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 vectors double integrals vector fields line integrals parametric surfaces greens theorem

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

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17calculus > vector fields > stokes' theorem

 Stokes' Theorem Example Meaning Deeper Explanation Practice

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

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 $$z=y^2$$, $$0\leq x\leq a$$, $$0\leq y\leq a$$.

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.

video by MIT OCW

video by MIT OCW

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

video by Evans Lawrence

### Stokes' Theorem Proofs Videos

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.

### Khan Academy - Stokes' Theorem Proof (7) [11mins-48secs]

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

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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.

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 $$x^2+y^2=1$$.

$$\pi$$

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 solution video

video by MIP4U

$$\pi$$

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 $$z=0$$.

$$-8\pi$$

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 solution video

video by MIP4U

$$-8\pi$$

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.

$$\pi$$

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 solution video

video by MIT OCW

$$\pi$$

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.

$$3 \pi/2$$

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 solution video

$$3 \pi/2$$

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.

$$20 \pi$$

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 solution video

video by MIP4U

$$20 \pi$$

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.

$$25 \pi$$

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 solution video

video by MIP4U

$$25 \pi$$

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

$$3 \pi/2$$

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 solution video

$$3 \pi/2$$

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.

$$-27 \pi$$

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 solution video

$$-27 \pi$$

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.

$$27 \pi$$

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 solution video

video by MIP4U

$$27 \pi$$

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

$$-18 \pi$$

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 solution video

video by Krista King Math

$$-18 \pi$$

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.

-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 solution video

video by Krista King Math