# 9. Line Integrals

## d. Line Integrals of Vectors

So far we have learned about line integrals of scalars. However, one can also compute line integrals of vectors. To do this, we must first generalize the notion of the (scalar) differential of arc length.

The vector differential of arc length, \(d\vec s\) or \(d\vec r\), is the vector whose components are the differentials of the coordinates. \[ d\vec s=d\vec r=\langle dx,dy,dz\rangle \] For a parametrized curve, \(\vec r(t)=\langle x(t),y(t),z(t)\rangle\), it can be expressed in terms of the velocity, \(\vec v=\left\langle\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right\rangle\), as \[ d\vec s=d\vec r =\left\langle\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right\rangle\,dt =\vec v\,dt \] Notice that the length of the vector differential of arc length is just the scalar differential of arclength: \[ |d\vec s| =\sqrt{dx^2+dy^2+dz^2\,} =ds=|\vec v|\,dt \]

The notation \(d\vec r\) emphasizes that it is the vector whose components are the differentials of the coordinates. The notation \(d\vec s\) emphasizes that it is the vector whose length is the scalar differential of arclength.

Using the vector differential of arc length, we can now define the line integral of a vector. First recall from here and here:

A vector field is function which assigns a vector to each point of space. Thus a vector field \(\vec F\) assigns to each point \((x,y,z)\), the vector \[ \vec F(x,y,z)=\langle F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)\rangle \] The \(3\) functions \(F_1(x,y,z)\), \(F_2(x,y,z)\) and \(F_3(x,y,z)\) are the component functions of \(\vec F\).

The line integral of a vector field \(\vec F\) along a parametric curve \(\vec r(t)=(x(t),y(t),z(t))\) between \(A=\vec r(a)\) and \(B=\vec r(b)\) is given by \[ \int_A^B \vec F\cdot d\vec s =\int_a^b \vec F(\vec r(t))\cdot\vec v\,dt \qquad (1) \] where \(\vec F(\vec r(t))\) is the composition of \(\vec F\) and \(\vec r\). i.e. the function \(\vec F(x,y,z)\) with \(x\), \(y\) and \(z\) replaced by \(x(t)\), \(y(t)\) and \(z(t)\). The composition \(\vec F(\vec r(t))\) is called the value of the vector field along the curve or the restriction of the vector field to the curve.

We sometimes write \(\vec F(t)\) to mean \(\vec F(\vec r(t))\), the value of \(\vec F\) at time \(t\), even though this is not mathematically correct. Consequently, this is sometimes called abuse of notation.

Notice that we again switch the limits on the integral from the abstract points \(A\) and \(B\) to the values \(a\) and \(b\) of the parameter \(t\) when we switch from the integral with the general differential \(d\vec s\) to the integral with the parameter differential \(dt\).

There are several other notations for the line integral of a vector.

- Since the integrand is the dot product \(\vec F\cdot d\vec s\), the integral can be rewritten as \[ \int_A^B \vec F\cdot d\vec s =\int_A^B F_1\,dx+F_2\,dy+F_3\,dz \qquad (2) \]
- If we introduce the unit tangent vector \(\hat{T}=\dfrac{\vec v}{|\vec v|}\), then the vector differential of arclength can be written as \(d\vec s=\vec v\,dt=\hat{T}\,|\vec v|\,dt=\hat{T}\,ds\) and the integral can be rewritten as \[ \int_A^B \vec F\cdot d\vec s =\int_A^B \vec F\cdot\hat{T}\,ds \qquad (3) \]

When computing a line integral of a vector field, always use the forms shown in (1) or (2) as demonstrated in the examples. Never use version (3) for computation. If you do, you end up unnecessarily computing \(|\vec v|\) since you divide by \(|\vec v|\) when forming \(\hat{T}=\dfrac{\vec v}{|\vec v|}\) and multiply by it when forming \(ds=|\vec v|\,dt\) so that they cancel out in the end. The version (3) is used mostly for theoretical purposes, since it writes the vector integral as the scalar integral of the tangential component of the vector field. This will become clear when we discuss applications starting on the next page.

Compute \(\displaystyle \int_{(0,0,0)}^{(1,1,2/3)} \vec F\cdot d\vec s\) along the twisted cubic \(\vec r(t)=\left(t,t^2,\dfrac{2}{3}t^3\right)\) for the vector field \(\vec F=\langle 3z,2y,x\rangle\).

The value of \(\vec F\) along the curve is: \[ \vec F(\vec r(t))=\langle 2t^3,2t^2,t\rangle \] The velocity is \(\vec v=(1,2t,2t^2)\). So \(\vec F(\vec r(t))\cdot\vec v=2t^3+4t^3+2t^3=8t^3\), and the integral is \[ \int_{(0,0,0)}^{(1,1,2/3)} \vec F\cdot d\vec s =\int_0^1 \vec F(\vec r(t))\cdot\vec v\,dt =\int_0^1 8t^3\,dt=\left[2t^4\rule{0pt}{10pt}\right]_0^1=2 \]

Compute \(\displaystyle \int_{(0,0,0)}^{(1,1,2/3)} 3z\,dx+2y\,dy+x\,dz\) along the twisted cubic \(\vec r(t)=\left(t,t^2,\dfrac{2}{3}t^3\right)\).

Looking at \(\vec r(t)=(x(t),y(t),z(t)) =\left(t,t^2,\dfrac{2}{3}t^3\right)\), the differentials are: \[ dx=dt,\qquad dy=2t\,dt,\qquad dz=2t^2\,dt \] So the integral is \[\begin{aligned} \int_{(0,0,0)}^{(1,1,2/3)} &3z\,dx+2y\,dy+x\,dz \\ &=\int_0^1 3\left(\dfrac{2}{3}t^3\right)(dt)+2(t^2)(2t\,dt)+(t)(2t^2\,dt) \\ &=\int_0^1 8t^3\,dt =\left[2t^4\right]_0^1=2 \end{aligned}\]

These two examples are the same, just stated and computed using versions (1) and (2) of the line integral. It is totally your choice which to use, partially based on how the problem is stated. You should also be able to go back and forth between them.

Find the line integral of the vector field \(\vec F=\langle -y,x,z^2\rangle\) along the helix \(\vec r(\theta)=(4\cos\theta,4\sin\theta,3\theta)\) for \(0 \le \theta \le 3\pi\).

\[ \int_A^B \vec F\cdot d\vec s =\int_a^b \vec F(\vec r(\theta))\cdot\vec v\,d\theta \]

\(\displaystyle \int_{\vec r(0)}^{\vec r(3\pi)} \vec F\cdot d\vec s =48\pi+243\pi^3\)

We are computing the line integral: \[ \int_{\vec r(0)}^{\vec r(3\pi)} \vec F\cdot d\vec s =\int_0^{3\pi} \vec F(\vec r(\theta))\cdot\vec v\,d\theta \] The velocity vector is \(\vec v=\langle -4\sin\theta,4\cos\theta,3\rangle\) and the vector field \(\vec F=\langle -y,x,z^2\rangle\) evaluated on the curve is \(\vec F(\vec r(\theta)) =\langle -4\sin\theta,4\cos\theta,9\theta^2\rangle\). So their dot product is: \[ \vec F(\vec r(\theta))\cdot\vec v =16\sin^2\theta+16\cos^2\theta+27\theta^2 =16+27\theta^2 \] and the line integral is: \[\begin{aligned} \int_0^{3\pi} \vec F(\vec r(\theta))\cdot\vec v\,d\theta &=\int_0^{3\pi} 16+27\theta^2\,d\theta =\left[16\theta+9\theta^3\right]_0^{3\pi} \\ &=16(3\pi)+9(3\pi)^3 =48\pi+243\pi^3 \end{aligned}\]

We look at applications starting on the next page.

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