Chapter 9: Vector Calculus
Section 9.10: Green's Theorem
Example 9.10.7
Apply the Stokes-form of Green's theorem to F=x−2 y i+x2y j, and R, the region inside the ellipse x2+4 y2=1.
Solution
Mathematical Solution
The expression gx−fy is the third component of ∇×F=ijk∂x∂y∂zx−2 yx2y0 = 002(x y+1), where a third-component of zero has been appended to F so that its curl could be calculated.
The integral of this expression over R can be obtained if the ellipse x2+4 y2=1 is given in polar coordinates by r=1/cos2θ+4 sin2θ. The left-hand side of the Stokes-form of Green's theorem is therefore
∫02 π∫01/cos2θ+4 sin2θ2 r r2cosθsinθ+1 dr dθ = π
The right-hand side of the Stokes-form of Green's theorem is the line integral of the tangential component of F around the ellipse x2+4 y2=1. Parametrize the ellipse as x=cost,y=sint/2, so that the line integral ∳Cf dx+g dy becomes
∫02 πF1⋅−sint+F2⋅cost/2x=a|f(x)x=cost,y=sint/2 dt = π
Note that this second parametrization of the ellipse does not use polar coordinates.
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
Define the Cartesian vector field F
Write the vector field as 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
x−2 y,x2y,0 = →to Vector Field →assign to a nameF
Using polar coordinates, integrate the third component of ∇×F over the region R
Context Panel: Assign to a Name≻E
1/cos2θ+4 sin2θ→assign to a nameE
Expression palette: Evaluation template
Common Symbols palette: Del and cross-product operators
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻q
∇×F3x=a|f(x)x=r cosθ,y=r sinθ = 2⁢r2⁢cos⁡θ⁢sin⁡θ+2→assign to a nameq
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Type the name q. Context Panel: Evaluate and Display Inline
Context Panel: Student Multivariate Calculus≻Integrate≻Iterated Complete the dialogs as per the figures below.
Context Panel: Evaluate Integral
q = 2⁢r2⁢cos⁡θ⁢sin⁡θ+2→MultiInt∫02⁢π∫0E2⁢r2⁢cos⁡θ⁢sin⁡θ+2⁢rⅆrⅆθ=π
Redefine F as a planar vector field
x−2 y,x2y = →to Vector Field →assign to a nameF
Form and evaluate the line integral ∳Cf dx+g dy
Write the name F. Context Panel: Evaluate and Display Inline
Context Panel: Student Vector Calculus≻Line Integral Complete dialog as per figure to the right.
F = →line integral∫02⁢π−cos⁡t−sin⁡t⁢sin⁡t+14⁢cos⁡t3⁢sin⁡tⅆt=π
Maple integrates around the ellipse x2+4 y2=1 by parametrizing the ellipse via the equations x=cost,y=sint/2.
Maple Solution - Coded
Install the Student VectorCalculus package.
withStudent:-VectorCalculus:
Use the BasisFormat command to set the display format for vectors.
BasisFormatfalse:
Define F with the VectorField command.
F≔VectorFieldx−2 y,x2y,0:
Use the int command to integrate the third component of ∇×F over the region R
intCurlF3,x,y=Ellipse x2+4 y2=1,r,θ,output=integral
∫02⁢π∫012⁢tan⁡θ2+114+tan⁡θ22⁢r2⁢cos⁡θ⁢sin⁡θ+2⁢rⅆrⅆθ
intCurlF3,x,y=Ellipse x2+4 y2=1,r,θ = π
Use the LineInt command to obtain ∳Cf dx+g dy
LineIntVectorFieldF1,F2,Ellipsex2+4 y2=1,output=integral
∫02⁢π−cos⁡t−sin⁡t⁢sin⁡t+14⁢cos⁡t3⁢sin⁡tⅆt
LineIntVectorFieldF1,F2,Ellipsex2+4 y2=1 = π
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