Chapter 9: Vector Calculus
Section 9.5: Line Integrals
Let C be that portion of the parabola y=x2 from the origin to the point 1,1. Obtain the line integral of the scalar function fx,y=x y, taken along C.
The line integral of the scalar fx,y along a path described parametrically by x=xt,y=yt, t∈a,b, is given by
∫abfxt,yt dsdt dt
where s is arc length, so dsdt=x.2+y.2=ρ = R., with Rt=xt i+yt j being the vector form of the parametric representation of the path.
A parametric representation of the given parabola is
ρ=dsdt=ddt t2+ddtt22 = 12+2 t2=1+4 t2
and the line integral is given by
∫01t t2 1+4 t2 dt= ∫01t31+4 t2 dt=5524+1120
Tools≻Load Package: Student Vector Calculus
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 9.5.9(a).
Context Panel: Evaluate Integral
x y→line integral∫01t3⁢4⁢t2+1ⅆt=524⁢5+1120
Figure 9.5.9(a) Path Integral Domain dialog
Form and evaluate the line integral via the PathInt command
PathIntx y,x,y=Pathx ,x2,x=0..1,output=integral
PathIntx y,x,y=Pathx ,x2,x=0..1 = 524⁢5+1120
A solution from first principles is also possible.
Obtain a parametric representation of the parabola
Position-vector form of parabola; t∈0,1
Calculus palette: Differentiation operator
Context Panel: Assign to a Name≻rho
ⅆⅆ t R = 4⁢t2+1→assign to a nameρ
Form the integrand fxt,yt⋅ρ and integrate with respect to t
Expression palette: Evaluation template
Press the Enter key.
Context Panel: Constructions≻Definite Integral≻t
(Complete dialog as per figure.)
→integrate w.r.t. t
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