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Line Integrals

Calculate a scalar line integral along a curve.

Scalar Line Integrals

A line integral gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. There are two types of line integrals: scalar line integrals and vector line integrals. Scalar line integrals are integrals of a scalar function over a curve in a plane or in space. Vector line integrals are integrals of a vector field over a curve in a plane or in space. Let’s look at scalar line integrals first.

A scalar line integral is defined just as a single-variable integral is defined, except that for a scalar line integral, the integrand is a function of more than one variable and the domain of integration is a curve in a plane or in space, as opposed to a curve on the x-axis.

For a scalar line integral, we let C be a smooth curve in a plane or in space and let \(f\) be a function with a domain that includes C. We chop the curve into small pieces. For each piece, we choose point P in that piece and evaluate \(f\) at P. (We can do this because all the points in the curve are in the domain of \(f.\)) We multiply \(f(P)\) by the arc length of the piece \(\text{\Delta }s,\) add the product \(f(P)\text{\Delta }s\) over all the pieces, and then let the arc length of the pieces shrink to zero by taking a limit. The result is the scalar line integral of the function over the curve.

You may have noticed a difference between this definition of a scalar line integral and a single-variable integral. In this definition, the arc lengths \(\text{\Delta }{s}_{1},\text{\Delta }{s}_{2}\text{,\ldots },\text{\Delta }{s}_{n}\) aren’t necessarily the same; in the definition of a single-variable integral, the curve in the x-axis is partitioned into pieces of equal length. This difference does not have any effect in the limit. As we shrink the arc lengths to zero, their values become close enough that any small difference becomes irrelevant.

If \(f\) is a continuous function on a smooth curve C, then \({\int }_{C}fds\) always exists. Since \({\int }_{C}fds\) is defined as a limit of Riemann sums, the continuity of \(f\) is enough to guarantee the existence of the limit, just as the integral \({\int }_{a}^{b}g(x)dx\) exists if g is continuous over \([a,b].\)

\[\text{length}({C}_{i})=\text{\Delta }{s}_{i}={\int }_{{t}_{i-1}}^{{t}_{i}}‖{r}^{'}(t)‖dt.\]\[{\int }_{{t}_{i-1}}^{{t}_{i}}‖{r}^{'}(t)‖dt\approx ‖{r}^{'}({t}_{i}^{*})‖\text{\Delta }{t}_{i},\]\[‖{r}^{'}(t)‖=\sqrt{{(x'(t))}^{2}+{(y'(t))}^{2}+{(z'(t))}^{2},}\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Vector Line Integrals

The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let

\[\text{F}(x,y,z)=P(x,y,z)\text{i}+Q(x,y,z)\text{j}+R(x,y,z)\text{k}\]

be a continuous vector field in \({ℝ}^{3}\) that represents a force on a particle, and let C be a smooth curve in \({ℝ}^{3}\) contained in the domain of \(\text{F}.\) How would we compute the work done by \(\text{F}\) in moving a particle along C?

To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve C; such a specified direction is called an orientation of a curve. The specified direction is the positive direction along C; the opposite direction is the negative direction along C. When C has been given an orientation, C is called an oriented curve (). The work done on the particle depends on the direction along the curve in which the particle is moving.

A closed curve is one for which there exists a parameterization \(\text{r}(t),\) \(a\le t\le b,\) such that \(\text{r}(a)=\text{r}(b),\) and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain \((a,b).\)

Let \(\text{r}(t)\) be a parameterization of C for \(a\le t\le b\) such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along C. Divide the parameter interval \([a,b]\) into n subintervals \([{t}_{i-1},{t}_{i}],0\le i\le n,\) of equal width. Denote the endpoints of \(\text{r}({t}_{0}),\text{r}({t}_{1})\text{,\ldots },\text{r}({t}_{n})\) by \({P}_{0}\text{,\ldots },{P}_{n}.\) Points P_i divide C into n pieces. Denote the length of the piece from P_i−1 to P_i by \(\text{\Delta }{s}_{i}.\) For each i, choose a value \({t}_{i}^{*}\) in the subinterval \([{t}_{i-1},{t}_{i}].\) Then, the endpoint of \(\text{r}({t}_{i}^{*})\) is a point in the piece of C between \({P}_{i-1}\) and P_i (). If \(\text{\Delta }{s}_{i}\) is small, then as the particle moves from \({P}_{i-1}\) to \({P}_{i}\) along C, it moves approximately in the direction of \(\text{T}({P}_{i}),\) the unit tangent vector at the endpoint of \(\text{r}({t}_{i}^{*}).\) Let \({P}_{i}^{*}\) denote the endpoint of \(\text{r}({t}_{i}^{*}).\) Then, the work done by the force vector field in moving the particle from \({P}_{i-1}\) to P_i is \(\text{F}({P}_{i}^{*})\cdot (\text{\Delta }{s}_{i}\text{T}({P}_{i}^{*})),\) so the total work done along C is

\[W={\int }_{C}\text{F}\cdot \text{T}ds,\]

which gives us the concept of a vector line integral.

Condensed — the full section is in OpenStax Calculus Volume 3.

Applications of Line Integrals

Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.

Suppose that a piece of wire is modeled by curve C in space. The mass per unit length (the linear density) of the wire is a continuous function \(\rho (x,y,z).\) We can calculate the total mass of the wire using the scalar line integral \({\int }_{C}\rho (x,y,z)ds.\) The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by \(\rho (x*,y*,z*)\text{\Delta }s\) for some point \((x*,y*,z*)\) in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral \({\int }_{C}\rho (x,y,z)ds.\)

Example

Try it.

Calculate the mass of a spring in the shape of a curve parameterized by \(〈t,2\ \text{cos}\ t,2\ \text{sin}\ t〉,\) \(0\le t\le \frac{\pi }{2},\) with a density function given by \(\rho (x,y,z)={e}^{x}+yz\) kg/m ().

Solution

To calculate the mass of the spring, we must find the value of the scalar line integral \({\int }_{C}({e}^{x}+yz)ds,\) where C is the given helix. To calculate this integral, we write it in terms of t using :

\[\begin{array}{ll}{\int }_{C}{e}^{x}+yzds & ={\int }_{0}^{\pi \text{/}2}(({e}^{t}+4\ \text{cos}\ t\ \text{sin}\ t)\sqrt{1+{(-2\ \text{cos}\ t)}^{2}+{(2\ \text{sin}\ t)}^{2}})dt \\ & ={\int }_{0}^{\pi \text{/}2}(({e}^{t}+4\ \text{cos}\ t\ \text{sin}\ t)\sqrt{5})dt \\ & =\sqrt{5}{[{e}^{t}+2\ {\text{sin}}^{2}t]}_{t=0}^{t=\pi \text{/}2} \\ & =\sqrt{5}({e}^{\pi \text{/}2}+1).\end{array}\]

Therefore, the mass is \(\sqrt{5}({e}^{\pi \text{/}2}+1)\) kg.

Condensed — the full section is in OpenStax Calculus Volume 3.

Flux and Circulation

We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used in applications to calculate fluid flow across a curve, and the concept of circulation is important for characterizing conservative gradient fields in terms of line integrals. Both these concepts are used heavily throughout the rest of this chapter. The idea of flux is especially important for Green’s theorem, and in higher dimensions for Stokes’ theorem and the divergence theorem.

Let C be a plane curve and let F be a vector field in the plane. Imagine C is a membrane across which fluid flows, but C does not impede the flow of the fluid. In other words, C is an idealized membrane invisible to the fluid. Suppose F represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing C?

Recall that the line integral of F along C is \({\int }_{C}\text{F}\cdot \text{T}ds\)—in other words, the line integral is the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector \(\text{N}(t)\) and instead compute integral \({\int }_{C}\text{F}\cdot \text{N}ds,\) we determine the flux across C. To be precise, the definition of integral \({\int }_{C}\text{F}\cdot \text{N}ds\) is the same as integral \({\int }_{C}\text{F}\cdot \text{T}ds,\) except the T in the Riemann sum is replaced with N. Therefore, the flux across C is defined as

\[{\int }_{C}\text{F}\cdot \text{N}ds=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\text{F}({P}_{i}^{*})\cdot \text{N}({P}_{i}^{*})\text{\Delta }{s}_{i},\]

where \({P}_{i}^{*}\) and \(\text{\Delta }{s}_{i}\) are defined as they were for integral \({\int }_{C}\text{F}\cdot \text{T}ds.\) Therefore, a flux integral is an integral that is perpendicular to a vector line integral, because N and T are perpendicular vectors.

If F is a velocity field of a fluid and C is a curve that represents a membrane, then the flux of F across C is the quantity of fluid flowing across C per unit time, or the rate of flow.

More formally, let C be a plane curve parameterized by \(\text{r}(t)=〈x(t),y(t)〉,\) \(a\le t\le b.\) Let \(\text{n}(t)=〈{y}^{'}(t),\text{-}{x}^{'}(t)〉\) be the vector that is normal to C at the endpoint of \(\text{r}(t)\) and points to the right as we traverse C in the positive direction (). Then, \(\text{N}(t)=\frac{\text{n}(t)}{‖\text{n}(t)‖}\) is the unit normal vector to C at the endpoint of \(\text{r}(t)\) that points to the right as we traverse C.

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

Line integrals generalize the notion of a single-variable integral to higher dimensions. The domain of integration in a single-variable integral is a line segment along the x-axis, but the domain of integration in a line integral is a curve in a plane or in space.
If C is a curve, then the length of C is \({\int }_{C}ds.\)
There are two kinds of line integral: scalar line integrals and vector line integrals. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field.
Scalar line integrals can be calculated using ; vector line integrals can be calculated using .
Two key concepts expressed in terms of line integrals are flux and circulation. Flux measures the rate that a field crosses a given line; circulation measures the tendency of a field to move in the same direction as a given closed curve.

Key Equations

Calculating a scalar line integral	\({\int }_{C}f(x,y,z)ds={\int }_{a}^{b}f(\text{r}(t))\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}+{({z}^{'}(t))}^{2}}dt\)
Calculating a vector line integral	\({\int }_{C}\text{F}\cdot dr={\int }_{C}\text{F}\cdot \text{T}ds={\int }_{a}^{b}\text{F}(\text{r}(t))\cdot {r}^{'}(t)dt\) or \({\int }_{C}Pdx+Qdy+Rdz={\int }_{a}^{b}(P(\text{r}(t))\frac{dx}{dt}+Q(\text{r}(t))\frac{dy}{dt}+R(\text{r}(t))\frac{dz}{dt})dt\)
Calculating flux	\({\int }_{C}\text{F}\cdot \frac{\text{n}(t)}{‖\text{n}(t)‖}\ ds={\int }_{a}^{b}\text{F}(\text{r}(t))\cdot \text{n}(t)dt\)

Line Integrals

For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path.

For the following exercises, find the work done.

For the following exercises, evaluate the line integrals.

In the following exercises, find the work done by force field F on an object moving along the indicated path.

For the following exercises, use a CAS to evaluate the given line integrals.

For the following exercises, find the flux.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the value of integral \({\int }_{C}2ds,\) where \(C\) is the upper half of the unit circle.

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The integrand is \(f(x,y)=2.\) shows the graph of \(f(x,y)=2,\) curve C, and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width \(\pi\) and length 2. Therefore, \({\int }_{C}2ds=2\pi .\)
To see that \({\int }_{C}2ds=2\pi\) using the definition of line integral, we let \(\text{r}(t)\) be a parameterization of C. Then, \(f(\text{r}({t}_{i}))=2\) for any number \({t}_{i}\) in the domain of r. Therefore,
\[\begin{array}{ll}{\int }_{C}fds & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f(\text{r}({t}_{i}^{*}))\text{\Delta }{s}_{i} \\ & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}2\text{\Delta }{s}_{i} \\ & =2\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\text{\Delta }{s}_{i} \\ & =2(\text{length of C}) \\ & =2\pi .\end{array}\]
Find the value of \({\int }_{C}(x+y)ds,\) where \(C\) is the curve parameterized by \(x=t,\) \(y=t,\) \(0\le t\le 1.\)

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\(\sqrt{2}\)
Find the value of integral \({\int }_{C}({x}^{2}+{y}^{2}+z)ds,\) where \(C\) is part of the helix parameterized by \(\text{r}(t)=〈\text{cos}\ t,\text{sin}\ t,t〉,\) \(0\le t\le 2\pi .\)

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To compute a scalar line integral, we start by converting the variable of integration from arc length s to t. Then, we can use to compute the integral with respect to t. Note that \(f(\text{r}(t))={\text{cos}}^{2}t+{\text{sin}}^{2}t+t=1+t\) and
\[\begin{array}{ll}\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}+{({z}^{'}(t))}^{2}} & =\sqrt{{(\text{-}\text{sin}(t))}^{2}+{\text{cos}}^{2}(t)+1} \\ & =\sqrt{2}.\end{array}\]
Therefore,
\[{\int }_{C}({x}^{2}+{y}^{2}+z)ds={\int }_{0}^{2\pi }(1+t)\sqrt{2}dt.\]
Notice that translated the original difficult line integral into a manageable single-variable integral. Since
\[\begin{array}{ll}{\int }_{0}^{2\pi }(1+t)\sqrt{2}dt & ={[\sqrt{2}t+\frac{\sqrt{2}{t}^{2}}{2}]}_{0}^{2\pi } \\ & =2\sqrt{2}\pi +2\sqrt{2}{\pi }^{2},\end{array}\]
we have
\[{\int }_{C}({x}^{2}+{y}^{2}+z)ds=2\sqrt{2}\pi +2\sqrt{2}{\pi }^{2}.\]
Evaluate \({\int }_{C}({x}^{2}+{y}^{2}+z)ds,\) where C is the curve with parameterization \(\text{r}(t)=⟨\text{sin}(3t),\text{cos}(3t)\text{,t}⟩,0\le t\le 2\pi .\)

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\(2\sqrt{10}\pi +2\sqrt{10}{\pi }^{2}\)
Find the value of integral \({\int }_{C}({x}^{2}+{y}^{2}+z)ds,\) where \(C\) is part of the helix parameterized by \(\text{r}(t)=〈\text{cos}(2t),\text{sin}(2t),2t〉,0\le t\le \pi .\) Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast.

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As with the previous example, we use to compute the integral with respect to t. Note that \(f(\text{r}(t))={\text{cos}}^{2}(2t)+{\text{sin}}^{2}(2t)+2t=2t+1\) and
\[\begin{array}{ll}\sqrt{{({x}^{'}(t))}^{2}+{({y}^{'}(t))}^{2}+{(z'(t))}^{2}} & =\sqrt{{\left(\text{-2sin}\left(\text{2}t\right)\right)}^{2}+{\left(2\text{cos}\left(\text{2}t\right)\right)}^{2}+{2}^{2}} \\ & \sqrt{4{\text{sin}}^{2}\left(2t\right)+4{\text{cos}}^{2}\left(2t\right)+4} \\ & =2\sqrt{2}\end{array}\]
so we have
\[\begin{array}{ll}{\int }_{C}({x}^{2}+{y}^{2}+z)ds & =2\sqrt{2}{\int }_{0}^{\pi }(1+2t)dt \\ & =2\sqrt{2}{[t+{t}^{2}]}_{0}^{\pi } \\ & =2\sqrt{2}(\pi +{\pi }^{2}).\end{array}\]
Notice that this agrees with the answer in the previous example. Changing the parameterization did not change the value of the line integral. Scalar line integrals are independent of parameterization, as long as the curve is traversed exactly once by the parameterization.
Evaluate line integral \({\int }_{C}({x}^{2}+yz)ds,\) where \(C\) is the line with parameterization \(\text{r}(t)=〈2t,5t,\text{-}t〉,0\le t\le 10.\) Reparameterize C with parameterization \(\text{s}(t)=〈4t,10t,-2t〉,0\le t\le 5,\) recalculate line integral \({\int }_{C}({x}^{2}+yz)ds,\) and notice that the change of parameterization had no effect on the value of the integral.

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Both line integrals equal \(-\frac{1000\sqrt{30}}{3}.\)
A wire has a shape that can be modeled with the parameterization \(\text{r}(t)=⟨\text{cos}\ t,\text{sin}\ t,\frac{2}{3}{t}^{3/2}⟩,0\le t\le 4\pi .\) Find the length of the wire.

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The length of the wire is given by \({\int }_{C}1ds,\) where C is the curve with parameterization r. Therefore,
\[\begin{array}{ll}\text{The length of the wire} & ={\int }_{C}1ds \\ & ={\int }_{0}^{4\pi }‖{r}^{'}(t)‖dt \\ & ={\int }_{0}^{4\pi }\sqrt{{(\text{-}\text{sin}\ t)}^{2}+{\text{cos}}^{2}t+t}dt \\ & ={\int }_{0}^{4\pi }\sqrt{1+t}dt \\ & ={[\frac{2{(1+t)}^{3\text{/}2}}{3}]}_{0}^{4\pi } \\ & =\frac{2}{3}({(1+4\pi )}^{3\text{/}2}-1).\end{array}\]
Find the length of a wire with parameterization \(\text{r}(t)=〈3t+1,4-2t,5+2t〉,0\le t\le 4.\)

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\(4\sqrt{17}\)
Find the value of integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(C\) is the semicircle parameterized by \(\text{r}(t)=〈\text{cos}\ t,\text{sin}\ t〉,\) \(0\le t\le \pi\) and \(\text{F}=〈\text{-}y,x〉.\)

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We can use to convert the variable of integration from r to t. We then have
\[\text{F}(\text{r}(t))=〈\text{-}\text{sin}\ t,\text{cos}\ t〉\ \text{and}\ {r}^{'}(t)=〈\text{-}\text{sin}\ t,\text{cos}\ t〉.\]
Therefore,
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{0}^{\pi }〈\text{-}\text{sin}\ t,\text{cos}\ t〉\cdot 〈\text{-}\text{sin}\ t,\text{cos}\ t〉dt \\ & ={\int }_{0}^{\pi }{\text{sin}}^{2}t+{\text{cos}}^{2}tdt \\ & ={\int }_{0}^{\pi }1dt=\pi .\end{array}\]
See .
Find the value of integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(C\) is the semicircle parameterized by \(\text{r}(t)=〈\text{cos}\ (t+\pi ),\text{sin}\ t〉,0\le t\le \pi\) and \(\text{F}=〈\text{-}y,x〉.\)

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Notice that this is the same problem as , except the orientation of the curve has been reversed. In this example, the parameterization starts at \(\text{r}(0)=〈-1,0〉\) and ends at \(\text{r}(\pi )=〈1,0〉.\) By ,
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{0}^{\pi }〈\text{-}\text{sin}\ t,\text{cos}\ (t+\pi )〉\cdot 〈\text{-}\text{sin}\ (t+\pi ),\text{cos}\ t〉dt \\ & ={\int }_{0}^{\pi }〈\text{-}\text{sin}\ t,\text{-}\text{cos}\ t〉\cdot 〈\text{sin}\ t,\text{cos}\ t〉dt \\ & ={\int }_{0}^{\pi }(\text{-}{\text{sin}}^{2}t-{\text{cos}}^{2}t)dt \\ & ={\int }_{0}^{\pi }-1dt \\ & =\text{-}\pi .\end{array}\]
Notice that this is the negative of the answer in . It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.
Let \(\text{F}=x\text{i}+y\text{j}\) be a vector field and let C be the curve with parameterization \(〈t,{t}^{2}〉\) for \(0\le t\le 2.\) Which is greater: \({\int }_{C}\text{F}\cdot \text{T}ds\) or \({\int }_{\text{-}C}\text{F}\cdot \text{T}ds?\)

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\({\int }_{C}\text{F}\cdot \text{T}ds\)
Find the value of integral \({\int }_{C}zdx+xdy+ydz,\) where C is the curve parameterized by \(\text{r}(t)=〈{t}^{2},\sqrt{t},t〉,1\le t\le 4.\)

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As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t. In this case, allows us to make this change:
\[\begin{array}{ll}{\int }_{C}zdx+xdy+ydz & ={\int }_{1}^{4}(t(2t)+{t}^{2}(\frac{1}{2\sqrt{t}})+\sqrt{t})dt \\ & ={\int }_{1}^{4}(2{t}^{2}+\frac{{t}^{3\text{/}2}}{2}+\sqrt{t})dt \\ & ={[\frac{2{t}^{3}}{3}+\frac{{t}^{5\text{/}2}}{5}+\frac{2{t}^{3\text{/}2}}{3}]}_{t=1}^{t=4} \\ & =\frac{793}{15}.\end{array}\]
Find the value of \({\int }_{C}4xdx+zdy+4{y}^{2}dz,\) where \(C\) is the curve parameterized by \(\text{r}(t)=〈4\ \text{cos}(2t),2\ \text{sin}(2t),3〉,0\le t\le \frac{\pi }{4}.\)

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\(-26\)
Find the value of integral \({\int }_{C}\text{F}\cdot \text{T}ds,\) where C is the rectangle (oriented counterclockwise) in a plane with vertices \((0,0),(2,0),(2,1),\ \text{and}\ (0,1),\) and where \(\text{F}=〈x-2y,y-x〉\) ().

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Note that curve C is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let \({C}_{1}\) represent the side from \((0,0)\) to \((2,0),\) let \({C}_{2}\) represent the side from \((2,0)\) to \((2,1),\) let \({C}_{3}\) represent the side from \((2,1)\) to \((0,1),\) and let \({C}_{4}\) represent the side from \((0,1)\) to \((0,0)\) (). Then,
\[{\int }_{C}\text{F}\cdot \text{T}d\text{r}={\int }_{{C}_{1}}\text{F}\cdot \text{T}d\text{r}+{\int }_{{C}_{2}}\text{F}\cdot \text{T}d\text{r}+{\int }_{{C}_{3}}\text{F}\cdot \text{T}d\text{r}+{\int }_{{C}_{4}}\text{F}\cdot \text{T}d\text{r}.\]
We want to compute each of the four integrals on the right-hand side using . Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse C counterclockwise):
\[\begin{array}{l}{C}_{1}:〈t,0〉,0\le t\le 2 \\ {C}_{2}:〈2,t〉,0\le t\le 1 \\ {C}_{3}:〈2-t,1〉,0\le t\le 2 \\ {C}_{4}:〈0,1-t〉,0\le t\le 1.\end{array}\]
Therefore,
\[\begin{array}{ll}{\int }_{{C}_{1}}\text{F}\cdot \text{T}d\text{r} & ={\int }_{0}^{2}\text{F}(\text{r}(t))\cdot {r}^{'}(t)dt \\ & ={\int }_{0}^{2}⟨t-2(0),0-t⟩\cdot ⟨1,0⟩dt={\int }_{0}^{2}tdt \\ & ={[\frac{{t}^{2}}{2}]}_{0}^{2}=2.\end{array}\]
Notice that the value of this integral is positive, which should not be surprising. As we move along curve C₁ from left to right, our movement flows in the general direction of the vector field itself. At any point along C₁, the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along C₁, and the line integral will have positive value.
The calculations for the three other line integrals are done similarly:
\[\begin{array}{ll}{\int }_{{C}_{2}}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}〈2-2t,t-2〉\cdot 〈0,1〉dt \\ & ={\int }_{0}^{1}(t-2)dt \\ & ={[\frac{{t}^{2}}{2}-2t]}_{0}^{1}=-\frac{3}{2},\end{array}\]\[\begin{array}{ll}{\int }_{{C}_{3}}\text{F}\cdot \text{T}ds & ={\int }_{0}^{2}〈(2-t)-2,1-(2-t)〉\cdot 〈-1,0〉dt \\ & ={\int }_{0}^{2}tdt=2,\end{array}\]
and
\[\begin{array}{ll}{\int }_{{C}_{4}}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}〈-2(1-t),1-t〉\cdot 〈0,-1〉dt \\ & ={\int }_{0}^{1}(t-1)dt \\ & ={[\frac{{t}^{2}}{2}-t]}_{0}^{1}=-\frac{1}{2}.\end{array}\]
Thus, we have \({\int }_{C}\text{F}\cdot d\text{r}=2.\)
Calculate line integral \({\int }_{C}\text{F}\cdot d\text{r},\) where F is vector field \(〈{y}^{2},2xy+1〉\) and C is a triangle with vertices \((0,0),\) \((4,0),\) and \((0,5),\) oriented counterclockwise.

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0
Calculate the mass of a spring in the shape of a curve parameterized by \(〈t,2\ \text{cos}\ t,2\ \text{sin}\ t〉,\) \(0\le t\le \frac{\pi }{2},\) with a density function given by \(\rho (x,y,z)={e}^{x}+yz\) kg/m ().

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To calculate the mass of the spring, we must find the value of the scalar line integral \({\int }_{C}({e}^{x}+yz)ds,\) where C is the given helix. To calculate this integral, we write it in terms of t using :
\[\begin{array}{ll}{\int }_{C}{e}^{x}+yzds & ={\int }_{0}^{\pi \text{/}2}(({e}^{t}+4\ \text{cos}\ t\ \text{sin}\ t)\sqrt{1+{(-2\ \text{cos}\ t)}^{2}+{(2\ \text{sin}\ t)}^{2}})dt \\ & ={\int }_{0}^{\pi \text{/}2}(({e}^{t}+4\ \text{cos}\ t\ \text{sin}\ t)\sqrt{5})dt \\ & =\sqrt{5}{[{e}^{t}+2\ {\text{sin}}^{2}t]}_{t=0}^{t=\pi \text{/}2} \\ & =\sqrt{5}({e}^{\pi \text{/}2}+1).\end{array}\]
Therefore, the mass is \(\sqrt{5}({e}^{\pi \text{/}2}+1)\) kg.
Calculate the mass of a spring in the shape of a helix parameterized by \(\text{r}(t)=〈\text{cos}\ t,\text{sin}\ t,t〉,0\le t\le 6\pi ,\) with a density function given by \(\rho (x,y,z)=x+y+z\) kg/m.

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\(18\sqrt{2}{\pi }^{2}\) kg
How much work is required to move an object in vector force field \(\text{F}=〈yz,xy,xz〉\) along path \(\text{r}(t)=〈{t}^{2},t,{t}^{4}〉,\) \(0\le t\le 1?\) See .

Fi àwọn àgbèwọlé hàn

Let C denote the given path. We need to find the value of \({\int }_{C}\text{F}\cdot d\text{r}.\) To do this, we use :
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}(〈{t}^{5},{t}^{3},{t}^{6}〉\cdot 〈2t,1,4{t}^{3}〉)dt \\ & ={\int }_{0}^{1}(2{t}^{6}+{t}^{3}+4{t}^{9})dt \\ & ={[\frac{2{t}^{7}}{7}+\frac{{t}^{4}}{4}+\frac{2{t}^{10}}{5}]}_{t=0}^{t=1}=\frac{131}{140}.\end{array}\]
Calculate the flux of \(\text{F}=〈2x,2y〉\) across a unit circle oriented counterclockwise ().

Fi àwọn àgbèwọlé hàn

To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization \(\text{r}(t)=〈\text{cos}\ t,\text{sin}\ t〉,\) \(0\le t\le 2\pi .\) The normal vector to a unit circle is \(〈\text{cos}\ t,\text{sin}\ t〉.\) Therefore, the flux is
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot \text{N}ds & ={\int }_{0}^{2\pi }〈2\ \text{cos}\ t,2\ \text{sin}\ t〉\cdot 〈\text{cos}\ t,\text{sin}\ t〉\ dt \\ & ={\int }_{0}^{2\pi }(2\ {\text{cos}}^{2}t+2\ {\text{sin}}^{2}t)\ dt=2{\int }_{0}^{2\pi }({\text{cos}}^{2}t+{\text{sin}}^{2}t)\ dt \\ & =2{\int }_{0}^{2\pi }dt=4\pi .\end{array}\]
Calculate the flux of \(\text{F}=〈x+y,2y〉\) across the line segment from \((0,0)\) to \((2,3),\) where the curve is oriented from left to right.

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3/2
Let \(\text{F}=〈-y,x〉\) be the vector field from and let C represent the unit circle oriented counterclockwise. Calculate the circulation of F along C.

Fi àwọn àgbèwọlé hàn

We use the standard parameterization of the unit circle: \(\text{r}(t)=〈\text{cos}\ t,\text{sin}\ t〉,0\le t\le 2\pi .\) Then, \(\text{F}(\text{r}(t))=〈\text{-}\text{sin}\ t,\text{cos}\ t〉\) and \({r}^{'}(t)=〈\text{-}\text{sin}\ t,\text{cos}\ t〉.\) Therefore, the circulation of F along C is
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot \text{T}ds & ={\int }_{0}^{2\pi }〈\text{-}\text{sin}\ t,\text{cos}\ t〉\cdot 〈\text{-}\text{sin}\ t,\text{cos}\ t〉dt \\ & ={\int }_{0}^{2\pi }({\text{sin}}^{2}t+{\text{cos}}^{2}t)\ dt \\ & ={\int }_{0}^{2\pi }dt=2\pi .\end{array}\]
Notice that the circulation is positive. The reason for this is that the orientation of C “flows” with the direction of F. At any point along the circle, the tangent vector and the vector from F form an angle of less than 90°, and therefore the corresponding dot product is positive.
Calculate the circulation of \(\text{F}(x,y)=〈-\frac{y}{{x}^{2}+{y}^{2}},\frac{x}{{x}^{2}+{y}^{2}}〉\) along a unit circle oriented counterclockwise.

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\(2\pi\)
Calculate the work done on a particle that traverses circle C of radius 2 centered at the origin, oriented counterclockwise, by field \(\text{F}(x,y)=〈-2,y〉.\) Assume the particle starts its movement at \((1,0).\)

Fi àwọn àgbèwọlé hàn

The work done by F on the particle is the circulation of F along C: \({\int }_{C}\text{F}\cdot \text{T}ds.\) We use the parameterization \(\text{r}(t)=〈2\ \text{cos}\ t,2\ \text{sin}\ t〉,0\le t\le 2\pi\) for C. Then, \({r}^{'}(t)=〈-2\ \text{sin}\ t,2\ \text{cos}\ t〉\) and \(\text{F}(\text{r}(t))=〈-2,2\ \text{sin}\ t〉.\) Therefore, the circulation of F along C is
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot \text{T}ds & ={\int }_{0}^{2\pi }〈-2,2\ \text{sin}\ t〉\cdot 〈-2\ \text{sin}\ t,2\ \text{cos}\ t〉dt \\ & ={\int }_{0}^{2\pi }(4\ \text{sin}\ t+4\ \text{sin}\ t\ \text{cos}\ t)dt \\ & ={[-4\ \text{cos}\ t+4\ {\text{sin}}^{2}t]}_{0}^{2\pi } \\ & =(-4\ \text{cos}(2\pi )+2\ {\text{sin}}^{2}(2\pi ))-(-4\ \text{cos}(0)+4\ {\text{sin}}^{2}(0)) \\ & =-4+4=0.\end{array}\]
The force field does zero work on the particle.
Notice that the circulation of F along C is zero. Furthermore, notice that since F is the gradient of \(f(x,y)=-2x+\frac{{y}^{2}}{2},\) F is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.
Calculate the work done by field \(\text{F}(x,y)=〈2x,3y〉\) on a particle that traverses the unit circle. Assume the particle begins its movement at \((-1,0).\)

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0
True or False? Line integral \({\int }_{C}^{}f(x,y)ds\) is equal to a definite integral if C is a smooth curve defined on \([a,b]\) and if function \(f\) is continuous on some region that contains curve C.

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True
True or False? Vector functions \({\text{r}}_{1}=t\text{i}+{t}^{2}\text{j},\) \(0\le t\le 1,\) and \({\text{r}}_{2}=(1-t)\text{i}+{(1-t)}^{2}\text{j},\) \(0\le t\le 1,\) define the same oriented curve.
True or False? \({\int }_{\text{-}C}^{}(Pdx+Qdy)={\int }_{C}^{}(Pdx-Qdy)\)

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False
True or False? A piecewise smooth curve C consists of a finite number of smooth curves that are joined together end to end.
True or False? If C is given by \(x(t)=t\text{,}\ y(t)=t\text{, 0}\le \text{t}\le 1,\) then \({\int }_{C}^{}xyds={\int }_{0}^{1}{t}^{2}dt.\)

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False
[T] \({\int }_{C}^{}(x+y)ds\)
\(C\text{:}\ x=t,y=(1-t)\text{,}\ z=0\) from (0, 1, 0) to (1, 0, 0)
[T] \({\int }_{C}^{}(x-y)ds\)
\(C\text{:}\ \text{r}(t)=4t\text{i}+3t\text{j}\) when \(0\le t\le 2\)

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\({\int }_{C}^{}(x-y)ds=10\)
[T] \({\int }_{C}^{}({x}^{2}+{y}^{2}+{z}^{2})ds\)
\(C\text{:}\ \text{r}(t)=\text{sin}\ t\text{i}+\text{cos}\ t\text{j}+8t\text{k}\) when \(0\le t\le \frac{\pi }{2}\)
[T] Evaluate \({\int }_{C}^{}x{y}^{4}ds,\) where C is the right half of circle \({x}^{2}+{y}^{2}=16\) and is traversed in the clockwise direction.

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\({\int }_{C}^{}x{y}^{4}ds=\frac{8192}{5}\)
[T] Evaluate \({\int }_{C}^{}4{x}^{3}ds,\) where C is the line segment from \((-2,-1)\) to (1, 2).
Find the work done by vector field \(\text{F}(x,y,z)=x\text{i}+3xy\text{j}-(x+z)\text{k}\) on a particle moving along a line segment that goes from \((1,4,2)\) to \((0,5,1).\)

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\(W=8\)
Find the work done by a person weighing 150 lb walking exactly one revolution up a circular, spiral staircase of radius 3 ft if the person rises 10 ft.
Find the work done by force field \(\text{F}(x,y,z)=-\frac{1}{2}x\text{i}-\frac{1}{2}y\text{j}+\frac{1}{4}\text{k}\) on a particle as it moves along the helix \(\text{r}(t)=\text{cos}\ t\text{i}+\text{sin}\ t\text{j}+t\text{k}\) from point \((1,0,0)\) to point \((-1,0,3\pi ).\)

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\(W=\frac{3\pi }{4}\)
Find the work done by vector field \(\text{F}(x,y)=y\text{i}+2x\text{j}\) in moving an object along path C, the straight line which joins points (1, 0) and (0, 1).
Find the work done by force \(\text{F}(x,y)=2y\text{i}+3x\text{j}+(x+y)\text{k}\) in moving an object along curve \(\text{r}(t)=\text{cos}(t)\text{i}+\text{sin}(t)\text{j}+\frac{1}{6}\text{k},\) where \(0\le t\le 2\pi .\)

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\(W=\pi\)
Find the mass of a wire in the shape of a circle of radius 2 centered at (3, 4) with linear mass density \(\rho (x,y)={y}^{2}.\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Line Integrals

Calculate a scalar line integral along a curve.
Calculate a vector line integral along an oriented curve in space.
Use a line integral to compute the work done in moving an object along a curve in a vector field.
Describe the flux and circulation of a vector field.
Suppose instead that
Line integrals generalize the notion of a single-variable integral to higher dimensions. The domain of integration in a single-variable integral is a line segment along the
If
There are two kinds of line integral: scalar line integrals and vector line integrals. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field.

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Wárá

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Diẹ̀ nínú Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems