Chapter 9: Vector Calculus
Section 9.5: Line Integrals
Example 9.5.7
Obtain the line integral of the scalar function fx,y=x y, taken around the ellipse whose center is 3,4 and whose semi-major and semi-minor axes, parallel to the coordinate axes, are 2 and 1, respectively.
Solution
Mathematical Solution
A parametric definition of the ellipse as a position vector is given by
R=3+2 cos(t)4+ sin(t)
(Indeed, x−3222+y−4212=cos2t+sin2t=1, the Cartesian description of the ellipse.)
For this ellipse, ρ=dsdt=R. is given by
ddt3+2 cost2+ddt4+ sint2
=4 sin2t+cos2t
=4−3 cos2t
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 around the ellipse is ρ fR. Hence, the line integral is
∫02 πρ fR dt = ∫02 π3+2 cost 4+ sint 4−3 cos2t dt ≐ 116.26
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 9.5.7(a).
Context Panel: Approximate≻10 (digits)
x y→line integral∫02⁢π2⁢cos⁡t+3⁢4+sin⁡t⁢4⁢sin⁡t2+cos⁡t2ⅆt→at 10 digits116.2613786
Figure 9.5.7(a) Path Integral Domain dialog
Form and evaluate the line integral via the PathInt command
PathIntx y,x,y=Ellipse3,4,2,1,output=integral
∫02⁢π2⁢cos⁡t+3⁢4+sin⁡t⁢4⁢sin⁡t2+cos⁡t2ⅆt
PathIntx y,x,y=Ellipse3,4,2,1. = 116.2613787
Maple can provide an exact evaluation of this line integral, but the resulting expression is at least five lines long. Hence, the integral is evaluated numerically. This is done interactively by setting the integral and using the Approximate option in the Context Panel. Alternatively, a single floating-point number is included in the PathInt command so that Maple evaluates the integral numerically. To this end, the length of the semi-minor axis is given as the number "1.", rather than as the exact number "1".
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 circle parametrically as the position vector R
Context Panel: Assign Name
R=3+2 cost,4+sint→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 = −3⁢cos⁡t2+4→assign to a nameρ
Form and evaluate the line integral ∫02 πfR ρ dt
Calculus palette: Definite integral operator
Context Panel: 2-D Math≻Convert To≻Inert Form
∫02 πρ fR ⅆt→at 10 digits116.2613786 116.2613786
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
−3⁢cos⁡t2+4⁢3+2⁢cos⁡t⁢4+sin⁡t
→integrate w.r.t. t
∫02⁢π−3⁢cos⁡t2+4⁢3+2⁢cos⁡t⁢4+sin⁡tⅆt
→at 10 digits
116.2613786
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