Back to Path-Independent Vector Fields and the Fundamental Theorem of Calculus for Line Integrals

Activity 13.5.4

Avg difficulty 2.5

Calculate each of the following line integrals.

  1. ∫CF⃗⋅dr⃗\int_C \vF\cdot d\vr if F⃗(x,y)=⟨2x,2y⟩\vF(x,y) = \langle 2x,2y\rangle and CC is the line segment from (1,2)(1,2) to (−1,0)(-1,0).

  2. ∫CG⃗⋅dr⃗\int_C \vG\cdot d\vr if G⃗(x,y)=⟨4x3−12ycos⁡(xy),9y2−12xcos⁡(xy)⟩\vG(x,y) = \langle 4x^3-12y\cos(xy),9y^2-12x\cos(xy)\rangle and CC is the portion of the unit circle from (0,−1)(0,-1) to (0,1)(0,1).

  3. ∫CH⃗⋅dr⃗\int_C \vH\cdot d\vr if H⃗(x,y,z)=⟨H1,H2,H3⟩\vH(x,y,z) = \langle H_1,H_2,H_3\rangle with H1(x,y,z)=ez2+2xy3z+cos⁡(x)−y3sin⁡(x)H2(x,y,z)=2yey2+3x2y2z+3y2z2+3y2cos⁡(x)H3(x,y,z)=x2y3+2xzez2+2y3z−4z3\begin{aligned} H_1(x,y,z) & = e^{z^2}+2xy^3z+\cos(x)-y^3\sin(x) \\ H_2(x,y,z) & = 2ye^{y^2}+3x^2y^2z+3y^2z^2+3y^2\cos(x) \\ H_3(x,y,z) & = x^2y^3+2xze^{z^2}+2y^3z-4z^3 \end{aligned} and CC is the curve consisting of the line segment from (1,1,1)(1,1,1) to (3,0,3)(3,0,3), followed by the line segment from (3,0,3)(3,0,3) to (1,5,−1)(1,5,-1), followed by the line segment from (1,5,−1)(1,5,-1) to (0,0,0)(0,0,0).

Long, Keller & Schlicker, Active Calculus Multivariable, §13.5 Path-Independent Vector Fields and the Fundamental Theorem of Calculus for Line Integrals, Activity 13.5.4

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Adapted from Active Calculus Multivariable © 2013–2026 Nicholas Long, Mitchel T. Keller, and Steven Schlicker. Licensed under CC BY-SA 4.0. Source: Active Calculus Multivariable. Imported at revision 692c03c.