Chapter 9: Vector Calculus
Section 9.5: Line Integrals
Example 9.5.8
Let C be the ellipse x2+4 y2=1 and let c be the arc subtended by an angle of π/4 radians measured counterclockwise from the positive x-axis. Obtain the line integral of the scalar function fx,y=x y, taken along c.
Solution
Mathematical Solution
A parametric definition of the ellipse as a position vector is given by
R= cos(t) sin(t)/2
since x2+4 y2=cos2t+4 sin2t/4=1.
For this ellipse, ρ=dsdt=R. is given by
ddt cost2+ddt sint/22
=sin2t+ cos2t/4
=4 sin2t+ cos2t/2
=4−3 cos2t/2
If fv=v1⋅v2 is the scalar-valued function of the vector argument v=v1 i+v2 j, then the integrand for the line integral along the ellipse is ρ fR. Before the requisite line integral can be formulated, however, the value of t that subtends a "polar angle" of π/4 radians must be determined. Where the ray y=x hits the ellipse in the first quadrant, cost=sint/2, so that t=arctan2. Hence, the line integral is
∫0arctan2ρ fR dt = 14∫0arctan2 costsint 4−3 cos2t dt = 1785900−136
Maple Solution
Initialize
Tools≻Load Package: Student Vector Calculus
Loading Student:-VectorCalculus
Access the PathInt command through the Context Panel
Write the scalar.
Context Panel: Student Vector Calculus≻Line Integral (2D) Complete the dialog as per Figure 9.5.8(a).
Context Panel: Evaluate Integral
x y→line integral∫0arctan⁡214⁢cos⁡t⁢sin⁡t⁢4⁢sin⁡t2+cos⁡t2ⅆt=−136+17900⁢85
Figure 9.5.8(a) Path Integral Domain dialog
Form and evaluate the line integral via the PathInt command
PathIntx y,x,y=ArcEllipsex2+4 y2−1,0,π/4,output=integral
∫0arctan⁡214⁢cos⁡t⁢sin⁡t⁢4⁢sin⁡t2+cos⁡t2ⅆt
PathIntx y,x,y=ArcEllipsex2+4 y2−1,0, π/4 = −136+17900⁢85
A solution from first principles is also possible.
Define f as a scalar-valued function of a vector argument
Context Panel: Assign Function
fv=v1⋅v2→assign as functionf
Define the ellipse parametrically as the position vector R
Context Panel: Assign Name
R= cost, sint/2→assign
Obtain ρ=dsdt=R.
Calculus palette: Differentiation operator
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻rho
ⅆⅆ t R = 12⁢−3⁢cos⁡t2+4→assign to a nameρ
Form and evaluate the line integral ∫0arctan2fR ρ dt
Calculus palette: Definite integral operator
∫0arctan2ρ fR ⅆt = −136+17900⁢85
Explicit formulation and evaluation of the line integral
Write the integrand and press the Enter key.
Context Panel: Constructions≻Definite Integral≻t (Complete dialog as per figure to the right.)
ρ fR
14⁢−3⁢cos⁡t2+4⁢cos⁡t⁢sin⁡t
→integrate w.r.t. t
∫0arctan⁡214⁢−3⁢cos⁡t2+4⁢cos⁡t⁢sin⁡tⅆt
=
−136+17900⁢85
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