at what point on the curve y = 9 + 2ex − 5x is the tangent line parallel to the line 5x − y = 7?

Learning Objectives

• 6.vii.1 Explain the meaning of Stokes' theorem.
• 6.vii.2 Utilise Stokes' theorem to evaluate a line integral.
• 6.7.3 Use Stokes' theorem to summate a surface integral.
• 6.7.4 Utilise Stokes' theorem to calculate a curl.

In this department, we study Stokes' theorem, a college-dimensional generalization of Greenish's theorem. This theorem, similar the Fundamental Theorem for Line Integrals and Dark-green's theorem, is a generalization of the Fundamental Theorem of Calculus to higher dimensions. Stokes' theorem relates a vector surface integral over surface Due south in space to a line integral around the boundary of S. Therefore, merely every bit the theorems before it, Stokes' theorem can be used to reduce an integral over a geometric object S to an integral over the purlieus of South.

In addition to allowing us to interpret between line integrals and surface integrals, Stokes' theorem connects the concepts of curl and circulation. Furthermore, the theorem has applications in fluid mechanics and electromagnetism. We apply Stokes' theorem to derive Faraday's law, an important consequence involving electric fields.

Stokes' Theorem

Stokes' theorem says nosotros tin summate the flux of curlicue F across surface S by knowing data only almost the values of F along the boundary of S. Conversely, we can calculate the line integral of vector field F forth the boundary of surface S by translating to a double integral of the curl of F over South.

Permit S be an oriented shine surface with unit of measurement normal vector N. Furthermore, suppose the boundary of South is a simple closed bend C. The orientation of S induces the positive orientation of C if, as yous walk in the positive management around C with your caput pointing in the direction of N, the surface is always on your left. With this definition in place, we can state Stokes' theorem.

Theorem half dozen.19

Stokes' Theorem

Let Southward be a piecewise smooth oriented surface with a boundary that is a simple closed curve C with positive orientation (Figure half dozen.79). If F is a vector field with component functions that have continuous partial derivatives on an open region containing S, then

$${\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_S\text{gyre}\text{F}·d\text{S}}.$$

Effigy 6.79 Stokes' theorem relates the flux integral over the surface to a line integral around the boundary of the surface. Annotation that the orientation of the curve is positive.

Suppose surface S is a flat region in the xy-plane with upward orientation. And then the unit normal vector is k and surface integral ${\displaystyle \underset{\mathrm{Due\; south}}{\iint }\text{ringlet}\text{F}}·d\text{S}$ is actually the double integral ${\displaystyle \underset{\mathrm{Due\; south}}{\iint }\text{whorl}\text{F}}·\text{k}dA.$ In this special example, Stokes' theorem gives ${\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_S\text{scroll}\text{F}·\text{k}dA.}$ Notwithstanding, this is the circulation form of Light-green's theorem, which shows us that Green's theorem is a special case of Stokes' theorem. Green's theorem tin can only handle surfaces in a aeroplane, but Stokes' theorem can handle surfaces in a plane or in infinite.

The complete proof of Stokes' theorem is beyond the scope of this text. Nosotros look at an intuitive explanation for the truth of the theorem and then meet proof of the theorem in the special case that surface Southward is a portion of a graph of a function, and South, the boundary of Due south, and F are all fairly tame.

Proof

Starting time, we look at an informal proof of the theorem. This proof is not rigorous, only it is meant to give a general feeling for why the theorem is true. Let South exist a surface and let D be a minor piece of the surface then that D does non share any points with the boundary of Southward. We cull D to be small enough so that it tin be approximated by an oriented foursquare E. Let D inherit its orientation from S, and give E the same orientation. This square has four sides; denote them ${\mathrm{Eastward}}_{\mathrm{50}},$ $E_r,$ $E_u,$ and ${\mathrm{Eastward}}_d$ for the left, right, upwards, and downwardly sides, respectively. On the square, we can use the flux form of Green's theorem:

$${\displaystyle {\int }_{E_l+{\mathrm{East}}_d+{\mathrm{Eastward}}_r+{\mathrm{Due\; east}}_u}\text{F}·d\text{r}}={\displaystyle {\iint }_E\text{coil}\text{F}·\text{Northward}d\mathrm{Southward}}={\displaystyle {\iint }_{\mathrm{Eastward}}\text{curlicue}\text{F}·d\text{S}}.$$

To approximate the flux over the entire surface, nosotros add the values of the flux on the pocket-sized squares approximating small pieces of the surface (Effigy 6.80). By Green's theorem, the flux beyond each approximating square is a line integral over its purlieus. Permit F exist an approximating square with an orientation inherited from S and with a correct side $E_l$ (so F is to the left of E). Let $F_r$ announce the correct side of $F$ ; then, $E_l=\text{−}{F}_r.$ In other words, the right side of $F$ is the aforementioned bend as the left side of E, but oriented in the opposite direction. Therefore,

$${\displaystyle {\int }_{E_l}\text{F}·d\text{r}}=\text{−}{\displaystyle {\int }_{F_r}\text{F}·d\text{r}}.$$

As we add up all the fluxes over all the squares approximating surface S, line integrals ${\displaystyle {\int }_{{\mathrm{Eastward}}_l}\text{F}·d\text{r}}$ and ${\displaystyle {\int }_{F_r}\text{F}·d\text{r}}$ abolish each other out. The same goes for the line integrals over the other three sides of Due east. These three line integrals abolish out with the line integral of the lower side of the foursquare in a higher place E, the line integral over the left side of the square to the right of E, and the line integral over the upper side of the foursquare below East (Figure 6.81). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of S. Therefore, the sum of all the fluxes (which, by Greenish'due south theorem, is the sum of all the line integrals effectually the boundaries of approximating squares) tin be approximated by a line integral over the purlieus of S. In the limit, equally the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.

Effigy 6.80 Chop the surface into small pieces. The pieces should be pocket-size enough that they can be approximated by a square.

Figure half dozen.81 (a) The line integral along $E_l$ cancels out the line integral along $F_r$ considering ${\mathrm{Due\; east}}_{\mathrm{50}}=\text{−}{F}_r.$ (b) The line integral along any of the sides of E cancels out with the line integral along a side of an next approximating square.

Let's at present look at a rigorous proof of the theorem in the special case that S is the graph of function $z=f\left(x,y\right),$ where 10 and y vary over a divisional, only connected region D of finite area (Figure half dozen.82). Furthermore, presume that $f$ has continuous 2d-lodge partial derivatives. Let C denote the boundary of Southward and let C′ announce the boundary of D. Then, D is the "shadow" of S in the aeroplane and C′ is the "shadow" of C. Suppose that Southward is oriented upward. The counterclockwise orientation of C is positive, as is the counterclockwise orientation of $C^{\prime }.$ Permit $\text{F}\left(x,y,z\right)=\langle P,Q,R\rangle$ be a vector field with component functions that have continuous partial derivatives.

Figure 6.82 D is the "shadow," or project, of S in the aeroplane and $C^{\prime }$ is the project of C.

We have the standard parameterization of $\mathrm{Due\; south}:\mathrm{10}=\mathrm{ten},y=y,z=g\left(x,y\right).$ The tangent vectors are ${\text{t}}_{\mathrm{ten}}=\langle 1,0,g_x\rangle$ and ${\text{t}}_y=\langle 0,\mathrm{ane},{\mathrm{one\; thousand}}_y\rangle ,$ and therefore, ${\text{t}}_x\times {\text{t}}_y=\langle \text{−}{\mathrm{yard}}_x,\text{−}{g}_y,1\rangle .$ By Equation 6.19,

$${\displaystyle \underset{\mathrm{Southward}}{\iint }\text{gyre}\text{F}·d\text{Due south}=}{\displaystyle \underset{D}{\iint }\left[\text{−}\left(R_y-Q_z\right)z_x-\left(P_z-R_{\mathrm{ten}}\right)z_y+\left(Q_x-P_y\right)\right]}dA,$$

where the partial derivatives are all evaluated at $\left(x,y,m\left(x,y\right)\right),$ making the integrand depend on x and y simply. Suppose $\langle x\left(t\right),y\left(t\right)\rangle ,a\le t\le b$ is a parameterization of $C^{\prime }.$ Then, a parameterization of C is $\langle x\left(t\right),y\left(t\right),g\left(\mathrm{10}\left(t\right),y\left(t\right)\right)\rangle ,a\le t\le b.$ Armed with these parameterizations, the Chain rule, and Light-green's theorem, and keeping in mind that P, Q, and R are all functions of x and y, we can evaluate line integral ${\displaystyle {\int }_C\text{F}·d\text{r}}\text{:}$

$$\begin{array}{cc}\hfill {\displaystyle {\int }_C\text{F}·d\text{r}}& ={\displaystyle {\int }_a^b\left(P{x}^{\prime }\left(t\right)+Q{y}^{\prime }\left(t\right)+R{z}^{\prime }\left(t\right)\right)}dt\hfill \\ & ={\displaystyle {\int }_a^b\left[P{\mathrm{10}}^{\prime }\left(t\right)+Q{y}^{\prime }\left(t\right)+R\left(\frac{\partial z}{\partial \mathrm{ten}}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\right)\right]}dt\hfill \\ & ={\displaystyle {\int }_a^b\left[\left(P+R\frac{\partial z}{\partial x}\right)x^{\prime }\left(t\right)+\left(Q+R\frac{\partial z}{\partial y}\right)y^{\prime }\left(t\right)\right]}dt\hfill \\ & ={\displaystyle \underset{C^{\prime }}{\int }\left(P+R\frac{\partial z}{\partial x}\right)d\mathrm{10}+\left(Q+R\frac{\partial z}{\partial y}\right)dy}\hfill \\ & ={\displaystyle \underset{D}{\iint }\left[\frac{\partial }{\partial \mathrm{10}}\left(Q+R\frac{\partial z}{\partial y}\right)-\frac{\partial }{\partial y}\left(P+R\frac{\partial z}{\partial \mathrm{10}}\right)\right]}dA\hfill \\ & =\begin{array}{c}{\displaystyle \underset{D}{\iint }\begin{array}{c}\left(\frac{\partial Q}{\partial x}+\frac{\partial Q}{\partial z}\frac{\partial z}{\partial \mathrm{10}}+\frac{\partial R}{\partial x}\frac{\partial z}{\partial y}+\frac{\partial R}{\partial z}\frac{\partial z}{\partial \mathrm{10}}\frac{\partial z}{\partial y}+R\frac{{\partial }^{\mathrm{two}}z}{\partial x\partial y}\right)\hfill \\ \text{−}(\frac{\partial P}{\partial y}+\frac{\partial P}{\partial z}\frac{\partial z}{\partial y}+\frac{\partial R}{\partial y}\frac{\partial z}{\partial x}+\frac{\partial R}{\partial z}\frac{\partial z}{\partial y}\frac{\partial z}{\partial \mathrm{ten}}+R\frac{{\partial }^{2}z}{\partial y\partial x})\end{array}}\hfill \end{array}dA.\hfill \end{array}$$

By Clairaut's theorem, $\frac{{\partial }^{2}z}{\partial \mathrm{ten}\partial y}=\frac{{\partial }^{2}z}{\partial y\partial \mathrm{10}}.$ Therefore, four of the terms disappear from this double integral, and nosotros are left with

$${\displaystyle \underset{D}{\iint }\left[\text{−}\left(R_y-Q_z\right)z_x-\left(P_z-R_x\right)z_y+\left(Q_x-P_y\right)\right]}dA,$$

which equals ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}·d\text{S}.}$

□

We accept shown that Stokes' theorem is truthful in the case of a function with a domain that is a simply connected region of finite area. Nosotros can chop-chop confirm this theorem for another important case: when vector field F is conservative. If F is conservative, the scroll of F is nada, so ${\displaystyle \underset{\mathrm{South}}{\iint }\text{coil}\text{F}·d\text{S}=0.}$ Since the boundary of South is a closed bend, ${\displaystyle {\int }_C\text{F}·d\text{r}}$ is also goose egg.

Example half dozen.73

Verifying Stokes' Theorem for a Specific Case

Verify that Stokes' theorem is true for vector field $\text{F}\left(\mathrm{10},y,z\right)=\langle y,\mathrm{two}z,x^2\rangle$ and surface Due south, where S is the paraboloid $z=4-x^2-y^{\mathrm{two}}$ . Assume the surface is outward oriented and $z\ge 0$ .

Effigy 6.83 Verifying Stokes' theorem for a hemisphere in a vector field.

Checkpoint 6.61

Verify that Stokes' theorem is true for vector field $\text{F}(x,y,z)=\langle y,x,\text{−}z\rangle$ and surface Due south, where Southward is the upwardly oriented portion of the graph of $f(x,y)=x^{2}y$ over a triangle in the xy-plane with vertices $(0,0),$ $(\mathrm{two},0),$ and $(0,2).$

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 u.s. 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 vi.74

Computing a Surface Integral

Summate surface integral ${\displaystyle {\iint }_S\text{scroll}\text{F}·d\text{South}},$ where S is the surface, oriented outward, in Figure 6.84 and $\text{F}=\langle z,2xy,\mathrm{ten}+y\rangle .$

Figure 6.84 A complicated surface in a vector field.

An amazing consequence of Stokes' theorem is that if Southward′ is whatsoever other smooth surface with boundary C and the same orientation equally S, so ${\displaystyle {\iint }_{\mathrm{Due\; south}}\text{scroll}\text{F}·d\text{Due south}}={\displaystyle {\int }_C\text{F}·d\text{r}}=0$ because Stokes' theorem says the surface integral depends on the line integral around the boundary only.

In Case half-dozen.74, we calculated a surface integral just past using information nigh the boundary of the surface. In general, allow ${\mathrm{South}}_1$ and $S_2$ exist shine surfaces with the aforementioned purlieus C and the same orientation. Past Stokes' theorem,

$${\displaystyle {\iint }_{{\mathrm{Southward}}_1}\text{whorl}\text{F}·d\text{S}}={\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_{{\mathrm{Due\; south}}_2}\text{scroll}\text{F}·d\text{S}}.$$

(6.23)

Therefore, if ${\displaystyle {\iint }_{S_{\mathrm{i}}}\text{curl}\text{F}·d\text{S}}$ is hard to summate merely ${\displaystyle {\iint }_{S_{\mathrm{two}}}\text{coil}\text{F}·d\text{S}}$ is easy to calculate, Stokes' theorem allows united states of america to calculate the easier surface integral. In Instance 6.74, we could accept calculated ${\displaystyle {\iint }_S\text{gyre}\text{F}·d\text{S}}$ by calculating ${\displaystyle {\iint }_{S^{\prime }}\text{curl}\text{F}·d\text{S}},$ where $S^{\prime }$ is the disk enclosed past purlieus bend C (a much more simple surface with which to work).

Equation 6.23 shows that flux integrals of curl vector fields are surface independent in the same way that line integrals of slope fields are path independent. Think that if F is a ii-dimensional conservative vector field defined on a simply continued domain, $f$ is a potential function for F, and C is a bend in the domain of F, then ${\displaystyle {\int }_C\text{F}·d\text{r}}$ depends only on the endpoints of C. Therefore if C′ is whatever other curve with the aforementioned starting indicate and endpoint as C (that is, C′ has the same orientation every bit C), then ${\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\int }_{C\text{′}}\text{F}·d\text{r}}.$ In other words, the value of the integral depends on the boundary of the path but; information technology does non really depend on the path itself.

Analogously, suppose that Due south and Southward′ are surfaces with the same purlieus and same orientation, and suppose that G is a 3-dimensional vector field that can be written as the curl of another vector field F (so that F is like a "potential field" of G). By Equation 6.23,

$${\displaystyle {\iint }_{\mathrm{Southward}}\text{G}·d\text{S}}={\displaystyle {\iint }_S\text{curl}\text{F}·d\text{South}}={\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_{\mathrm{Southward}\text{′}}\text{curl}\text{F}·d\text{S}}={\displaystyle {\iint }_{\mathrm{Due\; south}\text{′}}\text{Yard}·d\text{S}}.$$

Therefore, the flux integral of G does not depend on the surface, merely on the boundary of the surface. Flux integrals of vector fields that can be written as the ringlet of a vector field are surface independent in the same manner that line integrals of vector fields that can be written equally the gradient of a scalar function are path independent.

Checkpoint 6.62

Use Stokes' theorem to calculate surface integral ${\displaystyle {\iint }_{\mathrm{Southward}}\text{curl}\text{F}·d\text{Southward}},$ where $\text{F}=\langle z,x,y\rangle$ 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.

Example vi.75

Computing a Line Integral

Summate the line integral ${\displaystyle {\int }_C\text{F}·d\text{r}},$ where $\text{F}=\langle xy,x^2+y^2+z^{\mathrm{ii}},yz\rangle$ and C is the boundary of the parallelogram with vertices $\left(0,0,1\right),\left(0,1,0\right),\left(\mathrm{ii},0,\mathrm{-1}\right),$ and $\left(2,1,\mathrm{-2}\right).$

Checkpoint 6.63

Utilize Stokes' theorem to calculate line integral ${\displaystyle {\int }_C\text{F}·d\text{r}},$ where $\text{F}=\langle z,x,y\rangle$ and C is oriented clockwise and is the boundary of a triangle with vertices $\left(0,0,\mathrm{i}\right),\left(\mathrm{three},0,\mathrm{-two}\right),$ and $\left(0,\mathrm{i},2\right).$

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 curlicue and circulation, and nosotros use Stokes' theorem to state Faraday's law—an of import law in electricity and magnetism that relates the curl of an electric field to the charge per unit of modify of a magnetic field.

Think that if C is a closed curve and F is a vector field defined on C, then the circulation of F effectually C is line integral ${\displaystyle {\int }_C\text{F}·d\text{r}}.$ If F represents the velocity field of a fluid in infinite, so the circulation measures the tendency of the fluid to movement in the direction of C.

Let F be a continuous vector field and allow $D_r$ be a small-scale disk of radius r with eye $P_0$ (Figure half dozen.85). If $D_r$ is minor enough, then $(\text{whorl}\text{F})(P)\approx (\text{curl}\text{F})(P_0)$ for all points P in $D_r$ considering the whorl is continuous. Allow $C_r$ be the purlieus circumvolve of $D_r.$ By Stokes' theorem,

$${\displaystyle {\int }_{C_r}\text{F}·d\text{r}={\displaystyle {\iint }_{D_r}\text{roll}\text{F}·\text{N}dS}}\approx {\displaystyle {\iint }_{D_r}\left(\text{curl}\text{F}\right)\left(P_0\right)·\text{Northward}\left(P_0\right)d\mathrm{South}}.$$

Figure 6.85 Disk $D_r$ is a small disk in a continuous vector field.

The quantity $\left(\text{roll}\text{F}\right)\left(P_0\right)·\text{N}\left(P_0\right)$ is abiding, and therefore

$${\displaystyle {\iint }_{D_r}\left(\text{gyre}\text{F}\right)\left(P_0\right)·\text{N}\left(P_0\right)d\mathrm{South}}=\pi r^2\left[\left(\text{ringlet}\text{F}\right)\left(P_0\right)·\text{Due north}\left(P_0\right)\right].$$

Thus

$${\displaystyle {\int }_{C_r}\text{F}·d\text{r}}\approx \pi r^2\left[\left(\text{curl}\text{F}\right)\left(P_0\right)·\text{Northward}\left(P_0\right)\right],$$

and the approximation gets arbitrarily close as the radius shrinks to zero. Therefore Stokes' theorem implies that

$$\left(\text{curl}\text{F}\right)\left(P_0\right)·\text{Due north}\left(P_0\right)=\underset{r\to 0^{+}}{\text{lim}}\frac{1}{\pi r^2}{\displaystyle {\int }_{C_r}\text{F}·d\text{r}}.$$

This equation relates the curl of a vector field to the circulation. Since the surface area of the disk is $\pi r^{\mathrm{ii}},$ this equation says nosotros can view the curl (in the limit) as the circulation per unit surface area. Recall that if F is the velocity field of a fluid, and so apportionment ${\displaystyle {\oint }_{C_r}\text{F}·d\text{r}}={\displaystyle {\oint }_{C_r}\text{F}·\text{T}d\mathrm{south}}$ is a measure of the trend of the fluid to motion around $C_r.$ The reason for this is that $\text{F}·\text{T}$ is a component of F in the direction of T, and the closer the management of F is to T, the larger the value of $\text{F}·\text{T}$ (remember that if a and b are vectors and b is fixed, then the dot product $\text{a}·\text{b}$ is maximal when a points in the same management as b). Therefore, if F is the velocity field of a fluid, then $\text{roll}\text{F}·\text{N}$ is a measure of how the fluid rotates nearly axis Northward. The effect of the gyre is largest nearly the axis that points in the management of N, considering in this example $\text{curl}\text{F}·\text{N}$ is as large as possible.

To run into this effect in a more than concrete fashion, imagine placing a tiny paddlewheel at point $P_0$ (Figure half-dozen.86). The paddlewheel achieves its maximum speed when the axis of the bike points in the direction of curlF. This justifies the interpretation of the scroll nosotros have learned: curl is a measure of the rotation in the vector field well-nigh the axis that points in the direction of the normal vector N, and Stokes' theorem justifies this interpretation.

Figure 6.86 To visualize curl at a point, imagine placing a tiny paddlewheel at that point in the vector field.

Now that we have learned well-nigh Stokes' theorem, we tin talk over applications in the area of electromagnetism. In detail, nosotros examine how nosotros can apply Stokes' theorem to interpret between two equivalent forms of Faraday's law. Earlier stating the two forms of Faraday's law, we need some groundwork terminology.

Let C exist a closed curve that models a thin wire. In the context of electric fields, the wire may be moving over fourth dimension, so nosotros write $C(t)$ to represent the wire. At a given time t, curve $C(t)$ may be different from original curve C because of the movement of the wire, simply we assume that $C(t)$ is a closed curve for all times t. Let $D(t)$ be a surface with $C(t)$ every bit its boundary, and orient $C(t)$ so that $D(t)$ has positive orientation. Suppose that $C(t)$ is in a magnetic field $\text{B}(t)$ that can also change over time. In other words, B has the form

$$\text{B}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle ,$$

where P, Q, and R can all vary continuously over time. We can produce current forth the wire by changing field $\text{B}(t)$ (this is a consequence of Ampere'south constabulary). Flux $\varphi (t)={\displaystyle {\iint }_{D(t)}\text{B}(t)·d\text{S}}$ creates electric field $\text{E}(t)$ that does work. The integral grade of Faraday'due south police states that

$$\text{Work}={\displaystyle {\int }_{C(t)}\text{E}(t)}·d\text{r}=-\frac{\partial \varphi }{\partial t}.$$

In other words, the work done by E is the line integral around the boundary, which is also equal to the charge per unit of alter of the flux with respect to time. The differential course of Faraday's law states that

$$\text{coil}\text{E}=-\frac{\partial \text{B}}{\partial t}.$$

Using Stokes' theorem, we can show that the differential form of Faraday's law is a consequence of the integral course. By Stokes' theorem, we tin catechumen the line integral in the integral course into surface integral

$$-\frac{\partial \varphi }{\partial t}={\displaystyle {\int }_{C(t)}\text{E}(t)}·d\text{r}={\displaystyle {\iint }_{D(t)}\text{curl}\text{E}(t)}·d\text{S}.$$

Since $\varphi (t)={\displaystyle {\iint }_{D(t)}\text{B}(t)·d\text{S}},$ and then as long as the integration of the surface does not vary with time we also have

$$-\frac{\partial \varphi }{\partial t}={\displaystyle {\iint }_{D(t)}-\frac{\partial \text{B}}{\partial t}·d\text{Southward}}.$$

Therefore,

$${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{ringlet}\text{E}·d\text{S}}.$$

To derive the differential form of Faraday's police, nosotros would similar to conclude that $\text{ringlet}\text{Due east}=-\frac{\partial \text{B}}{\partial t}.$ In general, the equation

$${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{curl}\text{East}·d\text{S}}$$

is non enough to conclude that $\text{curl}\text{East}=-\frac{\partial \text{B}}{\partial t}.$ The integral symbols practice not just "abolish out," leaving equality of the integrands. To see why the integral symbol does not just cancel out in general, consider the two single-variable integrals ${\displaystyle {\int }_0^{1}xdx}$ and ${\displaystyle {\int }_0^{1}f(\mathrm{10})d\mathrm{ten}},$ where

$$f\left(x\right)=\{{}_{0,1\text{/}\mathrm{two}\le x\le 1.}^{1,0\le \mathrm{ten}\le 1\text{/}2}$$

Both of these integrals equal $\frac{\mathrm{i}}{2},$ so ${\displaystyle {\int }_0^{\mathrm{one}}\mathrm{10}dx}={\displaystyle {\int }_0^{\mathrm{i}}f(\mathrm{ten})d\mathrm{10}}.$ However, $x\ne f(\mathrm{10}).$ Analogously, with our equation ${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{ringlet}\text{Due east}·d\text{S}},$ we cannot simply conclude that $\text{gyre}\text{East}=-\frac{\partial \text{B}}{\partial t}$ just because their integrals are equal. However, in our context, equation ${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{roll}\text{Due east}·d\text{S}}$ is true for whatsoever region, however pocket-sized (this is in contrast to the single-variable integrals just discussed). If F and Thou are three-dimensional vector fields such that ${\displaystyle {\iint }_s\text{F}·d\text{S}=}{\displaystyle {\iint }_s\text{G}·d\text{S}}$ for any surface S, then it is possible to show that $\text{F}=\text{G}$ by shrinking the area of S to naught by taking a limit (the smaller the surface area of Due south, the closer the value of ${\displaystyle {\iint }_{\mathrm{due\; south}}\text{F}·d\text{Southward}}$ to the value of F at a point within Southward). Therefore, we can let surface area $D\left(t\right)$ shrink to zero by taking a limit and obtain the differential form of Faraday'southward law:

$$\text{gyre}\text{E}=-\frac{\partial \text{B}}{\partial t}.$$

In the context of electric fields, the curl of the electric field tin can be interpreted as the negative of the charge per unit of change of the corresponding magnetic field with respect to time.

Example 6.76

Using Faraday's Law

Calculate the curlicue of electric field E if the corresponding magnetic field is constant field $\text{B}\left(t\right)=\langle 1,\mathrm{-4},2\rangle .$

Assay

A event of Faraday's law is that the curl of the electric field respective to a abiding magnetic field is always zero.

Checkpoint 6.64

Summate the curl of electric field E if the corresponding magnetic field is $\text{B}(t)=\langle tx,ty,\mathrm{-2}tz\rangle ,0\le t<\infty .$

Notice that the gyre of the electric field does not modify over time, although the magnetic field does modify over time.

Section vi.seven Exercises

For the following exercises, without using Stokes' theorem, calculate directly both the flux of $\text{roll}\text{F}·\text{North}$ over the given surface and the apportionment integral around its boundary, assuming all boundaries are oriented clockwise as viewed from in a higher place.

326 .

$\text{F}(\mathrm{10},y,z)=y^2\text{i}+z^2\text{j}+x^2\text{k}\text{;}$ Due south is the starting time-octant portion of plane $x+y+z=1.$

327 .

$\text{F}(\mathrm{10},y,z)=z\text{i}+x\text{j}+y\text{k}\text{;}$ S is hemisphere $z={\left(a^2-{\mathrm{ten}}^2-y^{\mathrm{ii}}\right)}^{\mathrm{i}\text{/}2}.$

328 .

$\text{F}(x,y,z)=y^{\mathrm{two}}\text{i}+\mathrm{ii}x\text{j}+\mathrm{v}\text{k}\text{;}$ S is hemisphere $z={\left(4-x^2-y^2\right)}^{\mathrm{ane}\text{/}2}.$

329 .

$\text{F}(x,y,z)=z\text{i}+\mathrm{two}x\text{j}+\mathrm{iii}y\text{1000}\text{;}$ S is upper hemisphere $z=\sqrt{\mathrm{nine}-x^2-y^2}.$

330 .

$\text{F}(x,y,z)=\left(\mathrm{ten}+\mathrm{two}z\right)\text{i}+\left(y-x\right)\text{j}+\left(z-y\right)\text{k}\text{;}$ S is a triangular region with vertices (three, 0, 0), (0, 3/2, 0), and (0, 0, 3).

331 .

$\text{F}(x,y,z)=2y\text{i}-\mathrm{six}z\text{j}+3\mathrm{10}\text{k}\text{;}$ Southward is a portion of paraboloid $z=\mathrm{iv}-x^{\mathrm{ii}}-y^{\mathrm{ii}}$ and is above the xy-aeroplane.

For the following exercises, apply Stokes' theorem to evaluate ${\displaystyle {\iint }_{\mathrm{Southward}}(\text{curl}\text{F}·\text{Northward})dS}$ for the vector fields and surface.

332 .

$\text{F}(x,y,z)=\mathrm{ten}y\text{i}-z\text{j}$ and S is the surface of the cube $0\le x\le 1,0\le y\le \mathrm{one},0\le z\le 1,$ except for the face up where $z=0,$ and using the outward unit normal vector.

333 .

$\text{F}(x,y,z)=xy\text{i}+x^2\text{j}+z^{\mathrm{two}}\text{k}\text{;}$ and C is the intersection of paraboloid $z=x^2+y^2$ and plane $z=y,$ and using the outward normal vector.

334 .

$\text{F}(x,y,z)=4y\text{i}+z\text{j}+2y\text{grand}$ and C is the intersection of sphere $x^{\mathrm{two}}+y^{\mathrm{two}}+z^{\mathrm{two}}=\mathrm{four}$ with aeroplane $z=0,$ and using the outward normal vector

335 .

Use Stokes' theorem to evaluate ${\displaystyle \underset{C}{\int }\left[2x{y}^{\mathrm{two}}zdx+2x^{2}yzdy+\left(x^2{y}^2-2z\right)dz\right]},$ where C is the curve given by $x=\text{cos}t,y=\text{sin}t,z=\text{sin}t,0\le t\le \mathrm{two}\pi ,$ traversed in the management of increasing t.

336 .

[T] Use a estimator algebraic organization (CAS) and Stokes' theorem to judge line integral ${\displaystyle \underset{C}{\int }\left(yd\mathrm{10}+zdy+\mathrm{ten}dz\right)},$ where C is the intersection of aeroplane $\mathrm{10}+y=2$ and surface $x^2+y^2+z^2=2\left(x+y\right),$ traversed counterclockwise viewed from the origin.

337 .

[T] Use a CAS and Stokes' theorem to approximate line integral ${\displaystyle \underset{C}{\int }\left(3yd\mathrm{ten}+2zdy-5\mathrm{10}dz\right)},$ where C is the intersection of the xy-plane and hemisphere $z=\sqrt{1-x^{\mathrm{ii}}-y^2},$ traversed counterclockwise viewed from the top—that is, from the positive z-axis toward the xy-plane.

338 .

[T] Utilize a CAS and Stokes' theorem to gauge line integral ${\displaystyle \underset{C}{\int }\left[\left(1+y\right)zd\mathrm{ten}+\left(1+z\right)xdy+\left(1+\mathrm{ten}\right)ydz\right]},$ where C is a triangle with vertices $\left(1,0,0\right),$ $\left(0,\mathrm{ane},0\right),$ and $\left(0,0,1\right)$ oriented counterclockwise.

339 .

Use Stokes' theorem to evaluate ${\displaystyle {\iint }_S\text{scroll}\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=e^{xy}\text{cos}z\text{i}+x^{2}z\text{j}+\mathrm{ten}y\text{k},$ and S is half of sphere $x=\sqrt{\mathrm{one}-y^2-z^{\mathrm{two}}},$ oriented out toward the positive 10-axis.

340 .

[T] Use a CAS and Stokes' theorem to evaluate ${\displaystyle {\iint }_{\mathrm{Southward}}(\text{curl}\text{F}·\text{N})d\mathrm{Due\; south}},$ where $\text{F}(\mathrm{ten},y,z)=x^{\mathrm{ii}}y\text{i}+\mathrm{ten}{y}^{\mathrm{two}}\text{j}+z^{\mathrm{three}}\text{k}$ and C is the bend of the intersection of plane $3x+\mathrm{two}y+z=\mathrm{half\; dozen}$ and cylinder $x^2+y^{\mathrm{ii}}=4,$ oriented clockwise when viewed from above.

341 .

[T] Apply a CAS and Stokes' theorem to evaluate ${\displaystyle \underset{S}{\iint }\text{roll}\text{F}·d\text{Southward},}$ where $\text{F}(x,y,z)=\left(\text{sin}\left(y+z\right)-y{\mathrm{ten}}^{\mathrm{two}}-\frac{y^3}{3}\right)\text{i}+\mathrm{ten}\text{cos}\left(y+z\right)\text{j}+\text{cos}\left(\mathrm{two}y\right)\text{k}$ and S consists of the top and the iv sides simply not the lesser of the cube with vertices $\left(\mathrm{\pm 1},\mathrm{\pm i},\mathrm{\pm i}\right),$ oriented outward.

342 .

[T] Use a CAS and Stokes' theorem to evaluate ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}·d\text{S},}$ where $\text{F}(x,y,z)=z^{\mathrm{ii}}\text{i}-3xy\text{j}+x^3{y}^3\text{yard}$ and S is the top part of $z=5-{\mathrm{10}}^2-y^2$ above plane $z=1,$ and S is oriented upwardly.

343 .

Use Stokes' theorem to evaluate ${\displaystyle {\iint }_{\mathrm{Due\; south}}(\text{scroll}\text{F}·\text{N})dS},$ where $\text{F}(x,y,z)=z^2\text{i}+y^2\text{j}+\mathrm{ten}\text{m}$ and Due south is a triangle with vertices (i, 0, 0), (0, 1, 0) and (0, 0, ane) with counterclockwise orientation.

344 .

Use Stokes' theorem to evaluate line integral ${\displaystyle \underset{C}{\int }\left(zdx+xdy+ydz\right)},$ where C is a triangle with vertices (three, 0, 0), (0, 0, 2), and (0, half-dozen, 0) traversed in the given club.

345 .

Use Stokes' theorem to evaluate ${\displaystyle \underset{C}{\int }\left(\frac{1}{\mathrm{two}}{y}^{2}dx+zdy+xdz\right)},$ where C is the bend of intersection of airplane $x+z=\mathrm{ane}$ and ellipsoid $x^{\mathrm{ii}}+2y^2+z^2=1,$ oriented clockwise from the origin.

346 .

Use Stokes' theorem to evaluate ${\displaystyle {\iint }_S(\text{curl}\text{F}·\text{N})dS},$ where $\text{F}(x,y,z)=x\text{i}+y^2\text{j}+z{\mathrm{due\; east}}^{\mathrm{10}y}\text{thou}$ and S is the function of surface $z=\mathrm{ane}-x^2-2y^2$ with $z\ge 0\text{,}$ oriented counterclockwise.

347 .

Use Stokes' theorem for vector field $\text{F}(\mathrm{ten},y,z)=z\text{i}+\mathrm{iii}x\text{j}+\mathrm{two}z\text{k}$ where S is surface $z=1-{\mathrm{10}}^2-y^2,z\ge 0,$ C is boundary circle $x^2+y^2=\mathrm{ane},$ and S is oriented in the positive z-direction.

348 .

Utilize Stokes' theorem for vector field $\text{F}(x,y,z)=-\frac{\mathrm{iii}}{\mathrm{two}}{y}^2\text{i}-\mathrm{ii}\mathrm{10}y\text{j}+yz\text{m}\text{,}$ where South is that part of the surface of plane $\mathrm{10}+y+z=\mathrm{one}$ independent within triangle C with vertices (ane, 0, 0), (0, 1, 0), and (0, 0, 1), traversed counterclockwise equally viewed from higher up.

349 .

A sure closed path C in airplane $\mathrm{ii}x+\mathrm{two}y+z=\mathrm{ane}$ is known to project onto unit circumvolve $x^2+y^2=1$ in the xy-airplane. Let c exist a constant and permit $\text{R}(x,y,z)=x\text{i}+y\text{j}+z\text{k}.$ Use Stokes' theorem to evaluate ${\displaystyle {\int }_C^{}(c\text{1000}\times \text{R})·d\text{S}}.$

350 .

Employ Stokes' theorem and let C exist the boundary of surface $z={\mathrm{ten}}^2+y^2$ with $0\le x\le \mathrm{two}$ and $0\le y\le 1,$ oriented with upwardly facing normal. Define

$$\text{F}(x,y,z)=\left[\text{sin}\left(x^{\mathrm{three}}\right)+xz\right]\text{i}+(x-yz)\text{j}+\text{cos}\left(z^{\mathrm{four}}\right)\text{k}\text{and evaluate}{\displaystyle {\int }_C\text{F}·d\text{S}}.$$

351 .

Allow Southward be hemisphere ${\mathrm{10}}^{\mathrm{ii}}+y^2+z^2=\mathrm{iv}$ with $z\ge 0,$ oriented upward. Let $\text{F}(\mathrm{10},y,z)={\mathrm{ten}}^2{e}^{yz}\text{i}+y^2{e}^{xz}\text{j}+z^{\mathrm{two}}{e}^{xy}\text{k}$ exist a vector field. Use Stokes' theorem to evaluate ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}.$

352 .

Let $\text{F}(\mathrm{10},y,z)=xy\text{i}+\left(e^{z^2}+y\right)\text{j}+(\mathrm{ten}+y)\text{1000}$ and permit S exist the graph of function $y=\frac{x^2}{9}+\frac{z^2}{9}-1$ with $y\le 0$ oriented so that the normal vector S has a positive y component. Use Stokes' theorem to compute integral ${\displaystyle {\iint }_S\text{coil}\text{F}·d\text{South}}.$

353 .

Utilise Stokes' theorem to evaluate ${\displaystyle {\oint }_{}\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=y\text{i}+z\text{j}+x\text{one thousand}$ and C is a triangle with vertices (0, 0, 0), (two, 0, 0) and $(0,\mathrm{-2},2)$ oriented counterclockwise when viewed from above.

354 .

Use the surface integral in Stokes' theorem to calculate the apportionment of field F, $\text{F}(x,y,z)=x^2{y}^{\mathrm{three}}\text{i}+\text{j}+z\text{one thousand}$ effectually C, which is the intersection of cylinder $x^2+y^2=\mathrm{four}$ and hemisphere $x^2+y^{\mathrm{ii}}+z^{\mathrm{ii}}=16,z\ge 0,$ oriented counterclockwise when viewed from above.

355 .

Utilize Stokes' theorem to compute ${\displaystyle {\iint }_S\text{gyre}\text{F}·d\text{Due south}},$ where $\text{F}(x,y,z)=\text{i}+\mathrm{10}{y}^2\text{j}+\mathrm{ten}{y}^2\text{k}$ and South is a part of plane $y+z=\mathrm{ii}$ inside cylinder $x^2+y^2=1$ and oriented counterclockwise.

356 .

Use Stokes' theorem to evaluate ${\displaystyle {\iint }_{\mathrm{Southward}}\text{curl}\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=\text{−}{y}^2\text{i}+x\text{j}+z^2\text{thou}$ and S is the part of plane $x+y+z=1$ in the positive octant and oriented counterclockwise $x\ge 0\text{,}y\ge 0\text{,}z\ge 0.$

357 .

Permit $\text{F}(x,y,z)=xy\text{i}+2z\text{j}-2y\text{k}$ and allow C exist the intersection of airplane $x+z=\mathrm{five}$ and cylinder $x^{\mathrm{two}}+y^{\mathrm{two}}=9,$ which is oriented counterclockwise when viewed from the top. Compute the line integral of F over C using Stokes' theorem.

358 .

[T] Use a CAS and let $\text{F}(x,y,z)=x{y}^2\text{i}+(yz-x)\text{j}+e^{yxz}\text{k}.$ Utilise Stokes' theorem to compute the surface integral of curl F over surface S with inward orientation consisting of cube $[0,\mathrm{ane}]\times [0,1]\times [0,1]$ with the right side missing.

359 .

Let Southward exist ellipsoid $\frac{{\mathrm{ten}}^{\mathrm{two}}}{\mathrm{iv}}+\frac{y^2}{\mathrm{nine}}+z^2=1$ oriented counterclockwise and let F be a vector field with component functions that have continuous partial derivatives. $\int {\int }_s\text{curl}\mathbf{F}\mathbf{·}\mathbf{due\; north}$

360 .

Allow Southward be the part of paraboloid $z=9-{\mathrm{10}}^2-y^2$ with $z\ge 0.$ Verify Stokes' theorem for vector field $\text{F}(\mathrm{ten},y,z)=\mathrm{three}z\text{i}+4x\text{j}+2y\text{thou}.$

361 .

[T] Use a CAS and Stokes' theorem to evaluate ${\displaystyle {\oint }_C\text{F}·d\text{S}}\text{,}$ if $\text{F}(x,y,z)=\left(3z-\text{sin}x\right)\text{i}+\left(x^{\mathrm{ii}}+{\mathrm{east}}^y\right)\text{j}+\left(y^3-\text{cos}z\right)\text{thousand}\text{,}$ where C is the curve given past $x=\text{cos}t,y=\text{sin}t,z=\mathrm{ane};0\le t\le 2\pi .$

362 .

[T] Utilize a CAS and Stokes' theorem to evaluate $\text{F}(\mathrm{10},y,z)=2y\text{i}+{\mathrm{due\; east}}^z\text{j}-\text{arctan}x\text{chiliad}$ with South as a portion of paraboloid $z=4-{\mathrm{ten}}^2-y^2$ cut off past the xy-plane oriented counterclockwise.

363 .

[T] Apply a CAS to evaluate ${\displaystyle {\iint }_{\mathrm{South}}\text{curl(}\text{F})·d\text{South}\text{,}}$ where $\text{F}(x,y,z)=2z\text{i}+\mathrm{iii}x\text{j}+5y\text{k}$ and S is the surface parametrically by $\text{r}(r,\theta )=r\text{cos}\theta \text{i}+r\text{sin}\theta \text{j}+\left(4-r^2\right)\text{yard}$ $\left(0\le \theta \le 2\pi \text{,}0\le r\le 3\right).$

364 .

Let S be paraboloid $z=a\left(1-x^2-y^2\right),$ for $z\ge 0,$ where $a>0$ is a existent number. Let $\text{F}=\langle \mathrm{10}-y,y+z,z-x\rangle .$ For what value(southward) of a (if any) does ${\displaystyle {\iint }_S\left(\nabla \times \text{F}\right)}·\text{n}dS$ have its maximum value?

For the following application exercises, the goal is to evaluate $A={\displaystyle {\iint }_{\mathrm{Southward}}\left(\nabla \times \text{F}\right)}·\text{n}dS,$ where $\text{F}=\langle xz,\text{−}\mathrm{10}z,xy\rangle$ and S is the upper one-half of ellipsoid $x^{\mathrm{two}}+y^2+8z^2=1,\text{where}z\ge 0.$

365 .

Evaluate a surface integral over a more user-friendly surface to find the value of A.

366 .

Evaluate A using a line integral.

367 .

Take paraboloid $z=x^2+y^2,$ for $0\le z\le 4,$ and slice it with plane $y=0.$ Permit S be the surface that remains for $y\ge 0,$ including the planar surface in the xz-airplane. Let C be the semicircle and line segment that bounded the cap of S in plane $z=4$ with counterclockwise orientation. Let $\text{F}=\langle \mathrm{ii}z+y,2x+z,2y+x\rangle .$ Evaluate ${\displaystyle {\iint }_{\mathrm{Southward}}\left(\nabla \times \text{F}\right)}·\text{n}dS.$

For the post-obit exercises, permit South be the disk enclosed by bend

$C:\text{r}(t)=\langle \text{cos}\phi \text{cos}t,\text{sin}t,\text{sin}\phi \text{cos}t\rangle ,$ for $0\le t\le \mathrm{ii}\pi ,$ where $0\le \phi \le \frac{\pi }{2}$ is a fixed angle.

368 .

What is the length of C in terms of $\phi ?$

369 .

What is the apportionment of C of vector field $\text{F}=\langle \text{−}y,\text{−}z,x\rangle$ as a part of $\phi ?$

370 .

For what value of $\phi$ is the apportionment a maximum?

371 .

Circle C in plane $x+y+z=8$ has radius four and center (2, three, 3). Evaluate ${\displaystyle {\oint }_C\text{F}·d\text{r}}$ for $F=\langle 0,\text{−}z,2y\rangle ,$ where C has a counterclockwise orientation when viewed from to a higher place.

372 .

Velocity field $\text{v}=\langle 0,1-{\mathrm{ten}}^2,0\rangle ,$ for $\left|x\right|\le \mathrm{ane}\text{and}\left|z\right|\le 1,$ represents a horizontal flow in the y-management. Compute the ringlet of v in a clockwise rotation.

373 .

Evaluate integral ${\displaystyle {\iint }_S\left(\nabla \times \text{F}\right)}·\text{n}dS,$ where $\text{F}=\text{−}xz\text{i}+yz\text{j}+xy{e}^z\text{yard}$ and Southward is the cap of paraboloid $z=5-x^2-y^2$ higher up plane $z=\mathrm{iii},$ and n points in the positive z-direction on S.

For the following exercises, use Stokes' theorem to observe the circulation of the post-obit vector fields around whatever smooth, uncomplicated closed curve C.

374 .

$\text{F}=\nabla \left(\mathrm{10}\text{sin}y{e}^z\right)$

375 .

$\text{F}=\langle y^{\mathrm{ii}}{z}^3,z2xy{z}^3,\mathrm{iii}x{y}^2{z}^2\rangle$

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Source: https://openstax.org/books/calculus-volume-3/pages/6-7-stokes-theorem

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