maths.free › Multivariable Calculus › 6. Vector Calculus › Stokes’ Theorem
Stokes’ Theorem
Explain the meaning of Stokes’ theorem.
Stokes’ Theorem
Stokes’ theorem says we can calculate the flux of curl F across surface S by knowing information only about the values of F along the boundary of S. Conversely, we can calculate the line integral of vector field F along the boundary of surface S by translating to a double integral of the curl of F over S.
Let S be an oriented smooth surface with unit normal vector N. Furthermore, suppose the boundary of S is a simple closed curve C. The orientation of S induces the positive orientation of C if, as you walk in the positive direction around C with your head pointing in the direction of N, the surface is always on your left. With this definition in place, we can state Stokes’ theorem.
Suppose surface S is a flat region in the xy-plane with upward orientation. Then the unit normal vector is k and surface integral \(\underset{S}{∬}\text{curl}\ \text{F}\cdot d\text{S}\) is actually the double integral \(\underset{S}{∬}\text{curl}\ \text{F}\cdot \text{k}dA.\) In this special case, Stokes’ theorem gives \({\int }_{C}\text{F}\cdot d\text{r}={∬}_{S}\text{curl}\ \text{F}\cdot \text{k}dA.\) However, this is the circulation form of Green’s theorem, which shows us that Green’s theorem is a special case of Stokes’ theorem. Green’s theorem can only handle surfaces in a plane, but Stokes’ theorem can handle surfaces in a plane or in space.
The complete proof of Stokes’ theorem is beyond the scope of this text. We look at an intuitive explanation for the truth of the theorem and then see proof of the theorem in the special case that surface S is a portion of a graph of a function, and S, the boundary of S, and F are all fairly tame.
Condensed — the full section is in OpenStax Calculus Volume 3.
Applying Stokes’ Theorem
Stokes’ theorem translates between the flux integral of surface S to a line integral around the boundary of S. Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa. We now study some examples of each kind of translation.
Example
Try it.
Calculate surface integral \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S},\) where S is the surface, oriented outward, in and \(\text{F}=〈z,2xy,x+y〉.\)
Solution
Note that to calculate \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}\) without using Stokes’ theorem, we would need to use . Use of this equation requires a parameterization of S. Surface S is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes’ theorem, noting that the boundary C of the surface is merely a single circle with radius 1.
By Stokes’ theorem,
\[{∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}={\int }_{C}\text{F}\cdot d\text{r},\]where C has parameterization \(r(t)=〈\text{sin}\ t,0,1-\text{cos}\ t〉,0\le t<2\pi .\) By ,
\[\begin{array}{ll}{∬}_{S}\text{curl}\ \text{F}\cdot d\text{S} & ={\int }_{C}\text{F}\cdot d\text{r} \\ & ={\int }_{0}^{2\pi }⟨1-\text{cos}\ t\ ,0,+\text{sin}\ t⟩\cdot \ ⟨-\text{sin}\ t,0,\text{cos}\ t⟩dt\ \\ & ={\int }_{0}^{2\pi }(-\text{sin}\ t+2\text{sin}\ t\text{cos}\ t)dt \\ & ={\int }_{0}^{2\pi }(-\text{sin}\ t+\text{sin 2t) }\text{dt} \\ & ={\left[\text{+cos}t+\frac{1}{2}\text{cos}2t\right]}_{0}^{2\pi } \\ & =\left[\text{+cos}2\pi +\frac{1}{2}\text{cos}4\pi \right]-\left[\text{+cos}0+\frac{1}{2}\text{cos}0\right] \\ & =0\end{array}\]An amazing consequence of Stokes’ theorem is that if S′ is any other smooth surface with boundary C and the same orientation as S, then \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}={\int }_{C}\text{F}\cdot d\text{r}=0\) because Stokes’ theorem says the surface integral depends on the line integral around the boundary only.
In , we calculated a surface integral simply by using information about the boundary of the surface. In general, let \({S}_{1}\) and \({S}_{2}\) be smooth surfaces with the same boundary C and the same orientation. By Stokes’ theorem,
\[{∬}_{{S}_{1}}\text{curl}\ \text{F}\cdot d\text{S}={\int }_{C}\text{F}\cdot d\text{r}={∬}_{{S}_{2}}\text{curl}\ \text{F}\cdot d\text{S}.\]Condensed — the full section is in OpenStax Calculus Volume 3.
Interpretation of Curl
In addition to translating between line integrals and flux integrals, Stokes’ theorem can be used to justify the physical interpretation of curl that we have learned. Here we investigate the relationship between curl and circulation, and we use Stokes’ theorem to state Faraday’s law—an important law in electricity and magnetism that relates the curl of an electric field to the rate of change of a magnetic field.
Recall that if C is a closed curve and F is a vector field defined on C, then the circulation of F around C is line integral \({\int }_{C}\text{F}\cdot d\text{r}.\) If F represents the velocity field of a fluid in space, then the circulation measures the tendency of the fluid to move in the direction of C.
Let F be a continuous vector field and let \({D}_{r}\) be a small disk of radius r with center \({P}_{0}\) (). If \({D}_{r}\) is small enough, then \((\text{curl}\ \text{F})(P)\approx (\text{curl}\ \text{F})({P}_{0})\) for all points P in \({D}_{r}\) because the curl is continuous. Let \({C}_{r}\) be the boundary circle of \({D}_{r}.\) By Stokes’ theorem,
\[{\int }_{{C}_{r}}\text{F}\cdot d\text{r}={∬}_{{D}_{r}}\text{curl}\ \text{F}\cdot \text{N}dS\approx {∬}_{{D}_{r}}(\text{curl}\ \text{F})({P}_{0})\cdot \text{N}({P}_{0})dS.\]The quantity \((\text{curl}\ \text{F})({P}_{0})\cdot \text{N}({P}_{0})\) is constant, and therefore
\[{∬}_{{D}_{r}}(\text{curl}\ \text{F})({P}_{0})\cdot \text{N}({P}_{0})dS=\pi {r}^{2}[(\text{curl}\ \text{F})({P}_{0})\cdot \text{N}({P}_{0})].\]Thus
\[{\int }_{{C}_{r}}\text{F}\cdot d\text{r}\approx \pi {r}^{2}[(\text{curl}\ \text{F})({P}_{0})\cdot \text{N}({P}_{0})],\]and the approximation gets arbitrarily close as the radius shrinks to zero. Therefore Stokes’ theorem implies that
\[(\text{curl}\ \text{F})({P}_{0})\cdot \text{N}({P}_{0})=\underset{r\to {0}^{+}}{\text{lim}}\frac{1}{\pi {r}^{2}}{\int }_{{C}_{r}}\text{F}\cdot d\text{r}.\]\[\text{B}(x,y,z)=〈P(x,y,z),Q(x,y,z),R(x,y,z)〉,\]\[\text{Work}={\int }_{C(t)}\text{E}(t)\cdot d\text{r}=-\frac{∂ϕ}{∂t}.\]\[\text{curl}\ \text{E}=-\frac{∂\text{B}}{∂t}.\]\[-\frac{∂ϕ}{∂t}={\int }_{C(t)}\text{E}(t)\cdot d\text{r}={∬}_{D(t)}\text{curl}\ \text{E}(t)\cdot d\text{S}.\]\[-\frac{∂ϕ}{∂t}={∬}_{D(t)}-\frac{∂\text{B}}{∂t}\cdot d\text{S}.\]\[{∬}_{D(t)}-\frac{∂\text{B}}{∂t}\cdot d\text{S}={∬}_{D(t)}\text{curl}\ \text{E}\cdot d\text{S}.\]\[{∬}_{D(t)}-\frac{∂\text{B}}{∂t}\cdot d\text{S}={∬}_{D(t)}\text{curl}\ \text{E}\cdot d\text{S}\]\[f(x)=\{{}_{0,\ 1\text{/}2\le x\le 1.}^{1,\ 0\le x\le 1\text{/}2}\]\[\text{curl}\ \text{E}=-\frac{∂\text{B}}{∂t}.\]Condensed — the full section is in OpenStax Calculus Volume 3.
Key Concepts
- Stokes’ theorem relates a flux integral over a surface to a line integral around the boundary of the surface. Stokes’ theorem is a higher dimensional version of Green’s theorem, and therefore is another version of the Fundamental Theorem of Calculus in higher dimensions.
- Stokes’ theorem can be used to transform a difficult surface integral into an easier line integral, or a difficult line integral into an easier surface integral.
- Through Stokes’ theorem, line integrals can be evaluated using the simplest surface with boundary C.
- Faraday’s law relates the curl of an electric field to the rate of change of the corresponding magnetic field. Stokes’ theorem can be used to derive Faraday’s law.
Stokes’ Theorem
For the following exercises, without using Stokes’ theorem, calculate directly both the flux of \(\text{curl}\ \text{F}\cdot \text{N}\) over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.
For the following exercises, use Stokes’ theorem to evaluate \({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS\) for the vector fields and surface.
For the following application exercises, the goal is to evaluate \(A={∬}_{S}\text{curl}\text{F}\cdot dS\) where \(\text{F}=〈xz,\text{-}xz,xy〉\) and S is the upper half of ellipsoid \({x}^{2}+{y}^{2}+8{z}^{2}=1,\ \text{where}\ z\ge 0.\)
For the following exercises, let S be the disk enclosed by curve
\(C:\text{r}(t)=〈\text{cos}\ \phi \ \text{cos}\ t,\text{sin}\ t,\text{sin}\ \phi \ \text{cos}\ t〉,\) for \(0\le t\le 2\pi ,\) where \(0\le \phi \le \frac{\pi }{2}\) is a fixed angle.
For the following exercises, use Stokes’ theorem to find the circulation of the following vector fields around any smooth, simple closed curve C.
Stokes' Theorem
This section relies heavily on understanding flux integrals as well as the calculation of circulation around a closed curve (from and ). gives some reminders about the different ways to parameterize surfaces, which was first introduced in .
Introduction
When we studied in , we saw how integrating the circulation density over a region in the plane bounded by a simple closed curve is equivalent to calculating the circulation along the boundary curve. When we consider simple closed curves in \(\R^3\), the situation gets more complicated. However, there is an interesting, and perhaps surprising, generalization of for us to examine.
Exploration
In this activity, we will look at how we can apply the ideas about circulation along overlapping curves from the beginning of to curves in space.
For this part, consider the curves in Figure, where the yellow curve is \(Y\), the blue curve is \(B\), and the magenta curve is \(M\).
You should also go back and refamiliarize yourself with our notation for combining paths (as used in line integrals) from . In our convention, \(Y+M\) would be closed loop, but \(Y-M\) would not make sense because the segment \(-M\) does not begin where \(Y\) begins.
Using the three segments in , write out at least four different closed curves in terms of \(B\), \(Y\), and \(M\). (Remember to consider orientation!)
Let \(C_1=M+B\) and \(C_2=Y-B\). Describe the curve given by \(C_1+C_2\).
Write a couple of sentences explaining how the circulation around \(C_1+C_2\) would compare to the circulation around \(C_1\) and the circulation around \(C_2\). Write an equation in terms of \(\int_{C_1} \vF \cdot d\vr\), \(\int_{C_2} \vF \cdot d\vr\), and \(\int_{C_1+C_2} \vF \cdot d\vr\).
Explain how your arguments or equations from any of the parts above would or would not change if you considered the curves depicted in .
Let \(C\) be the simple closed curve consisting of the yellow and magenta curves in . You can see \(C\) plotted in red in . The drop-down allows you to select three different surfaces. You can visually verify that each of the three surfaces contains \(C\). Notice that the scale on the \(z\)-axis changes as you select different surfaces.
The simple closed curve consisting of the yellow and magenta curves in can be parameterized by \(\langle \cos(t), \sin(t), \cos(2t)\rangle\) with \(0\leq t\leq 2 \pi\). Let \(C_3 = Y+M\). Use the given parameterization of \(C_3\) to show that \(C_3\) is on each of the following surfaces:
- \(x^2-y^2=z\)
- \(z=x^4-y^4\)
- \(z=1-2y^2\)
- \(z=-\cos(\pi \sqrt{x^2+y^2})(x^2-y^2)\)
Circulation in three dimensions and Stokes' Theorem
In , we saw that a simple closed curve in \(\R^3\) can bound many different surfaces. For now, however, we want to focus on a smooth surface \(S\) in \(\R^3\) that has a well-defined normal vector \(\vn\) at every point and a boundary curve \(C\). We will use the normal vector to define an orientation of \(C\) so that if a person were to walk along \(C\) in the direction of the orientation with the top of their head pointing in the direction of \(\vn\), their left arm would be over the surface \(S\). Notice that this is the same convention that we used with if we assume that the normal vector being used is \(\vk\).
In , we show the curve \(C\) from in magenta as well as a surface \(S\) that has \(C\) as its boundary. The chosen normal vector \(\vn\) to \(S\) is shown, as is the orientation of \(C\) that matches \(\vn\).
Thinking back to , our main idea was that we could calculate the circulation around a simple closed curve in \(\R^2\) by taking the double integral of the circulation density over the region bounded by the curve. As we saw in , we can break up \(\oint_C\vF\cdot d\vr\) into line integrals around other simple closed curves so that overlapping portions are oriented oppositely just as we did with the square grid for Green's Theorem. To find a three-dimensional analog of Green's Theorem, we require that a simple closed curve \(C\) in three dimensions bound a smooth surface \(S\) with a normal vector \(\vn\). In doing this, we can choose our smaller curves similar to the squares we used in Green's Theorem to lie on the surface \(S\). This gives us almost all the ingredients used in Green's Theorem, but we still need to find a suitable replacement for the circulation density.
As we saw in , the curl of a vector field in \(\R^3\) measures the rotation of the vector field. says that for a unit vector \(\vv\), the scalar \((\curl(\vF)(a,b,c))\cdot \vv\) measures the rotational strength of \(\vF\) at the point \((a,b,c)\) around the axis defined by \(\vv\). When \(\vv\) is the normal vector to the surface \(S\) at the point \((a,b,c)\), we have the appropriate analog for the circulation density of \(\vF\) on \(S\) at \((a,b,c)\). Thus, the equivalent idea to integrating the circulation density of a two-dimensional vector field over a region in the plane is calculating the flux integral \(\iint_D \curl(\vF)\cdot (\vr_s\times \vr_t)\, dA\), where \(\vr(s,t)\) on the domain \(D\) that gives a parameterization of the smooth surface \(S\).
A rigorous proof of the following theorem is beyond the scope of this text. However, and our discussion of provide an intuitive description of why this theorem is true.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Verifying and Applying Stokes' Theorem
In this subsection, we will look at some examples and activities that will verify Stokes' Theorem by calculation both side for a few different situations.
We close this subsection with a pair of activities. The first focuses on calculating both of the integrals in . The second asks you to calculate some line integrals along simple closed curves and gives you the discretion to choose the best method to use for this (as well as the best surface to use, if you choose Stokes' Theorem).
Activity
In this activity, we will verify by calculating both a line integral and a flux integral.
Consider the vector field \(\vF = \langle x^2 ,y^2 ,z^2 \rangle\) and the circle \(C_1\) parameterized as \(\vr(t) =\langle \sqrt{2}\cos(t), \sqrt{2}\cos(t), 2\sin(t)\rangle\) for \(0\leq t\leq 2\pi\).
Calculate \(\oint_{C_1} \vF\cdot d\vr\) directly using the given parametrization.
Let \(S_1\) be the hemisphere of the sphere of radius \(2\) centered at the origin with \(y\leq x\). Calculate the flux of \(\curl(\vF)\) through \(S_1\).
What could you have observed about \(\vF\) that would have gotten you the same answer without doing either of the above calculations?
Consider the vector field \(\vG = x\vi + y^2z\vj + x^2\vk\) and the curve \(C_2\), which is the triangle with vertices \((1,0,0)\), \((0,1,0)\), and \((0,0,1)\) with orientation corresponding to the order the points are listed here.
Find the circulation of \(\vG\) along \(C_2\) by calculating the appropriate line integrals.
The vertices of \(C_2\) lie in a plane. Let \(S_2\) be the portion of this plane lying in the first octant, i.e., the portion with \(x,y,z\geq 0\). Find the flux of \(\curl(\vG)\) through \(S_2\).
Write a sentence to explain why the sign of your answer to the previous two parts makes sense.
Activity
Find the circulation of \(\vF = \langle 3yz, xz, -xy\rangle\) along the curve \(C\) consisting of (given in order of the orientation) the quarter-circle of radius \(1\) centered at \((0,-2,0)\) in the plane \(y=-2\) from \((0,-2,1)\) to \((1,-2,0)\), the line segment from \((1,-2,0)\) to \((1,5,0)\), the quarter-circle of radius \(1\) centered at \((0,5,0)\) in the plane \(y=5\) from \((1,5,0)\) to \((0,5,1)\), and the line segment from \((0,5,1)\) to \((0,-2,1)\).
Find the circulation of \(\vG = 3z^2 \vi -(z^2 + 2x) \vj +zy \vk\) along the circle in the \(xy\)-plane of radius \(3\) centered at the origin. Assume the counterclockwise orientation of the circle.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice with Surfaces and their Boundaries
When we looked , it was generally most useful when we were given a line integral and we calculated it using a double integral. In fact, except in the circumstances described in and of , we did not use Green's Theorem to rewrite a double integral as a line integral because of the difficulty of finding a suitable vector field. The situation for will be similar, with the exception of in this section. However, Stokes' Theorem gives us an interesting additional piece of freedom: selecting the surface \(S\) through which we calculate the flux of \(\curl(\vF)\) from amongst possibly several reasonable surfaces with boundary \(C\). The next two activities focus on the relationships between surfaces and their boundary.
Activity
Because requires us to consider a surface (with normal vector) and the boundary of the surface, this activity will give you a chance to practice identifying the boundary of some surfaces in \(\R^3\). For each surface below:
- Describe the boundary in words.
- Find a parametrization for the boundary.
- Ensure that a person walking along the boundary in the direction of your parametrization with head pointing in the direction of the surface's normal vector would hold their left hand over the surface.
The surface \(S_1\) is the portion of the sphere \(x^2+y^2+z^2=4\) with \(z\geq x\). Assume the outward orientation on the sphere.
The surface \(S_2\) is the portion of the sphere \(x^2+y^2+z^2=4\) with \(z\geq 0\). Assume the outward orientation on the sphere.
The surface \(S_3\) is the portion of the hyperbolic paraboloid \(z=x^2-y^2\) with \(x^2+y^2\leq 1\). Assume the upward orientation, e.g., the normal vector at \((0,0,0)\) is \(\vk\).
The surface \(S_4\) is the portion of the cylinder \(x^2+y^2=4\) for which \(-2\leq z\leq 2\), assuming the outward orientation.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Verify that Stokes’ theorem is true for vector field \(\text{F}(x,y,z)=〈y,2z,{x}^{2}〉\) and surface S, where S is the paraboloid \(z=4-{x}^{2}-{y}^{2}\). Assume the surface is outward oriented and \(z\ge 0\).
答えを明らかにしろ
As a surface integral, you have \(g(x,y)=4-{x}^{2}-{y}^{2},{g}_{x}=-2x\) and \({g}_{y}=-2y\)
\[\begin{array}{llllllll}\text{curl}F\ & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z} \\ y & 2z & {x}^{2}\end{array}|\end{array}=⟨-2,-2x,-1⟩\text{.}\]By ,
\[\begin{array}{ll}{∬}_{S}\text{curl}\ \text{F}\cdot d\text{S} & ={∬}_{D}\text{curl}\ \text{F}(\text{r}(ϕ,\theta ))\cdot ({\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta })dA \\ & ={∬}_{D}〈-2,-2x,-1〉\cdot 〈2x,2y,1〉dA \\ & ={\int }_{-2}^{2}{\int }_{\sqrt{4-{x}^{2}}}^{\sqrt{4-{x}^{2}}}(-4x-4xy-1)dydx \\ & ={\int }_{-2}^{2}(-8x\sqrt{4-{x}^{2}}-2\sqrt{4-{x}^{2}})dx \\ & =-4\pi \end{array}\]As a line integral, you can parameterize C by \(\text{r}(t)=〈2\ \text{cos}\ t,2\ \text{sin}\ t,0〉\ 0\le t\le 2\pi\). By ,
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{0}^{2\pi }〈2\text{sin}\ t,0,4{\text{cos}}^{2}t〉\cdot 〈-2\text{sin}\ t,2\text{cos}\ t,0〉dt \\ & ={\int }_{0}^{2\pi }-4{\text{sin}}^{2}tdt=-4\pi \end{array}\]Therefore, we have verified Stokes' theorem for this example.
-
Verify that Stokes’ theorem is true for vector field \(\text{F}(x,y,z)=〈y,x,\text{-}z〉\) and surface S, where S is the upwardly oriented portion of the graph of \(f(x,y)={x}^{2}y\) over a triangle in the xy-plane with vertices \((0,0),\) \((2,0),\) and \((0,2).\)
答えを明らかにしろ
Both integrals give \(0\)
-
Calculate surface integral \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S},\) where S is the surface, oriented outward, in and \(\text{F}=〈z,2xy,x+y〉.\)
答えを明らかにしろ
Note that to calculate \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}\) without using Stokes’ theorem, we would need to use . Use of this equation requires a parameterization of S. Surface S is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes’ theorem, noting that the boundary C of the surface is merely a single circle with radius 1.
By Stokes’ theorem,
\[{∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}={\int }_{C}\text{F}\cdot d\text{r},\]where C has parameterization \(r(t)=〈\text{sin}\ t,0,1-\text{cos}\ t〉,0\le t<2\pi .\) By ,
\[\begin{array}{ll}{∬}_{S}\text{curl}\ \text{F}\cdot d\text{S} & ={\int }_{C}\text{F}\cdot d\text{r} \\ & ={\int }_{0}^{2\pi }⟨1-\text{cos}\ t\ ,0,+\text{sin}\ t⟩\cdot \ ⟨-\text{sin}\ t,0,\text{cos}\ t⟩dt\ \\ & ={\int }_{0}^{2\pi }(-\text{sin}\ t+2\text{sin}\ t\text{cos}\ t)dt \\ & ={\int }_{0}^{2\pi }(-\text{sin}\ t+\text{sin 2t) }\text{dt} \\ & ={\left[\text{+cos}t+\frac{1}{2}\text{cos}2t\right]}_{0}^{2\pi } \\ & =\left[\text{+cos}2\pi +\frac{1}{2}\text{cos}4\pi \right]-\left[\text{+cos}0+\frac{1}{2}\text{cos}0\right] \\ & =0\end{array}\] -
Use Stokes’ theorem to calculate surface integral \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S},\) where \(\text{F}=〈z,x,y〉\) and S is the surface as shown in the following figure. The boundary curve, C, is oriented clockwise when looking along the positive y-axis.
答えを明らかにしろ
\(\text{-}\pi\)
-
Calculate the line integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(\text{F}=〈xy,{x}^{2}+{y}^{2}+{z}^{2},yz〉\) and C is the boundary of the parallelogram with vertices \((0,0,1),(0,1,0),(2,1,-2),\) and \((2,0,-1).\)
答えを明らかにしろ
To calculate the line integral directly, we need to parameterize each side of the parallelogram separately, calculate four separate line integrals, and add the result. This is not overly complicated, but it is time-consuming.
By contrast, let’s calculate the line integral using Stokes’ theorem. Let S denote the surface of the parallelogram. Note that S is the portion of the graph of \(z=1-x-y\) for \((x,y)\) varying over the rectangular region with vertices \((0,0),\) \((0,1),\) \((2,0),\) and \((2,1)\) in the xy-plane. Therefore, a parameterization of S is \(〈x,y,1-x-y〉,0\le x\le 2,0\le y\le 1.\) The curl of F is \(〈-z,0,x〉,\) and Stokes’ theorem and give
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={∬}_{S}\text{curl}\ \text{F}\cdot d\text{S} \\ & ={\int }_{0}^{2}{\int }_{0}^{1}\text{curl}\ \text{F}(x,y)\cdot ({\text{t}}_{x}\times {\text{t}}_{y})dydx \\ & ={\int }_{0}^{2}{\int }_{0}^{1}〈\text{-}(1-x-y)\text{,0,x}〉\cdot (〈1,0,-1〉\times 〈0,1,-1〉)dydx \\ & ={\int }_{0}^{2}{\int }_{0}^{1}〈x+y-1,0,x〉\cdot 〈1,1,1〉dydx \\ & \ {\int }_{0}^{2}{\int }_{0}^{1}2x+y-1dydx \\ & =3.\end{array}\] -
Use Stokes’ theorem to calculate line integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(\text{F}=〈z,x,y〉\) and C is oriented clockwise and is the boundary of a triangle with vertices \((0,0,1),(3,0,-2),\) and \((0,1,2).\)
答えを明らかにしろ
\(\frac{3}{2}\)
-
Calculate the curl of electric field E if the corresponding magnetic field is constant field \(\text{B}(t)=〈1,-4,2〉.\)
答えを明らかにしろ
Since the magnetic field does not change with respect to time, \(-\frac{∂\text{B}}{∂t}=0.\) By Faraday’s law, the curl of the electric field is therefore also zero.
-
Calculate the curl of electric field E if the corresponding magnetic field is \(\text{B}(t)=〈tx,ty,-2tz〉,0\le t<\infty .\)
答えを明らかにしろ
\(\text{curl}\ \text{E}=〈x,y,-2z〉\)
-
\(\text{F}(x,y,z)={y}^{2}\text{i}+{z}^{2}\text{j}+{x}^{2}\text{k}\text{;}\) S is the first-octant portion of plane \(x+y+z=1.\)
-
\(\text{F}(x,y,z)=z\text{i}+x\text{j}+y\text{k}\text{;}\) S is hemisphere \(z={({a}^{2}-{x}^{2}-{y}^{2})}^{1\text{/}2}.\)
答えを明らかにしろ
\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=\pi {a}^{2}\)
-
\(\text{F}(x,y,z)={y}^{2}\text{i}+2x\text{j}+5\text{k}\text{;}\) S is hemisphere \(z={(4-{x}^{2}-{y}^{2})}^{1\text{/}2}.\)
-
\(\text{F}(x,y,z)=z\text{i}+2x\text{j}+3y\text{k}\text{;}\) S is upper hemisphere \(z=\sqrt{9-{x}^{2}-{y}^{2}}.\)
答えを明らかにしろ
\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=18\pi\)
-
\(\text{F}(x,y,z)=(x+2z)\text{i}+(y-x)\text{j}+(z-y)\text{k}\text{;}\) S is a triangular region with vertices (3, 0, 0), (0, 3/2, 0), and (0, 0, 3).
-
\(\text{F}(x,y,z)=2y\text{i}-6z\text{j}+3x\text{k}\text{;}\) S is a portion of paraboloid \(z=4-{x}^{2}-{y}^{2}\) and is above the xy-plane.
答えを明らかにしろ
\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=-8\pi\)
-
\(\text{F}(x,y,z)=xy\text{i}-z\text{j}\) and S is the surface of the cube \(0\le x\le 1,0\le y\le 1,0\le z\le 1,\) except for the face where \(z=0,\) and using the outward unit normal vector.
-
\(\text{F}(x,y,z)=xy\text{i}+{x}^{2}\text{j}+{z}^{2}\text{k}\text{;}\) and S is the part of paraboloid \(z={x}^{2}+{y}^{2}\) below plane \(z=y,\) and using the outward normal vector.
答えを明らかにしろ
\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=0\)
-
\(\text{F}(x,y,z)=4y\text{i}+z\text{j}+2y\text{k}\) and S is the part of sphere \({x}^{2}+{y}^{2}+{z}^{2}=4\) above plane \(z=0,\) and using the outward normal vector
-
Use Stokes’ theorem to evaluate \(\underset{C}{\int }[2x{y}^{2}zdx+2{x}^{2}yzdy+({x}^{2}{y}^{2}-2z)dz],\) where C is the curve given by \(x=\text{cos}\ t,y=\text{sin}\ t,z=\text{sin}\ t,0\le t\le 2\pi ,\) traversed in the direction of increasing t.
答えを明らかにしろ
\({\int }_{C}\text{F}\cdot dr=0\)
-
[T] Use a computer algebraic system (CAS) and Stokes’ theorem to approximate line integral \(\underset{C}{\int }(ydx+zdy+xdz),\) where C is the intersection of plane \(x+y=2\) and surface \({x}^{2}+{y}^{2}+{z}^{2}=2(x+y),\) traversed counterclockwise viewed from the origin.
-
[T] Use a CAS and Stokes’ theorem to approximate line integral \(\underset{C}{\int }(3ydx+2zdy-5xdz),\) where C is the intersection of the xy-plane and hemisphere \(z=\sqrt{1-{x}^{2}-{y}^{2}},\) traversed counterclockwise viewed from the top—that is, from the positive z-axis toward the xy-plane.
答えを明らかにしろ
\({\int }_{s}\text{F}\cdot dr=-3\pi \approx -9.4248\)
-
[T] Use a CAS and Stokes’ theorem to approximate line integral \(\underset{C}{\int }[(1+y)zdx+(1+z)xdy+(1+x)ydz],\) where C is a triangle with vertices \((1,0,0),\) \((0,1,0),\) and \((0,0,1)\) oriented counterclockwise.
-
Use Stokes’ theorem to evaluate \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)={e}^{xy}\text{cos}\ z\text{i}+{x}^{2}z\text{j}+xy\text{k},\) and S is half of sphere \(x=\sqrt{1-{y}^{2}-{z}^{2}},\) oriented out toward the positive x-axis.
答えを明らかにしろ
\(\underset{S}{∬}\text{curl}\ \text{F}\cdot d\text{S}=0\)
-
[T] Use a CAS and Stokes’ theorem to evaluate \({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS,\) where \(\text{F}(x,y,z)={x}^{2}y\text{i}+x{y}^{2}\text{j}+{z}^{3}\text{k}\) and S is the curve of the part of plane \(3x+2y+z=6\) above cylinder \({x}^{2}+{y}^{2}=4,\) oriented clockwise when viewed from above.
-
[T] Use a CAS and Stokes’ theorem to evaluate \(\underset{S}{∬}\text{curl}\ \text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=(\text{sin}(y+z)-y{x}^{2}-\frac{{y}^{3}}{3})\text{i}+x\ \text{cos}(y+z)\text{j}+\text{cos}(2y)\text{k}\) and S consists of the top and the four sides but not the bottom of the cube with vertices \((\pm 1,\pm 1,\pm 1),\) oriented outward.
答えを明らかにしろ
\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=2.6667\)
-
[T] Use a CAS and Stokes’ theorem to evaluate \(\underset{S}{∬}\text{curl}\ \text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)={z}^{2}\text{i}-3xy\text{j}+{x}^{3}{y}^{3}\text{k}\) and S is the top part of \(z=5-{x}^{2}-{y}^{2}\) above plane \(z=1,\) and S is oriented upward.
-
Use Stokes’ theorem to evaluate \({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS,\) where \(\text{F}(x,y,z)={z}^{2}\text{i}+{y}^{2}\text{j}+x\text{k}\) and S is a triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) with upward orientation.
答えを明らかにしろ
\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=-\frac{1}{6}\)
-
Use Stokes’ theorem to evaluate line integral \(\underset{C}{\int }(zdx+xdy+ydz),\) where C is a triangle with vertices (3, 0, 0), (0, 0, 2), and (0, 6, 0) traversed in the given order.
-
Use Stokes’ theorem to evaluate \(\underset{C}{\int }(\frac{1}{2}{y}^{2}dx+zdy+xdz),\) where C is the curve of intersection of plane \(x+z=1\) and ellipsoid \({x}^{2}+2{y}^{2}+{z}^{2}=1,\) oriented clockwise from the origin.
答えを明らかにしろ
\(\underset{C}{\int }(\frac{1}{2}{y}^{2}dx+zdy+xdz)=-\frac{\pi }{4}\)
-
Use Stokes’ theorem to evaluate \({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS,\) where \(\text{F}(x,y,z)=x\text{i}+{y}^{2}\text{j}+z{e}^{xy}\text{k}\) and S is the part of surface \(z=1-{x}^{2}-2{y}^{2}\) with \(z\ge 0\text{,}\\) oriented upward.
-
Use Stokes’ theorem to evaluate \({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS,\) for vector field \(\text{F}(x,y,z)=z\text{i}+3x\text{j}+2z\text{k}\) where S is surface \(z=1-{x}^{2}-{y}^{2},z\ge 0,\) C is boundary circle \({x}^{2}+{y}^{2}=1,\) and S is oriented in the positive z-direction.
答えを明らかにしろ
\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=3\pi\)
-
Use Stokes’ theorem to evaluate \({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS,\) for vector field \(\text{F}(x,y,z)=-\frac{3}{2}{y}^{2}\text{i}-2xy\text{j}+yz\text{k}\text{,}\) where S is that part of the surface of plane \(x+y+z=1\) contained within triangle C with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), traversed counterclockwise as viewed from above.
-
A certain closed path C in plane \(2x+2y+z=1\) is known to project onto unit circle \({x}^{2}+{y}^{2}=1\) in the xy-plane. Let c be a constant and let \(\text{R}(x,y,z)=x\text{i}+y\text{j}+z\text{k}.\) Use Stokes’ theorem to evaluate \({\int }_{C}^{}(c\text{k}\ \times \ \text{R})\cdot d\text{r}.\)
答えを明らかにしろ
\({\int }_{C}^{}(c\text{k}\ \times \ \text{R})\cdot d\text{r}=2\pi c\)
-
Use Stokes’ theorem and let C be the boundary of surface \(z={x}^{2}+{y}^{2}\) with \(0\le x\le 2\) and \(0\le y\le 1,\) oriented with upward facing normal. Define
\[\text{F}(x,y,z)=[\text{sin}({x}^{3})+xz]\text{i}+(x-yz)\text{j}+\text{cos}({z}^{4})\text{k}\ \text{and evaluate}\ {\int }_{C}\text{F}\cdot d\text{r}.\] -
Let S be hemisphere \({x}^{2}+{y}^{2}+{z}^{2}=4\) with \(z\ge 0,\) oriented upward. Let \(\text{F}(x,y,z)={x}^{2}{e}^{yz}\text{i}+{y}^{2}{e}^{xz}\text{j}+{z}^{2}{e}^{xy}\text{k}\) be a vector field. Use Stokes’ theorem to evaluate \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}.\)
答えを明らかにしろ
\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=0\)
-
Let \(\text{F}(x,y,z)=xy\text{i}+({e}^{{z}^{2}}+y)\text{j}+(x+y)\text{k}\) and let S be the graph of function \(y=\frac{{x}^{2}}{9}+\frac{{z}^{2}}{9}-1\) with \(y\le 0\) oriented so that the normal vector of S has a positive j component. Use Stokes’ theorem to compute integral \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}.\)
-
Use Stokes’ theorem to evaluate \({\int }_{C}F\cdot dr\) where \(\text{F}(x,y,z)=y\text{i}+z\text{j}+x\text{k}\) and C is a triangle with vertices (0, 0, 0), (2, 0, 0) and \((0,-2,2)\) oriented counterclockwise when viewed from above.
答えを明らかにしろ
\({\int }_{C}F\cdot dr=-4\)
-
Use the surface integral in Stokes’ theorem to calculate the circulation of field F, \(\text{F}(x,y,z)={x}^{2}{y}^{3}\text{i}+\text{j}+z\text{k}\) around C, which is the intersection of cylinder \({x}^{2}+{y}^{2}=4\) and hemisphere \({x}^{2}+{y}^{2}+{z}^{2}=16,z\ge 0,\) oriented counterclockwise when viewed from above.
-
Use Stokes’ theorem to compute \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=\text{i}+x{y}^{2}\text{j}+x{y}^{2}\text{k}\) and S is a part of plane \(y+z=2\) inside cylinder \({x}^{2}+{y}^{2}=1\) and oriented upward.
答えを明らかにしろ
\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=0\)
-
Use Stokes’ theorem to evaluate \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=\text{-}{y}^{2}\text{i}+x\text{j}+{z}^{2}\text{k}\) and S is the part of plane \(x+y+z=1\) in the first octant and oriented upward \(x\ge 0\text{,}\ y\ge 0\text{,}\ z\ge 0.\)
-
Let \(\text{F}(x,y,z)=xy\text{i}+2z\text{j}-2y\text{k}\) and let C be the intersection of plane \(x+z=5\) and cylinder \({x}^{2}+{y}^{2}=9,\) which is oriented counterclockwise when viewed from the top. Compute the line integral of F over C using Stokes’ theorem.
答えを明らかにしろ
\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=-36\pi\)
Symbols used here
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Not a number: "grows without bound" in limits and intervals.
Ratio of a circle's circumference to its diameter, 3.14159…
Chance of A; chance of A given that B happened.
i² = −1.
Equal to the precision shown, not exactly.
Inequalities that allow equality; < and > exclude it.
Least upper bound, greatest lower bound.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Stokes’ Theorem
- Explain the meaning of Stokes’ theorem.
- Use Stokes’ theorem to evaluate a line integral.
- Use Stokes’ theorem to calculate a surface integral.
- Use Stokes’ theorem to calculate a curl.
- Stokes’ theorem relates a flux integral over a surface to a line integral around the boundary of the surface. Stokes’ theorem is a higher dimensional version of Green’s theorem, and therefore is another version of the Fundamental Theorem of Calculus in higher dimensions.
- Stokes’ theorem can be used to transform a difficult surface integral into an easier line integral, or a difficult line integral into an easier surface integral.
- Through Stokes’ theorem, line integrals can be evaluated using the simplest surface with boundary
- Faraday’s law relates the curl of an electric field to the rate of change of the corresponding magnetic field. Stokes’ theorem can be used to derive Faraday’s law.
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
あなた自身を試してみてください
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
ここに Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems