Chapter 9: Vector Calculus
Section 9.5: Line Integrals
Obtain the line integral of the scalar function fx,y=x y, taken along the line segment from 1,2 to 3,5.
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 line segment is
R=x(t)y(t)=12+t (35−12) = 1+2 t2+3 t,0≤t≤1
ρ=dsdt=ddt1+2 t2+ddt2+3 t2 = 22+32=13
and the line integral is given by
∫011+2 t 2+3 t 13 dt=13 ∫016⁢t2+7⁢t+2 dt=1513/2
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 9.5.1(a).
Context Panel: Evaluate (from inert)
x y→line integral∫011+2⁢t⁢2+3⁢t⁢13ⅆt=152⁢13
Figure 9.5.1(a) Path Integral Domain dialog
Form and evaluate the line integral via the PathInt command
PathIntx y,x,y=Line1,2,3,5 = 152⁢13
A solution from first principles is also possible.
Obtain a parametric representation of the line segment
Define the points as position vectors.
Position-vector form of line; t∈0,1
Calculus palette: Differentiation operator
Context Panel: Assign to a Name≻rho
ⅆⅆ t R = 13→assign to a nameρ
Form the integrand fxt,yt⋅ρ and integrate with respect to t
Expression palette: Evaluation template
Press the Enter key.
(Complete dialog as per figure.)
Context Panel: Evaluate Integral
→integrate w.r.t. t
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