Chapter 9: Vector Calculus
Section 9.7: Conservative and Solenoidal Fields
Example 9.7.1
Show that the line integral of F=2 x y−y3 i+x2−3 x y2 j over any circle in the plane is zero.
Solution
Mathematical Solution
To evaluate the line integral ∫CF·dr, where C is an arbitrary planar circle, parametrize the circle as
x=a+r cost,y=b+r sint
and evaluate F on that circle. Then F·dr is
2(a+r cost)(b+r sint)−(b+r sint)3(a+r cost)2−3(a+r cost)(b+r sint)2·−r sin(t)r cos(t)
or
Φ=−r sint(2a+r costb+r sint−(b+r sint)3)+r cost((a+r cost)2−3a+r cost(b+r sint)2)
and the line integral becomes ∫02 πΦ dt =0.
Maple Solution - Interactive
Table 9.7.1(a) provides a solution in which the LineInt command is accessed through the Context Panel.
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
Define the vector field F
Enter the components of F in a free vector. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Conversions≻To Vector Field
Context Panel: Assign to a Name≻F
2 x y−y3,x2−3 x y2 = −y3+2⁢x⁢y−3⁢x⁢y2+x2→to Vector Field−y3+2⁢x⁢y−3⁢x⁢y2+x2→assign to a nameF
Form and evaluate the requisite line integral
Write the name F. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Line Integral Figure 9.7.1(a) shows the dialog for obtaining the inert integral; Figure 9.7.1(b), its value.
Figure 9.7.1(a) Obtaining the inert integral
Figure 9.7.1(b) Evaluating the integral
F = →line integral∫02⁢π−r⁢−b+r⁢sin⁡t3+2⁢a+r⁢cos⁡t⁢b+r⁢sin⁡t⁢sin⁡t+r⁢−3⁢a+r⁢cos⁡t⁢b+r⁢sin⁡t2+a+r⁢cos⁡t2⁢cos⁡tⅆt
F = →line integral0
Table 9.7.1(a) Context Panel access to the LineInt command
Table 9.7.1(b) provides a solution from first principles.
Parametric definition of the general circle in the plane
Context Panel: Assign to a Name≻X
a+r cost→assign to a nameX
Context Panel: Assign to a Name≻Y
b+r sint→assign to a nameY
Form and evaluate the line integral ∫Cf dx+g dy
Calculus palette: Definite-integral operator
Expression palette: Evaluation template
Calculus palette: Derivative operator
Context Panel: Evaluate and Display Inline
∫02 πF1x=a|f(x)x=X,y=Y⋅ⅆⅆ t X ⅆt+∫02 πF2x=a|f(x)x=X,y=Y⋅ⅆⅆ t Y ⅆt = 0
Table 9.7.1(b) Solution from first principles
Maple Solution - Coded
Table 9.7.1(c) provides a solution via an explicit call to the LineInt command.
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
Use the VectorField command to define F.
F≔VectorField2 x y−y3,x2−3 x y2:
Form and evaluate the requisite line integral with the LineInt command
LineIntF,Circlea,b,r,output=integral
∫02⁢π−r⁢−b+r⁢sin⁡t3+2⁢a+r⁢cos⁡t⁢b+r⁢sin⁡t⁢sin⁡t+r⁢−3⁢a+r⁢cos⁡t⁢b+r⁢sin⁡t2+a+r⁢cos⁡t2⁢cos⁡tⅆt
LineIntF,Circlea,b,r = 0
Table 9.7.1(c) Solution via the LineInt command
Table 9.7.1(d) provides a solution from first principles.
Parametric definition of the general circle.
X,Y≔a+r cost,b+r sint:
Evaluate F on the circle with the eval command.
Obtain dx and dy by means of the diff command.
Write the inert line integral of f dx+g dy with the Int command.
IntevalF1,x=X,y=Y⋅diffX,t,t=0..2 π+IntevalF2,x=X,y=Y⋅diffY,t,t=0..2 π
∫02⁢π−−b+r⁢sin⁡t3+2⁢a+r⁢cos⁡t⁢b+r⁢sin⁡t⁢r⁢sin⁡tⅆt+∫02⁢π−3⁢a+r⁢cos⁡t⁢b+r⁢sin⁡t2+a+r⁢cos⁡t2⁢r⁢cos⁡tⅆt
Evaluate the line integral by writing it with the top-level form of int, accessed via the colon-dash prefix :-
:-intevalF1,x=X,y=Y⋅diffX,t,t=0..2 π+ :-intevalF2,x=X,y=Y⋅diffY,t,t=0..2 π
0
Table 9.7.1(d) Solution from first principles
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