian cylindrical spherical bipolarcylindrical bispherical cardioidal cardioidcylindrical casscylindrical conical ellcylindrical hypercylindrical invcasscylindrical logcylindrical logcoshcylindrical oblatespheroidal paraboloidal paracylindrical prolatespheroidal rosecylindrical sixsphere tangentcylindrical tangentsphere toroidal
Table 9.9.5(d) Task template used to evaluate ∫ ∫ S 2 ∇ × F · N dσ
To evaluate ∫ ∫ S 3 ∇ × F · N dσ , use the task template in Table 9.9.5(e). 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 Parametric Surface
Flux through a Parametrically Defined Surface
For the Vector Field:
Select Coordinate System Cartesian [x,y,z] Cartesian - other cylindrical spherical bipolarcylindrical bispherical cardioidal cardioidcylindrical casscylindrical conical ellcylindrical hypercylindrical invcasscylindrical logcylindrical logcoshcylindrical oblatespheroidal paraboloidal paracylindrical prolatespheroidal rosecylindrical sixsphere tangentcylindrical tangentsphere toroidal
Select Coordinate System Cartesian cylindrical spherical bipolarcylindrical bispherical cardioidal cardioidcylindrical casscylindrical conical ellcylindrical hypercylindrical invcasscylindrical logcylindrical logcoshcylindrical oblatespheroidal paraboloidal paracylindrical prolatespheroidal rosecylindrical sixsphere tangentcylindrical tangentsphere toroidal
Table 9.9.5(e) Task template used to evaluate ∫ ∫ S 3 ∇ × F · N dσ
Unfortunately, Maple takes N = i , not N = − i on S 3 . Maple's default normal on the open surface described by the position vector R = R y , z is the normalized version of R y × R z . Since R y , z = y j + z k ,
R y × R z = i j k 0 1 0 0 0 1 = 1 0 0 = i
Hence, the value of ∫ ∫ S 3 ∇ × F · N dσ is actually 64, not − 64 , and
∫ ∫ S ∇ × F · N dσ = − 328 + 72 + 64 = − 192
Table 9.9.4(f) accesses the LineInt command through the Context Panel.
Form and evaluate the line integral of F around C
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Write the name F . Context Panel: Evaluate and Display Inline
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Context Panel: Student Vector Calculus≻Line Integral (Complete the dialog as per Figure 9.9.5(b).)
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Context Panel: Evaluate Integral
Figure 9.9.5(b) Line Integral Domain dialog
F =
→ line integral 2 ∫ 0 1 0 ⅆ t + ∫ 0 1 − 96 12 − 12 t t ⅆ t = − 192
Table 9.9.5(f) 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
Initialize
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Install the Student VectorCalculus package.
with Student :- VectorCalculus :
BasisFormat false :
F ≔ VectorField x y , y z , x z :
Use the Curl and Flux commands to obtain the flux of ∇ × F through S 1
Flux Curl F , Surface x , y , 6 − x / 2 − 3 y / 4 , x , y = Triangle 0 , 0 , 0 , 8 , 12 , 0 , output = integral
∫ 0 12 ∫ 0 − 2 3 x + 8 1 16 y − 9 2 − 5 8 x ⅆ y ⅆ x
Flux Curl F , Surface x , y , 6 − x / 2 − 3 y / 4 , x , y = Triangle 0 , 0 , 0 , 8 , 12 , 0 = − 328
Use the Curl and Flux commands to obtain the flux of ∇ × F through S 2
Flux Curl F , Surface x , 0 , z , x = 0 .. 12 , z = 0 .. 6 − x / 2 , output = integral
∫ 0 12 ∫ 0 6 − 1 2 x z ⅆ z ⅆ x
Flux Curl F , Surface x , 0 , z , x = 0 .. 12 , z = 0 .. 6 − x / 2 = 72
Use the Curl and Flux commands to obtain the flux of ∇ × F through S 3
Flux Curl F , Surface 0 , y , z , y = 0 .. 8 , z = 0 .. 6 − 3 y / 4 , output = integral
∫ 0 8 ∫ 0 6 − 3 4 y − y ⅆ z ⅆ y
Flux Curl F , Surface 0 , y , z , y = 0 .. 8 , z = 0 .. 6 − 3 y / 4 = − 64
Unfortunately, Maple takes N = i , not N = − i on S 3 . Maple's default normal on the open surface described by the position vector R = R y , z is the normalized version of R y × R z . Since R y , z = y j + z k ,
R y × R z = i j k 0 1 0 0 0 1 = 1 0 0 = i
Hence, the value of ∫ ∫ S 3 ∇ × F · N dσ is actually 64, not − 64 , and
∫ ∫ S ∇ × F · N dσ = − 328 + 72 + 64 = − 192
Table 9.9.5(g) uses the LineInt command to obtain the value of ∳ C F · dr , where C is the triangle whose vertices are 0 , 0 , 20 , 30 , 0 , 0 , 0 , 24 , 0 .
LineInt F , LineSegments 0 , 0 , 0 , 12 , 0 , 0 , 0 , 8 , 0 , 0 , 0 , 0 , output = integral
2 ∫ 0 1 0 ⅆ t + ∫ 0 1 − 96 12 − 12 t t ⅆ t
LineInt F , LineSegments 0 , 0 , 0 , 12 , 0 , 0 , 0 , 8 , 0 , 0 , 0 , 0 = − 192
Table 9.9.5(g) Line integral of the tangential component 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.
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