This page contains lots of videos from various sources. Stokes' Theorem is not something that you can just study once and expect to understand right away. You need time to think about it, studying explanations from multiple sources with the concepts explained in several ways. In addition, you need to work plenty of practice problems. Stokes' Theorem has deep implications and uses that can only be understood with lots of work. So take your time to work through this page and the videos.
Stokes' Theorem is the threedimensional version of the circulation form of Green's Theorem and it relates the line integral around a closed curve C to the curl of a vector field over the surface. Remember that the \(curl ~ \vec{F} = \nabla \times \vec{F}\) indicates the tendency of \(\vec{F}\) to circulate around the surface S or cause S to turn. We use the counterclockwise direction as the basemark for the positive direction.
Stokes' Theorem [proof] 

For a surface S bounded by a closed curve C with the unit normal vector \(\vec{N}\) 
\(\displaystyle{ \oint\limits_C { \vec{F} \cdot d\vec{r} = \iint\limits_S { ( \nabla \times \vec{F} ) \cdot \vec{N} ~ dS } } }\) 
Okay, so what does Stokes' Theorem say? Before looking at the conditions and integrals in more detail, let's watch a video clip that will help us understand what we are about to read.
video by Krista King Math 

Figure 1 [source: Pauls Online Notes  Stokes' Theorem] 

First, we must have a surface S that is bounded by a closed curve C. An example is shown in Figure 1. Notice that there is an important relationship between S and C. The surface S must be bounded by C and C must be closed (if it wasn't closed, it couldn't bound S).
Next, the first integral is a line integral around the closed curve C. This line integral is calculating how much of the vector field \(\vec{F}\) is in the same direction as the curve C. Another thing to note about C is that it must be oriented such that the inside of the curve and the surface must be on your left if you were to imagine yourself walking on the curve. This is important since this will make the normal, \(\vec{N}\), in the other integral the outward pointing normal.
Now, the second integral is the curl of \(\vec{F}\) at each outward pointing normal vector over the surface S. Now think about this. The vector field along the curve C is equal to the sum of all the curl values around each normal vector over the surface. And the shape of the surface does not matter. As long as C bounds S, the equality holds. Is that cool or what?! Okay, let's watch a quick video going over the statement of Stokes' Theorem.
video by Michael Hutchings 

This next video contains a couple of general examples of Stokes' Theorem to start to give you a feel for how it works.
video by Michael Hutchings 

When doing actual calculations, one of the integrals is usually easier to set up and evaluate than the other. For example, if the curve C has multiple segments, then the surface integral may be easier to work with. But if the curve has one or maybe two segments where you need to evaluate the line integral separately, then the line integral may be easier to evaluate. You will get a better feel for how to use theorem with more experience.
Before we go on, let's do a specific example.
Stokes' Theorem Example
Verify Stokes' Theorem for \(\vec{F}(x,y,z)=z^2\hat{i}+x^2\hat{j}+y^2\hat{k}\) for the surface \(z=y^2\), \(0\leq x\leq a\), \(0\leq y\leq a\).
Verify Stokes' Theorem for \(\vec{F}(x,y,z)=z^2\hat{i}+x^2\hat{j}+y^2\hat{k}\) for the surface \(z=y^2\), \(0\leq x\leq a\), \(0\leq y\leq a\).
Solution 

Plot 1  plotted with 3d Grapher 

Verify Stokes' Theorem for \(\vec{F}(x,y,z)=z^2\hat{i}+x^2\hat{j}+y^2\hat{k}\) for the surface \(z=y^2\), \(0\leq x\leq a\), \(0\leq y\leq a\).
When the problem says to verify Stokes' Theorem, it means to calculate both integrals and confirm that they are equal. Plot 1 shows the surface that we are asked to work with. Let's start with the surface integral (in this case, it is the easier of the two but this is not always the case). We show the outline of the solution. It will help you a lot to fill in the details.
\(curl~\vec{F}=2y\hat{i}+2z\hat{j}+2x\hat{k}\)
For the equation, \(z=y^2\), we subtract \(y^2\) from both sides and call this new function \(G=zy^2\). We need to do this to calculate the gradient since \(\vec{N}~dS=\vec{\nabla}G~dA\).
\(\vec{\nabla}G=2y\hat{j}+\hat{k}\)
The area is the shadow of the surface in the xyplane, so our integral at this point is
\(\displaystyle{\int_{0}^{a}{\int_{0}^{a}{ 4yz+2x~dy~dx } }}\)
We know that \(z=y^2\), so our integral \(\displaystyle{\int_{0}^{a}{\int_{0}^{a}{ 4y^3+2x~dy~dx } }}\) is ready to integrate.
The result of the final integration is \(a^3a^5\). This should be what we get when we integrate along the curve. So let's do that now.
We will use the line integral \(\int_{C}{M~dx+N~dy+P~dz}\).
Notice that we have 4 sides, so we need 4 integrals. We need to make sure that the curve is oriented anticlockwise. So we start at \((0,0,0)\), numbering the curves as we go.
1. \(C_1\) is the line from \((0,0,0)\), along the xaxis to \((a,0,0)\).
In this case, \(y=0 \to dy=0\) and \(z=0 \to dz=0\) giving us \(\int_{0}^{a}{z^2~dx} \to \int_{0}^{a}{0~dx}=0\)
2. \(C_2\) is the line from \((a,0,0)\) along the line \(z=y^2\) to \((a,a,a^2)\).
In this case, \(x=a \to dx=0\) and \(z=y^2 \to dz=2y~dy\) giving us \(\int_{C}{x^2~dy+y^2~dz} \to \)
\(\int_{0}^{a}{a^2~dy+y^2(2y~dy) = }\) \(a^3+a^4/2\)
3. \(C_3\) is the line from \((a,a,a^2)\) parallel to the xaxis to \((0,a,a^2)\).
Since y and z are constant, \(dy=dz=0\). So the only term we need to concern ourselves with is the xterm, \(\int_{C}{z^2~dx}\). We need to be careful here to integrate in the right direction, starting from \(x=a\) and ending at \(x=0\) giving us \(\int_{a}^{0}{a^4~dx}=a^5\)
4. \(C_4\) is the line from \((0,a,a^2)\) to \((0,0,0)\)
Since \(x=0\), we are left with \(\int_{C}{y^2~dz}\). Again, we need to be careful to integrate in the right direction, \(y=a \to y=0\) giving us \(\int_{a}^{0}{y^2(2y~dy)}=a^4/2\)
Adding up the results from all 4 integrals gives us \(0+a^3+a^4/2a^5a^4/2=a^3a^5\). This is the same result as the first integral and so Stokes' Theorem is verified in this case.
Note  As is usual in higher mathematics, we have left out a lot of detail in this solution. We do not expect you to be able to sit and look at this solution and visualize all the steps in your head. You need to get out a pencil and a piece of paper and fill in the details yourself in order to understand where everything comes from. Get used to doing this all the time with examples and solutions. If you do, then you will understand and be able to use calculus and higher math.
Meaning of Stokes' Theorem
Okay, so what does Stokes' Theorem actually mean? Here is a quick video clip explaining in more detail what Stokes' Theorem is actually saying and under what conditions we can apply the theorem.
video by MIP4U 

Stokes' Theorem Deeper Explanation
Okay, so, now you should have a beginning understanding of Stokes' Theorem. But there is a lot more to it than you have read so far. Here are a couple of good lectures explaining the theorem at a deeper level. Watch as many of them as you can so that you have a very good grasp on how to use Stokes' Theorem. It will also help you understand this theorem better if you go over the proof of Stokes' Theorem found on this separate page.
video by MIT OCW 

video by MIT OCW 

video by Evans Lawrence 

Okay, you are now ready for some practice problems. After that, the next logical step is the Divergence Theorem, if you haven't already learned it.
Practice
Unless otherwise instructed, solve these problems using Stokes' Theorem.
 If you are given a curve, use Stokes' Theorem by evaluating the surface integral. Remember that Stokes' Theorem will hold for any surface. So you can choose the simplest surface in order to work out the surface integral.
 Similarly, if you are given a surface, use Stokes' Theorem by evaluating the line integral. Remember the curve must bound the surface.
 When the problem statement says to verify Stokes' Theorem, you need to evaluate both integrals to show that they are equal.
Evaluate \( \oint\limits_C { \vec{F} \cdot d\vec{r}} \) where \( \vec{F} = \langle y^2/2, x, z^2 \rangle\) and C is the intersection of the plane \(y+z=4\) and the cylinder \(x^2+y^2=1\).
Problem Statement 

Evaluate \( \oint\limits_C { \vec{F} \cdot d\vec{r}} \) where \( \vec{F} = \langle y^2/2, x, z^2 \rangle\) and C is the intersection of the plane \(y+z=4\) and the cylinder \(x^2+y^2=1\).
Final Answer 

\( \pi \)
Problem Statement 

Evaluate \( \oint\limits_C { \vec{F} \cdot d\vec{r}} \) where \( \vec{F} = \langle y^2/2, x, z^2 \rangle\) and C is the intersection of the plane \(y+z=4\) and the cylinder \(x^2+y^2=1\).
Solution 

video by MIP4U 

Final Answer 

\( \pi \) 
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Evaluate the surface integral for \( \vec{F} = \langle z,2x,xy \rangle \) and S is \(z=4x^2y^2\) above \(z=0\).
Problem Statement 

Evaluate the surface integral for \( \vec{F} = \langle z,2x,xy \rangle \) and S is \(z=4x^2y^2\) above \(z=0\).
Final Answer 

\( 8\pi \)
Problem Statement 

Evaluate the surface integral for \( \vec{F} = \langle z,2x,xy \rangle \) and S is \(z=4x^2y^2\) above \(z=0\).
Solution 

video by MIP4U 

Final Answer 

\( 8\pi \) 
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Verify Stokesâ€™ Theorem for \( \vec{F} = \langle 2z, x, y \rangle \) where S is the top half of the unit sphere.
Problem Statement 

Verify Stokesâ€™ Theorem for \( \vec{F} = \langle 2z, x, y \rangle \) where S is the top half of the unit sphere.
Final Answer 

\( \pi \)
Problem Statement 

Verify Stokesâ€™ Theorem for \( \vec{F} = \langle 2z, x, y \rangle \) where S is the top half of the unit sphere.
Solution 

video by MIT OCW 

Final Answer 

\( \pi \) 
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Calculate \( \int_{C}{ \vec{F} \cdot d\vec{r} }\) where \( \vec{F} = \langle \sin(\sin z), z^3, y^3 \rangle \) and C is the curve \(x^2+y^2=1, z=x\) oriented counterclockwise when viewed from above.
Problem Statement 

Calculate \( \int_{C}{ \vec{F} \cdot d\vec{r} }\) where \( \vec{F} = \langle \sin(\sin z), z^3, y^3 \rangle \) and C is the curve \(x^2+y^2=1, z=x\) oriented counterclockwise when viewed from above.
Final Answer 

\( 3 \pi/2 \)
Problem Statement 

Calculate \( \int_{C}{ \vec{F} \cdot d\vec{r} }\) where \( \vec{F} = \langle \sin(\sin z), z^3, y^3 \rangle \) and C is the curve \(x^2+y^2=1, z=x\) oriented counterclockwise when viewed from above.
Solution 

video by Michael Hutchings 

Final Answer 

\( 3 \pi/2 \) 
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Evaluate the surface integral for \( \vec{F} = \langle yz,xz,xy \rangle \) where S is the part of the paraboloid \(z=11x^2y^2\) that lies above the plane \(z=10\), oriented upwards.
Problem Statement 

Evaluate the surface integral for \( \vec{F} = \langle yz,xz,xy \rangle \) where S is the part of the paraboloid \(z=11x^2y^2\) that lies above the plane \(z=10\), oriented upwards.
Final Answer 

\( 20 \pi \)
Problem Statement 

Evaluate the surface integral for \( \vec{F} = \langle yz,xz,xy \rangle \) where S is the part of the paraboloid \(z=11x^2y^2\) that lies above the plane \(z=10\), oriented upwards.
Solution 

video by MIP4U 

Final Answer 

\( 20 \pi \) 
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Evaluate the surface integral for \( \vec{F} = \langle x^2e^{yz},xe^{xz},z^2e^{xy} \rangle \) where S is the hemisphere \( x^2+y^2+z^2=25, z\geq 0 \), oriented upwards.
Problem Statement 

Evaluate the surface integral for \( \vec{F} = \langle x^2e^{yz},xe^{xz},z^2e^{xy} \rangle \) where S is the hemisphere \( x^2+y^2+z^2=25, z\geq 0 \), oriented upwards.
Final Answer 

\( 25 \pi \)
Problem Statement 

Evaluate the surface integral for \( \vec{F} = \langle x^2e^{yz},xe^{xz},z^2e^{xy} \rangle \) where S is the hemisphere \( x^2+y^2+z^2=25, z\geq 0 \), oriented upwards.
Solution 

video by MIP4U 

Final Answer 

\( 25 \pi \) 
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Calculate \( \int_{C}{ \vec{F} \cdot d\vec{r} }\) where C is \( \vec{r}(t) = \langle \cos t,0,\sin t \rangle \), \(0\leq t\leq 2\pi\) for \( \vec{F} = \langle \sin(x^3)+z^3, \sin(y^3), \sin(z^3)x^3 \rangle \).
Problem Statement 

Calculate \( \int_{C}{ \vec{F} \cdot d\vec{r} }\) where C is \( \vec{r}(t) = \langle \cos t,0,\sin t \rangle \), \(0\leq t\leq 2\pi\) for \( \vec{F} = \langle \sin(x^3)+z^3, \sin(y^3), \sin(z^3)x^3 \rangle \).
Final Answer 

\( 3 \pi/2 \)
Problem Statement 

Calculate \( \int_{C}{ \vec{F} \cdot d\vec{r} }\) where C is \( \vec{r}(t) = \langle \cos t,0,\sin t \rangle \), \(0\leq t\leq 2\pi\) for \( \vec{F} = \langle \sin(x^3)+z^3, \sin(y^3), \sin(z^3)x^3 \rangle \).
Solution 

video by Michael Hutchings 

Final Answer 

\( 3 \pi/2 \) 
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Evaluate \( \iint_S{ curl \vec{F } \cdot d\vec{S} } \) for \( \vec{F}(x,y,z) = \langle 3y,4z,6x \rangle \) where S is the paraboloid \(z=9x^2y^2\) above the xyplane, oriented upward.
Problem Statement 

Evaluate \( \iint_S{ curl \vec{F } \cdot d\vec{S} } \) for \( \vec{F}(x,y,z) = \langle 3y,4z,6x \rangle \) where S is the paraboloid \(z=9x^2y^2\) above the xyplane, oriented upward.
Final Answer 

\( 27 \pi \)
Problem Statement 

Evaluate \( \iint_S{ curl \vec{F } \cdot d\vec{S} } \) for \( \vec{F}(x,y,z) = \langle 3y,4z,6x \rangle \) where S is the paraboloid \(z=9x^2y^2\) above the xyplane, oriented upward.
Solution 

video by Michael Hutchings 

Final Answer 

\( 27 \pi \) 
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Evaluate \( \int_C{ \vec{F}\cdot d\vec{r} } \) for \( \vec{F} = \langle xy,2z,5y \rangle \) where C is the curve of intersection of the parabolic cylinder \(z=y^2x\) and the circular cylinder \(x^2+y^2=9\), oriented counterclockwise when viewed from above.
Problem Statement 

Evaluate \( \int_C{ \vec{F}\cdot d\vec{r} } \) for \( \vec{F} = \langle xy,2z,5y \rangle \) where C is the curve of intersection of the parabolic cylinder \(z=y^2x\) and the circular cylinder \(x^2+y^2=9\), oriented counterclockwise when viewed from above.
Final Answer 

\( 27 \pi \)
Problem Statement 

Evaluate \( \int_C{ \vec{F}\cdot d\vec{r} } \) for \( \vec{F} = \langle xy,2z,5y \rangle \) where C is the curve of intersection of the parabolic cylinder \(z=y^2x\) and the circular cylinder \(x^2+y^2=9\), oriented counterclockwise when viewed from above.
Solution 

video by MIP4U 

Final Answer 

\( 27 \pi \) 
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Find \( \iint_S{ curl \vec{F}\cdot dS } \) where S is the hemisphere \(x^2+y^2+z^2=9\) with \(z\geq0\), oriented upward and \(\vec{F}(x,y,z) = 2y\cos z\hat{i} + e^x\sin z\hat{j} + xe^y\hat{k}\).
Problem Statement 

Find \( \iint_S{ curl \vec{F}\cdot dS } \) where S is the hemisphere \(x^2+y^2+z^2=9\) with \(z\geq0\), oriented upward and \(\vec{F}(x,y,z) = 2y\cos z\hat{i} + e^x\sin z\hat{j} + xe^y\hat{k}\).
Final Answer 

\( 18 \pi \)
Problem Statement 

Find \( \iint_S{ curl \vec{F}\cdot dS } \) where S is the hemisphere \(x^2+y^2+z^2=9\) with \(z\geq0\), oriented upward and \(\vec{F}(x,y,z) = 2y\cos z\hat{i} + e^x\sin z\hat{j} + xe^y\hat{k}\).
Solution 

video by Krista King Math 

Final Answer 

\( 18 \pi \) 
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Find \(\int_C{ \vec{F}\cdot dr }\) where C is the triangle with vertices \((1,0,0), (0,1,0), (0,0,1)\) and \( \vec{F}(x,y,z) = (x+y^2)\hat{i} + (y+z^2)\hat{j} + (z+x^2)\hat{k} \). C is counterclockwise when viewed from above.
Problem Statement 

Find \(\int_C{ \vec{F}\cdot dr }\) where C is the triangle with vertices \((1,0,0), (0,1,0), (0,0,1)\) and \( \vec{F}(x,y,z) = (x+y^2)\hat{i} + (y+z^2)\hat{j} + (z+x^2)\hat{k} \). C is counterclockwise when viewed from above.
Final Answer 

1
Problem Statement 

Find \(\int_C{ \vec{F}\cdot dr }\) where C is the triangle with vertices \((1,0,0), (0,1,0), (0,0,1)\) and \( \vec{F}(x,y,z) = (x+y^2)\hat{i} + (y+z^2)\hat{j} + (z+x^2)\hat{k} \). C is counterclockwise when viewed from above.
Solution 

video by Krista King Math 

Final Answer 

1 
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You CAN Ace Calculus
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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Practice Instructions
Unless otherwise instructed, solve these problems using Stokes' Theorem.
 If you are given a curve, use Stokes' Theorem by evaluating the surface integral. Remember that Stokes' Theorem will hold for any surface. So you can choose the simplest surface in order to work out the surface integral.
 Similarly, if you are given a surface, use Stokes' Theorem by evaluating the line integral. Remember the curve must bound the surface.
 When the problem statement says to verify Stokes' Theorem, you need to evaluate both integrals to show that they are equal.