17Calculus - Line Integrals
Line Integrals
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Here is an important introduction to line integrals that explains very well the difference between line integrals and path integrals. He also explains the notation and how to calculate them. Notation is critical at this point since line integrals can be written several different ways. So be especially aware of that section of this video.
Dr Chris Tisdell - Intro to line integrals [27mins-35secs]
video by Dr Chris Tisdell |
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Similar to the path integrals video, this video gives the equations used to calculate the line integral, which are summarized below.
\(\displaystyle{ \int_{\mathcal{C}}{\vec{F} \cdot d\vec{s}} = \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \vec{c}'(t) ~dt } }\) |
This line integral form is called the vector form. Make sure you see the difference between this equation for the line integral and the equation given earlier for the path integral.
Line Integrals Using The Unit Tangent Vector
For a parameterized vector function \(\vec{c}(t)\), it's unit tangent vector is \(\displaystyle{ \hat{T} = \frac{\vec{c}'(t)}{\| \vec{c}'(t) \|} }\). Since we are dividing by \(\| \vec{c}'(t) \|\), we require that \(\vec{c}'(t) \neq \vec{0}\). Notice that we are dividing the vector by it's length, giving us a unit vector, i.e. \(\| \hat{T} \| = 1\).
Since the parameterized vector function \(\vec{c}(t)\) sweeps out the curve in a certain direction, the unit tangent vector will be the tangent vector in the forward direction, i.e. in the same direction that the curve is being swept out.
The unit tangent vector comes into the equation for the line integral given above. Here are the intermediate equations.
\( \int_{\mathcal{C}}{\vec{F} \cdot d\vec{s}} \) |
\( \int_{\mathcal{C}}{\vec{F} \cdot \hat{T} ds} \) |
\( \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \hat{T}(\vec{c}(t)) \| \vec{c}'(t) \|~dt } \) |
\( \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \displaystyle{ \frac{\vec{c}'(t)}{\| \vec{c}'(t) \|} } \| \vec{c}'(t) \|~dt } \) |
\( \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \vec{c}'(t) ~dt } \) |
These equations came from the last video above. We highly recommend that you download the notes for this topic from
Dr Chris Tisdell's site. The video also describes the geometric meaning of the unit tangent vector. If you haven't watched it yet, now would be a good time.
You will probably find in your class, textbook and some of these videos, the boundary between path integrals and line integrals is difficult to see. So don't be too worried. The techniques are very similar. You just need to get some practice with the notation.
Practice - Line Integrals, Unit Tangent Vector
Evaluate the line integral \( \vec{F} = \langle -y/(x^2+y^2), x/(x^2+y^2) \rangle \)
over the curve \( \mathcal{C}: \langle \cos(t), \sin(t) \rangle; 0 \leq t \leq 2\pi \)
Problem Statement |
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Evaluate the line integral \( \vec{F} = \langle -y/(x^2+y^2), x/(x^2+y^2) \rangle \)
over the curve \( \mathcal{C}: \langle \cos(t), \sin(t) \rangle; 0 \leq t \leq 2\pi \)
Solution |
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857 video
video by Dr Chris Tisdell |
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Evaluate the line integral \(\displaystyle{ \vec{F} = zx\hat{i} + zy\hat{j} + x\hat{k} }\)
\( \mathcal{C}: \langle \cos(t),\sin(t),t \rangle; 0 \leq t \leq \pi \)
Problem Statement |
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Evaluate the line integral \(\displaystyle{ \vec{F} = zx\hat{i} + zy\hat{j} + x\hat{k} }\)
\( \mathcal{C}: \langle \cos(t),\sin(t),t \rangle; 0 \leq t \leq \pi \)
Solution |
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1558 video
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Determine the work done by the conservative vector field \(\displaystyle{ \vec{F} = yz\hat{i} + xz\hat{j} + xy\hat{k} }\) along any smooth curve from \(A(0,0,0)\) to \(B(1,1,1)\).
Problem Statement |
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Determine the work done by the conservative vector field \(\displaystyle{ \vec{F} = yz\hat{i} + xz\hat{j} + xy\hat{k} }\) along any smooth curve from \(A(0,0,0)\) to \(B(1,1,1)\).
Solution |
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1557 video
video by Dr Chris Tisdell |
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Evaluate the line integral \( \vec{F} = \langle x^3, -z, 2xy \rangle \) over the curve \( \mathcal{C}: \langle t^2, \sqrt{t}, \sqrt{t} \rangle; 2 \leq t \leq 4 \)
Problem Statement |
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Evaluate the line integral \( \vec{F} = \langle x^3, -z, 2xy \rangle \) over the curve \( \mathcal{C}: \langle t^2, \sqrt{t}, \sqrt{t} \rangle; 2 \leq t \leq 4 \)
Solution |
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856 video
video by Dr Chris Tisdell |
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Consider the force field \(\vec{F}(x,y) = \langle y^2-x^3, x^5 \rangle\) Newtons, where \(x\) and \(y\) are in meters. Let \(C_1\) be the oriented curve from \((0,0)\) to \((1,1)\), along the graph of \(y=x\). Let \(C_2\) be the oriented curve along \((0,0)\) to \((1,1)\), along the graph of \(y=x^2\). Calculate the work done by \(\vec{F}\) along \(C_1\) and \(C_2\) and show that they are not equal.
Problem Statement |
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Consider the force field \(\vec{F}(x,y) = \langle y^2-x^3, x^5 \rangle\) Newtons, where \(x\) and \(y\) are in meters. Let \(C_1\) be the oriented curve from \((0,0)\) to \((1,1)\), along the graph of \(y=x\). Let \(C_2\) be the oriented curve along \((0,0)\) to \((1,1)\), along the graph of \(y=x^2\). Calculate the work done by \(\vec{F}\) along \(C_1\) and \(C_2\) and show that they are not equal.
Solution |
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He has several mistakes when writing this problem.
1. He says the curves start at \((0,1)\) but they actually start at \((0,0)\). He works the problem as if they start at \((0,0)\).
2. His second integral should be \(\int{ F(\hat{r}(t)) \cdot \hat{r}’(t)~dt }\)
Here is the link to the blog entry for this problem: blog entry
2213 video
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Calculate the work done by the vector field \( \vec{F}(x,y) = xy\hat{i} + 3y^2\hat{j} \) along the path \(x=11t^4\), \(y=t^3\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \) on \( 0 \leq t \leq 1 \).
Problem Statement |
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Calculate the work done by the vector field \( \vec{F}(x,y) = xy\hat{i} + 3y^2\hat{j} \) along the path \(x=11t^4\), \(y=t^3\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \) on \( 0 \leq t \leq 1 \).
Final Answer |
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45
Problem Statement |
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Calculate the work done by the vector field \( \vec{F}(x,y) = xy\hat{i} + 3y^2\hat{j} \) along the path \(x=11t^4\), \(y=t^3\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \) on \( 0 \leq t \leq 1 \).
Solution |
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2227 video
video by Michel vanBiezen |
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Final Answer |
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45 |
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The force exerted by an electric charge at the origin on a charged particle at the point \((x,y,z)\) with position vector \(\vec{r} = \langle x,y,z \rangle\) is \(\displaystyle{ \vec{F}(\vec{r}) = \frac{K\vec{r}}{|\vec{r}|^3}}\), where K is constant. Assume \(K=10\). Find the work done as the particle moves along a straight line from \((2,0,0)\) to \((2,2,3)\).
Problem Statement |
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The force exerted by an electric charge at the origin on a charged particle at the point \((x,y,z)\) with position vector \(\vec{r} = \langle x,y,z \rangle\) is \(\displaystyle{ \vec{F}(\vec{r}) = \frac{K\vec{r}}{|\vec{r}|^3}}\), where K is constant. Assume \(K=10\). Find the work done as the particle moves along a straight line from \((2,0,0)\) to \((2,2,3)\).
Solution |
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2299 video
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For these next 3 problems, notice that the vector function \(\vec{F}\) and the beginning and ending points are the same in all 3 problems. However, the line integrals are along 3 different paths.
Calculate the work done by the vector field \(\vec{F}(x,y)=(x+y)\hat{i} + (y-x)\hat{j}\) along the path \(x=y^2\) from \((1,1)\) to \((4,2)\) by evaluating the integral \(W=\int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} }\).
Problem Statement |
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Calculate the work done by the vector field \(\vec{F}(x,y)=(x+y)\hat{i} + (y-x)\hat{j}\) along the path \(x=y^2\) from \((1,1)\) to \((4,2)\) by evaluating the integral \(W=\int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} }\).
Solution |
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2233 video
video by Michel vanBiezen |
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Calculate the work done by the vector field \(\vec{F}(x,y)=(x+y)\hat{i} + (y-x)\hat{j}\) along the path \(x=2t^2+t+1\), \(y=1+t^2\) from \((1,1)\) to \((4,2)\) by evaluating the integral \(W=\int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} }\).
Problem Statement |
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Calculate the work done by the vector field \(\vec{F}(x,y)=(x+y)\hat{i} + (y-x)\hat{j}\) along the path \(x=2t^2+t+1\), \(y=1+t^2\) from \((1,1)\) to \((4,2)\) by evaluating the integral \(W=\int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} }\).
Solution |
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2234 video
video by Michel vanBiezen |
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Calculate the work done by the vector field \( \vec{F}(x,y) = (x+y)\hat{i} + (y-x)\hat{j} \) along \(y=1\) from \((1,1)\) to \((4,1)\) then along \(x=4\) from \((4,1)\) to \((4,2)\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \).
Problem Statement |
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Calculate the work done by the vector field \( \vec{F}(x,y) = (x+y)\hat{i} + (y-x)\hat{j} \) along \(y=1\) from \((1,1)\) to \((4,1)\) then along \(x=4\) from \((4,1)\) to \((4,2)\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \).
Solution |
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2235 video
video by Michel vanBiezen |
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Line Integrals in Differential Form
Another way to write line integrals of vector fields is in differential form.
If we are given a vector field \(\vec{F}(x,y,z) = M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k} \) and we need to find the line integral of this vector field over the curve \( \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} \), we can do the following, where \( \vec{r}'(t) = x'(t)\hat{i} + y'(t)\hat{j} + z'(t)\hat{k} \).
\(\displaystyle{ \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r}} }\) |
\(\displaystyle{ \int_{\mathcal{C}}{ \vec{F} \cdot \frac{d\vec{r}} {dt}~dt } }\) |
\(\displaystyle{ \int_{a}^{b}{ \left( M\hat{i} + N\hat{j} + P\hat{k} \right) \cdot \vec{r}'(t) ~dt } }\) |
\(\displaystyle{ \int_{a}^{b}{ \left[ M\frac{dx}{dt} + N\frac{dy}{dt} + P\frac{dz}{dt} \right] ~dt } }\) |
\(\displaystyle{ \int_{\mathcal{C}}{ M~dx + N~dy + P~dz } }\) |
Some general comments are in order.
1. The above equations relate to vector fields in (3-dim) space. For vector fields in the plane, we just drop the \(dz\) term to get \(\displaystyle{ \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r}} = \int_{\mathcal{C}}{ M~dx + N~dy } }\).
2. The final equation that we use to calculate the line integral does not seem to involve a vector field. However, you can see that M, N and P are the components of the vector field \(\vec{F}\). So keep in mind that this is a line integral of a vector field (not a path integral of a scalar function).
Practice - Line Integrals, Differential Form
Evaluate the line integral \( \int_{\mathcal{C}}{ xy~dx+y~dy } \) along the curve \( \mathcal{C}: \vec{r}(t)=2t\vec{i} +10t\vec{j}; 0 \leq t \leq 1 \)
Problem Statement |
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Evaluate the line integral \( \int_{\mathcal{C}}{ xy~dx+y~dy } \) along the curve \( \mathcal{C}: \vec{r}(t)=2t\vec{i} +10t\vec{j}; 0 \leq t \leq 1 \)
Final Answer |
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\(\displaystyle{\frac{190}{3}}\)
Problem Statement |
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Evaluate the line integral \( \int_{\mathcal{C}}{ xy~dx+y~dy } \) along the curve \( \mathcal{C}: \vec{r}(t)=2t\vec{i} +10t\vec{j}; 0 \leq t \leq 1 \)
Solution |
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\(x=2t\) | \(y=10t\) | |
\(dx=2~dt\) | \(dy=10~dt\) |
\(\displaystyle{ \int_{\mathcal{C}}{ xy~dx+y~dy } }\) |
\(\displaystyle{ \int_{0}^{1}{ (2t)(10t)2~dt + 10t(10~dt) } }\) |
\(\displaystyle{ \int_{0}^{1}{ 40t^2 + 100t ~ dt } }\) |
\(\displaystyle{ \left[ \frac{40t^3}{3} + 100\frac{t^2}{2} \right]_{0}^{1} }\) |
\(\displaystyle{ \frac{40}{3} +50 = \frac{190}{3} }\) |
Final Answer |
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\(\displaystyle{\frac{190}{3}}\) |
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Evaluate the line integral \( \int_{\mathcal{C}}{ (2x-y)~dx+(x+3y)~dy } \)
\( \mathcal{C}: \) along the x-axis from 0 to 5
Problem Statement |
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Evaluate the line integral \( \int_{\mathcal{C}}{ (2x-y)~dx+(x+3y)~dy } \)
\( \mathcal{C}: \) along the x-axis from 0 to 5
Final Answer |
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25
Problem Statement |
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Evaluate the line integral \( \int_{\mathcal{C}}{ (2x-y)~dx+(x+3y)~dy } \)
\( \mathcal{C}: \) along the x-axis from 0 to 5
Solution |
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To parameterize the curve, we let \(x=t\) which gives us the interval \(0\leq t\leq 5\). The y-value is just zero since the line is along the x-axis. So we have \(\vec{r}(t)=t\vec{i}+0\vec{j}\)
\(x=t\) | \(y=0\) | |
\(dx=dt\) | \(dy=0~dt\) |
\(\displaystyle{ \int_{\mathcal{C}}{ (2x-y)~dx+(x+3y)~dy } }\) |
\(\displaystyle{ \int_{0}^{5}{ 2t~dt } }\) |
\(\displaystyle{ \left[ \frac{2t^2}{2} \right]_{0}^{5} = 25 }\) |
Final Answer |
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25 |
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Evaluate the line integral \( \int_{\mathcal{C}}{ 9x~dy-9y~dx }\) along the curve \( \mathcal{C}: \langle 3\cos(t), 2\sin(t) \rangle; 0 \leq t \leq \pi \)
Problem Statement |
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Evaluate the line integral \( \int_{\mathcal{C}}{ 9x~dy-9y~dx }\) along the curve \( \mathcal{C}: \langle 3\cos(t), 2\sin(t) \rangle; 0 \leq t \leq \pi \)
Solution |
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855 video
video by Dr Chris Tisdell |
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Evaluate the line integral \( \int_{\mathcal{C}}{(2x+yz)dx +xz dy + xy dz} \) along the curve \( \mathcal{C}: \) straight line \( (0,0,0) \) to \( (1,2,-1) \)
Problem Statement |
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Evaluate the line integral \( \int_{\mathcal{C}}{(2x+yz)dx +xz dy + xy dz} \) along the curve \( \mathcal{C}: \) straight line \( (0,0,0) \) to \( (1,2,-1) \)
Solution |
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863 video
video by PatrickJMT |
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Compute \(\displaystyle{ \int_{(1,1)}^{(4,2)}{ (x+y)dx+(y-x)dy } }\) along the curve \(x=y^2\).
Problem Statement |
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Compute \(\displaystyle{ \int_{(1,1)}^{(4,2)}{ (x+y)dx+(y-x)dy } }\) along the curve \(x=y^2\).
Solution |
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1774 video
video by Dr Chris Tisdell |
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Fundamental Theorem of Line Integrals
Similar to the Fundamental Theorem of Calculus we have the Fundamental Theorem of Line Integrals. Basically, this says, that for a conservative vector field, the line integral is independent of the path and depends only on the endpoints.
Here is a great video that explains this theorem in detail and includes examples.
Dr Chris Tisdell - Line integrals: Fundamental theorem [19mins-36secs]
video by Dr Chris Tisdell |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
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external links you may find helpful |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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