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17calculus > vector fields > stokes theorem

Stokes' Theorem

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, explanation 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 three-dimensional 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 counter-clockwise direction as the basemark for the positive direction.

Stokes' Theorem

For a surface S bounded by a closed curve C with the unit normal vector \(\vec{N}\)
determined by the orientation of S, this equation holds,

\(\displaystyle{ \oint\limits_C { \vec{F} \cdot d\vec{r} = \iint\limits_S { ( \nabla \times \vec{F} ) \cdot \vec{N} ~ dS } } }\)

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.

Krista King Math - What is Stokes' Theorem? [2min-38secs]

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.

Michael Hutchings - Statement of Stokes' Theorem [5min-53secs]

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

Michael Hutchings - General Examples of Stokes' Theorem [9min-50secs]

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

Example 1 - Verify Stokes' Theorem for \(\vec{F}(x,y,z)=z^2\hat{i}+x^2\hat{j}+y^2\hat{k}\) for the surface \(z=y^2\), \(0\leq x\leq a\), \(0\leq y\leq a\).

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.

MIP4U - Stokes' Theorem [3min-6secs]

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.

MIT OCW - Lec 31-32 | MIT 18.02 Multivariable Calculus, Fall 2007 [2 videos 98min-28secs total]

Evans Lawrence - Multivariable Calculus: Lecture 33 - Big Picture of Integration, Stokes Theorem [40min]

Stokes' Theorem Proof

These next sets of videos show the proof of Stokes' Theorem. Although not necessary in order to use Stokes' Theorem, these videos will give you a deeper understanding and appreciation for the theorem.

Michael Hutchings - Proof of Stokes' Theorem [16min-57secs]

Khan Academy - Stokes' Theorem Proof [7 videos]

Okay, you are now ready for some practice problems.
The next logical step is the Divergence Theorem, if you haven't already learned it.

Divergence Theorem →

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Practice Problems

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 from the above discussion 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, so choose the curve carefully.
- When the problem statement says to verify Stokes' Theorem, you need to evaluate both integrals to show that they are equal, like we did in Example 1 above.

Practice 1

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 \(x^2+y^2=1\).

answer

solution

Practice 2

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

answer

solution

Practice 3

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

answer

solution

Practice 4

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 counter-clockwise when viewed from above.

answer

solution

Practice 5

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

answer

solution

Practice 6

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.

answer

solution

Practice 7

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 \(\vec{F}=\langle\sin(x^3)+z^3,\sin(y^3),\sin(z^3)-x^3\rangle\).

answer

solution

Practice 8

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 \(z=9-x^2-y^2\) above the xy-plane, oriented upward.

answer

solution

Practice 9

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 \(z=y^2-x\) and the circular cylinder \(x^2+y^2=9\), oriented counter-clockwise when viewed from above.

answer

solution

Practice 10

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 \(\vec{F}(x,y,z) = 2y\cos z\hat{i} + e^x\sin z\hat{j} + xe^y\hat{k}\).

answer

solution

Practice 11

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 counter-clockwise when viewed from above.

answer

solution

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