Multivariable Calculus · free preview

5.4 Using Parameterizations to Calculate Line Integrals

How can we use a parametrization of an oriented curve $C$ to calculate $\int_C\vF\cdot d\vr$?

1. Using Parameterizations to Calculate Line Integrals

Using Parameterizations to Calculate Line Integrals

2. Introduction

Introduction

3. Parameterizations in the Definition of

Parameterizations in the Definition of

4. Alternative Notation for Line Integrals

Alternative Notation for Line Integrals

5. Let $\vF=\langle xy…$

Let \vF=⟨xy,y2⟩\vF=\langle xy,y^2\rangle, let C1C_1 be the line segment from (1,1)(1,1) to (4,1)(4,1), let C2C_2 be the line segment from (4,1)(4,1) to (4,3)(4,3), and let C3C_3 be the line segment from (1,1)(1,1) to (4,3)(4,3). Also let C=C1+C2C = C_1 + C_2. This vector field and the curves are shown in .

Every point along C1C_1 has y=1y=1. Therefore, along C1C_1, the vector field \vF\vF can be viewed purely as a function of xx. In particular, along C1C_1, we have \vF(x,1)=⟨x,1⟩\vF(x,1) = \langle x,1\rangle. Since every point along C2C_2 has the same xx-value, write \vF\vF as a function of yy only (for the points on C2C_2).

6. Work through this ex…

Work through this exercise and explain your reasoning step by step.

7. Work through this ex…

Work through this exercise and explain your reasoning step by step.

8. Let $\vF(x…$

Let \vF(x,y)=x\vi+y2\vj\vF(x,y) = x\vi + y^2\vj and let CC be the quarter of the circle of radius 33 from (0,3)(0,3) to (3,0)(3,0). This vector field and curve are shown in . By properties of line integrals, we know that ∫C\vF⋅d\vr=−∫−C\vF⋅d\vr\int_C \vF\cdot d\vr = -\int_{-C}\vF\cdot d\vr, and we will use this property since −C-C is the usual clockwise orientation of a circle, meaning we can parametrize −C-C by \vr(t)=⟨3cos⁡(t),3sin⁡(t)⟩\vr(t) = \langle 3\cos(t),3\sin(t)\rangle for 0≤t≤π/20\leq t\leq \pi/2.

To evaluate ∫−C\vF⋅d\vr\int_{-C}\vF\cdot d\vr using this parametrization, we need to note that

\vF(\vr(t))=⟨3cos⁡(t),9sin⁡2(t)⟩ and \vr′(t)=⟨−3sin⁡(t),3cos⁡(t)⟩\vF(\vr(t)) = \langle 3\cos(t) , 9\sin^2(t)\rangle\qquad\text{ and } \qquad \vr'(t) = \langle -3\sin(t),3\cos(t)\rangle

Thus, we have

∫C\vF⋅d\vr=−∫−C\vF⋅d\vr=−∫0π/2⟨3cos⁡(t),9sin⁡2(t)⟩⋅⟨−3sin⁡(t),3cos⁡(t)⟩ dt=−∫0π/2(−9sin⁡(t)cos⁡(t)+27sin⁡2(t)cos⁡(t)) dt=−∫01(−9u+27u2) du=−[−92u2+9u3]01=−(−92+9)=−92\begin{aligned} \int_C\vF\cdot d\vr & = -\int_{-C}\vF\cdot d\vr \\ & = -\int_0^{\pi/2} \langle 3\cos(t),9\sin^2(t)\rangle\cdot\langle-3\sin(t),3\cos(t)\rangle\, dt \\ & = -\int_0^{\pi/2} \left(-9\sin(t)\cos(t) + 27\sin^2(t)\cos(t)\right)\, dt \\ & = -\int_0^1 (-9 u + 27u^2)\, du = -\left[ -\frac{9}{2}u^2 + 9u^3\right]_0^1 \\ & = -\left(-\frac{9}{2} + 9\right) = -\frac{9}{2} \end{aligned}

. Note that we have used the substitution u=sin⁡(t)u =\sin(t) in evaluating the definite integral here.

9. Find the work done b…

Find the work done by the vector field \vF(x,y,z)=6x2z\vi+3y2\vj+x\vk\vF(x,y,z) = 6x^2z\vi + 3y^2\vj + x\vk on a particle that moves from the point (3,0,0)(3,0,0) to the point (3,0,6π)(3,0,6\pi) along the helix given by \vr(t)=⟨3cos⁡(t),3sin⁡(t),t⟩\vr(t) = \langle 3\cos(t),3\sin(t),t\rangle.

10. Let $\vF(x…$

Let \vF(x,y)=⟨0,x⟩\vF(x,y) = \langle 0,x\rangle. Let CC be the closed curve consisting of the top half of the circle of radius 22 centered at the origin and the portion of the xx-axis from (2,0)(2,0) to (−2,0)(-2,0), oriented clockwise. Find the circulation of \vF\vF around CC.

11. Let $C_1$ be the por…

Let C1C_1 be the portion of the graph of y=2x3+3x2−12x−15y=2x^3+3x^2-12x-15 from (−2,5)(-2,5) to (3,30)(3,30). Calculate ∫C1\vF⋅d\vr\int_{C_1}\vF\cdot d\vr.

12. Let $C_2$ be the lin…

Let C2C_2 be the line segment from (−2,5)(-2,5) to (3,30)(3,30). Calculate ∫C2\vF⋅d\vr\int_{C_2}\vF\cdot d\vr.

13. Let $C_3$ be the cir…

Let C3C_3 be the circle of radius 33 centered at the origin, oriented counterclockwise. Calculate ∮C3\vF⋅d\vr\oint_{C_3} \vF\cdot d\vr.

14. Let $\vF(x…$

Let \vF(x,y)=⟨y2,2xy+3⟩\vF(x,y) = \langle y^2,2xy+3\rangle.

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