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17Calculus - Stokes' Theorem

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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 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 [proof]

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? [2mins-39secs]

video by Krista King Math

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 [5mins-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 [9mins-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

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

example 1

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 \(G=z-y^2\). 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 xy-plane, 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 \(a^3-a^5\). 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 anti-clockwise. 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 x-axis 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 x-axis 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 x-term, \(\int_{C}{z^2~dx}\). 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 \(0+a^3+a^4/2-a^5-a^4/2=a^3-a^5\). 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.

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

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.

MIT OCW - Lec 31 | MIT 18.02 Multivariable Calculus, Fall 2007 [48mins-20secs]

video by MIT OCW

MIT OCW - Lec 32 | MIT 18.02 Multivariable Calculus, Fall 2007 [50mins-8secs]

video by MIT OCW

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

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

1975 video

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 \(z=4-x^2-y^2\) above \(z=0\).

Problem Statement

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

Final Answer

\( -8\pi \)

Problem Statement

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

Solution

1976 video

video by MIP4U

Final Answer

\( -8\pi \)

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

Problem Statement

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

Solution

1977 video

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 counter-clockwise 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 counter-clockwise 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 counter-clockwise when viewed from above.

Solution

1978 video

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 \(z=11-x^2-y^2\) 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 \(z=11-x^2-y^2\) 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 \(z=11-x^2-y^2\) that lies above the plane \(z=10\), oriented upwards.

Solution

1979 video

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

1980 video

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

1981 video

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

Solution

1982 video

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

Solution

1983 video

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

2146 video

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 counter-clockwise 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 counter-clockwise 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 counter-clockwise when viewed from above.

Solution

2147 video

video by Krista King Math

Final Answer

-1

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You CAN Ace Calculus

Trig Formulas

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 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle 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

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

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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Stokes' Theorem

Example

Meaning

Deeper Explanation

Practice

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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.

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