Chapter 9: Vector Calculus
Section 9.5: Line Integrals
Example 9.5.2
Obtain the line integral of the scalar function fx,y,z=x y z, taken along the line segment from 1,2,3 to 5,3,2.
Solution
Mathematical Solution
The line integral of the scalar fx,y,z along a path described parametrically by x=xt,y=yt,z=zt, t∈a,b, is given by
∫abfxt,yt,zt dsdt dt
where s is arc length, so dsdt=x.2+y.2+z.2=ρ = R., with Rt=xt i+yt j+zt k 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)z(t)=123+t (532−123) = 1+4 t2+ t3−t,0≤t≤1
so that
ρ=dsdt=ddt1+4 t2+ddt2+ t2+ddt3−t2 = 42+12+12=32
and the line integral is given by
∫011+4 t 2+ t 3−t 32 dt=32 ∫01−4⁢t3+3⁢t2+25⁢t+6 dt=1112/2
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 (3D) Complete the dialog as per Figure 9.5.2(a).
Context Panel: Evaluate (from inert)
x y z→line integral∫013⁢1+4⁢t⁢2+t⁢3−t⁢2ⅆt=1112⁢2
Figure 9.5.2(a) Path Integral Domain dialog
Form and evaluate the line integral via the PathInt command
PathIntx y z,x,y,z=Line1,2,3,5,3,2,output=integral
∫013⁢1+4⁢t⁢2+t⁢3−t⁢2ⅆt
PathIntx y z,x,y,z=Line1,2,3,5,3,2 = 1112⁢2
A solution from first principles is also possible.
Obtain a parametric representation of the line segment
Define the points as position vectors.
P1,P2≔1,2,3,5,3,2:
Position-vector form of line; t∈0,1
R≔P1+t P2−P1:
Obtain ρ=x.2+y.2+z.2=R.
Calculus palette: Differentiation operator
Context Panel: Assign to a Name≻rho
ⅆⅆ t R = 3⁢2→assign to a nameρ
Form the integrand fxt,yt,zt⋅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.)
Context Panel: Evaluate Integral
x y zx=a|f(x)x=R1,y=R2,z=R3⋅ρ
3⁢1+4⁢t⁢2+t⁢3−t⁢2
→integrate w.r.t. t
=
1112⁢2
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