maths.free › Multivariable Calculus › Vector Calculus › Stokes' theorem

Stokes' theorem

Q: 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.

Q: 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.

Stokes' theorem, also known as the Kelvin–Stokes theorem, is a theorem in vector calculus that relates the behavior of a vector field along the edge of a surface to the behavior of its curl on the surface itself.

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.

Stokes' theorem

Stokes' theorem, also known as the Kelvin-Stokes theorem, is a theorem in vector calculus that relates the behavior of a vector field along the edge of a surface to the behavior of its curl on the surface itself. In its usual three-dimensional form, it says that the total circulation of a vector field around a closed curve is equal to the total curl of the field through a surface bounded by that curve.

If Σ is an oriented surface with boundary ∂Σ, Stokes' theorem is commonly written as

\[\oint_{\partial\Sigma} \mathbf F\cdot d\mathbf r = \iint_{\Sigma} (\nabla\times \mathbf F)\cdot \mathbf n\,dS.\]

Here the left side is the line integral of the vector field around the boundary curve, while the right side is the surface integral of its curl over the surface. Informally, the theorem says that adding up the local rotation of a vector field across a surface gives the net circulation around its edge.

The theorem is also called the fundamental theorem for curls, the curl theorem, or the rotor theorem. It is a special case of the generalized Stokes theorem. In the language of differential forms, the vector field corresponds to a 1-form and its curl corresponds to the exterior derivative of that form.

Theorem

Let \(\Sigma\) be a smooth oriented surface in \(\R^3\), parametrized by \(\mathbf\Sigma(u,v)\), with boundary \(\partial \Sigma \equiv \Gamma\), parametrized by \(\mathbf\Gamma(t)\). If a vector field

\(\mathbf{F}(x,y,z) = (F_x(x, y, z), F_y(x, y, z), F_z(x, y, z))\)

has continuous first-order partial derivatives in \(\Sigma\), then \[\iint_\Sigma (\nabla \times \mathbf{F}) \cdot d\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot d\mathbf{\Gamma}\] with the shorthands for the line element \(d\mathbf\Gamma = \frac{d\mathbf\Gamma}{dt}dt\) and the surface element \[d\mathbf{\Sigma} = \mathbf{n} d\Sigma = \left( \frac{\partial \mathbf\Sigma}{\partial u} \times \frac{\partial \mathbf\Sigma}{\partial v} \right) du dv\] where \(\mathbf{n}(u,v)\) is the vector orthogonal to the surface at the point \(\mathbf\Sigma(u,v)\).

The equality can be expressed in terms of differential forms, with \(\wedge\) being the wedge product and \(\text{d}\) the exterior derivative: \[\begin{aligned} &\iint_\Sigma \left(\left(\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z} \right)\,\mathrm{d}y\wedge \mathrm{d}z +\left(\frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\right)\, \mathrm{d}z\wedge \mathrm{d}x +\left (\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\, \mathrm{d}x\wedge \mathrm{d}y\right) \\ & = \oint_{\partial\Sigma} \Bigl(F_x\, \mathrm{d}x+F_y\, \mathrm{d}y+F_z\, \mathrm{d}z\Bigr). \end{aligned}\]

The main challenge in a precise statement of Stokes' theorem is in defining the notion of a boundary. Surfaces such as the Koch snowflake, for example, are well-known not to exhibit a Riemann-integrable boundary, and the notion of surface measure in Lebesgue theory cannot be defined for a non-Lipschitz surface. One (advanced) technique is to pass to a weak formulation and then apply the machinery of geometric measure theory; for that approach see the coarea formula. In this article, we instead use a more elementary definition, based on the fact that a boundary can be discerned for full-dimensional subsets of \(\R^2\).

A more detailed statement will be given for subsequent discussions. Let \(\gamma:[a,b]\to\R^2\) be a piecewise smooth Jordan plane curve: a simple closed curve in the plane. The Jordan curve theorem implies that \(\gamma\) divides \(\R^2\) into two components, a compact one and another that is non-compact. Let \(D\) denote the compact part; then \(D\) is bounded by \(\gamma\). It now suffices to transfer this notion of boundary along a continuous map to our surface in \(\R^3\). But we already have such a map: the parametrization of \(\Sigma\).

Suppose \(\psi:D\to\R^3\) is piecewise smooth at the neighborhood of \(D\)　, with \(\Sigma=\psi(D)\). If \(\Gamma\) is the space curve defined by \(\Gamma(t)=\psi(\gamma(t))\) then we call \(\Gamma\) the boundary of \(\Sigma\), written \(\partial\Sigma\)

With the above notation, if \(\mathbf{F}\) is any smooth vector field on \(\R^3\), then\[\oint_{\partial\Sigma} \mathbf{F}\, \cdot\, d{\mathbf{\Gamma}} = \iint_{\Sigma} \nabla\times\mathbf{F}\, \cdot\, d\mathbf{\Sigma}.\]

Condensed: the full section is in Wikipedia.

Special case of a more general theorem

Stokes' theorem can be viewed as a special case of the following identity: \[\oint_{\partial\Sigma} (\mathbf{F}\, \cdot\, d{\mathbf{\Gamma}})\,\mathbf{g} = \iint_{\Sigma}\left[ d\mathbf{\Sigma}\cdot\left( \nabla\times\mathbf{F}- \mathbf{F}\times\nabla\right)\right]\mathbf{g},\] where \(\mathbf{g}\) is any smooth vector or scalar field in \(\mathbb{R}^3\). When \(\mathbf{g}\) is a uniform scalar field, the standard Stokes' theorem is recovered.

Proof

The proof of the theorem consists of 4 steps. We assume Green's theorem, so what is of concern is how to boil down the three-dimensional complicated problem (Stokes' theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally deduce it as a special case of a more general result, which is stated in terms of differential forms, and proved using more sophisticated machinery. While powerful, these techniques require substantial background, so the proof below avoids them, and does not presuppose any knowledge beyond a familiarity with basic vector calculus and linear algebra. At the end of this section, a short alternative proof of Stokes' theorem is given, as a corollary of the generalized Stokes' theorem.

Proof via differential forms

The functions \(\R^3\to\R^3\) can be identified with the differential 1-forms on \(\R^3\) via the map \[F_x\mathbf{e}_1+F_y\mathbf{e}_2+F_z\mathbf{e}_3 \mapsto F_x\,\mathrm{d}x + F_y\,\mathrm{d}y + F_z\,\mathrm{d}z .\]

Write the differential 1-form associated to a function F as ω_F. Then one can calculate that \[\star\omega_{\nabla\times\mathbf{F}}=\mathrm{d}\omega_{\mathbf{F}},\] where ★ is the Hodge star and \(\mathrm{d}\) is the exterior derivative. Thus, by generalized Stokes' theorem, \[\oint_{\partial\Sigma}{\mathbf{F}\cdot\,\mathrm{d}\mathbf{\gamma}} =\oint_{\partial\Sigma}{\omega_{\mathbf{F}}} =\int_{\Sigma}{\mathrm{d}\omega_{\mathbf{F}}} =\int_{\Sigma}{\star\omega_{\nabla\times\mathbf{F}}} =\iint_{\Sigma}{\nabla\times\mathbf{F}\cdot\,\mathrm{d}\mathbf{\Sigma}}\]

Irrotational fields

In this section, we will discuss the irrotational field (lamellar vector field) based on Stokes' theorem.

Definition 2-1 (irrotational field). A smooth vector field F on an open set \(U\subseteq\R^3\) is irrotational (lamellar vector field) if ∇ × F = 0.

This concept is very fundamental in mechanics; as we'll prove later, if F is irrotational and the domain of F is simply connected, then F is a conservative vector field.

Conservative forces

Above Helmholtz's theorem gives an explanation as to why the work done by a conservative force in changing an object's position is path independent. First, we introduce the Lemma 2-2, which is a corollary of and a special case of Helmholtz's theorem.

Lemma 2-2. Let \(U\subseteq\R^3\) be an open subset, with a Lamellar vector field F and a piecewise smooth loop c₀: [0, 1] → U. Fix a point p ∈ U, if there is a homotopy H: [0, 1] × [0, 1] → U such that

[SC0] H is piecewise smooth,
[SC1] H(t, 0) = c₀(t) for all t ∈ [0, 1],
[SC2] H(t, 1) = p for all t ∈ [0, 1],
[SC3] H(0, s) = H(1, s) = p for all s ∈ [0, 1].

Then, \[\int_{c_0} \mathbf{F} \, \mathrm{d}c_0=0\]

Above Lemma 2-2 follows from theorem 2-1. In Lemma 2-2, the existence of H satisfying [SC0] to [SC3] is crucial;the question is whether such a homotopy can be taken for arbitrary loops. If U is simply connected, such H exists. The definition of simply connected space follows:

Definition 2-2 (simply connected space). Let \(M\subseteq\R^n\) be non-empty and path-connected. M is called simply connected if and only if for any continuous loop, c: [0, 1] → M there exists a continuous tubular homotopy H: [0, 1] × [0, 1] → M from c to a fixed point p ∈ c; that is,

[SC0'] H is continuous,
[SC1] H(t, 0) = c(t) for all t ∈ [0, 1],
[SC2] H(t, 1) = p for all t ∈ [0, 1],
[SC3] H(0, s) = H(1, s) = p for all s ∈ [0, 1].

The claim that "for a conservative force, the work done in changing an object's position is path independent" might seem to follow immediately if the M is simply connected.　However, recall that simple-connection only guarantees the existence of a continuous homotopy satisfying [SC1-3]; we seek a piecewise smooth homotopy satisfying those conditions instead.

Fortunately, the gap in regularity is resolved by the Whitney's approximation theorem. In other words, the possibility of finding a continuous homotopy, but not being able to integrate over it, is actually eliminated with the benefit of higher mathematics. We thus obtain the following theorem.

Condensed: the full section is in Wikipedia.

Maxwell's equations

In the physics of electromagnetism, Stokes' theorem provides the justification for the equivalence of the differential form of the Maxwell-Faraday equation and the Maxwell-Ampère equation and the integral form of these equations. For Faraday's law, Stokes theorem is applied to the electric field, \(\mathbf{E}\): \[\oint_{\partial\Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{l}= \iint_\Sigma \mathbf{\nabla}\times \mathbf{E} \cdot \mathrm{d} \mathbf{S} .\]

For Ampère's law, Stokes' theorem is applied to the magnetic field, \(\mathbf{B}\): \[\oint_{\partial\Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{l}= \iint_\Sigma \mathbf{\nabla}\times \mathbf{B} \cdot \mathrm{d} \mathbf{S} .\]

Indi sen Hiç bir hasapçy buny çözmez, emma onuň bölekleri hasaplanmak üçindir. Aşakda birini synla ýa-da özüňi ýaz.

Tälim (40)

Her birini kagyzda synlap gör. Çözgüni barlap görmek üçin aç; barlap görlenler her bir ädim üçin çözgüççide açylyp bilner.

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 \(z\ge 0\).

Jawaby görkez

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 \((0,2).\)

Jawaby görkez

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 \(\text{F}=〈z,2xy,x+y〉.\)

Jawaby görkez

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.

Jawaby görkez

\(\text{-}\pi\)
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 \((2,0,-1).\)

Jawaby görkez

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 \((0,1,2).\)

Jawaby görkez

\(\frac{3}{2}\)
Calculate the curl of electric field E if the corresponding magnetic field is constant field \(\text{B}(t)=〈1,-4,2〉.\)

Jawaby görkez

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 \(\text{B}(t)=〈tx,ty,-2tz〉,0\le t<\infty .\)

Jawaby görkez

\(\text{curl}\ \text{E}=〈x,y,-2z〉\)
\(\text{F}(x,y,z)={y}^{2}\text{i}+{z}^{2}\text{j}+{x}^{2}\text{k}\text{;}\) 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 \(z={({a}^{2}-{x}^{2}-{y}^{2})}^{1\text{/}2}.\)

Jawaby görkez

\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=\pi {a}^{2}\)
\(\text{F}(x,y,z)={y}^{2}\text{i}+2x\text{j}+5\text{k}\text{;}\) 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 \(z=\sqrt{9-{x}^{2}-{y}^{2}}.\)

Jawaby görkez

\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=18\pi\)
\(\text{F}(x,y,z)=(x+2z)\text{i}+(y-x)\text{j}+(z-y)\text{k}\text{;}\) 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.

Jawaby görkez

\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=-8\pi\)
\(\text{F}(x,y,z)=xy\text{i}-z\text{j}\) 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.

Jawaby görkez

\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=0\)
\(\text{F}(x,y,z)=4y\text{i}+z\text{j}+2y\text{k}\) 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.

Jawaby görkez

\({\int }_{C}\text{F}\cdot dr=0\)
[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.

Jawaby görkez

\({\int }_{s}\text{F}\cdot dr=-3\pi \approx -9.4248\)
[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.

Jawaby görkez

\(\underset{S}{∬}\text{curl}\ \text{F}\cdot d\text{S}=0\)
[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.

Jawaby görkez

\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=2.6667\)
[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.

Jawaby görkez

\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=-\frac{1}{6}\)
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.

Jawaby görkez

\(\underset{C}{\int }(\frac{1}{2}{y}^{2}dx+zdy+xdz)=-\frac{\pi }{4}\)
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.

Jawaby görkez

\({∬}_{S}(\text{curl}\ \text{F}\cdot \text{N})dS=3\pi\)
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 \({\int }_{C}^{}(c\text{k}\ \times \ \text{R})\cdot d\text{r}.\)

Jawaby görkez

\({\int }_{C}^{}(c\text{k}\ \times \ \text{R})\cdot d\text{r}=2\pi c\)
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 \({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}.\)

Jawaby görkez

\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=0\)
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.

Jawaby görkez

\({\int }_{C}F\cdot dr=-4\)
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.

Jawaby görkez

\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=0\)
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.

Jawaby görkez

\({∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}=-36\pi\)

Özüň işle

Mugt hasap her bir ders üçin notlar goşýar, tamamlanan zadyňyzyň ýazgysyny, çözen problemaňyzy bir ýerde, we bu sahypa barada sorap biljek mugallymy. Matematika özüçe her kim üçin açyk, girseň ýa girmeseň.

_Gün _Giriş

Şu ýerde ulanylan simwollar

Eň täze düşündirme, surat, we her harpyň manysy üçin her bir simwoly basyň.

Adamlar soraýan soraglar

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.

Bu sahypanyň käbir bölegi şundan alyndy Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0), Wikipedia (CC BY-SA 4.0). Bu ýerde gysgaldylan we tekrar beýan edilen; hatalar biziňkidir. Bu dersiň çeşmesi kommersiýa däl ulanmak üçin lisenziyalanandygy üçin bu ders mugt, hiç zat satmaýar we şol bir lisenziya astynda paýlaşylýar.

_Ýaşa Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems