Evaluate Sc(X^2+y^2)ds, Where C Is the Line Segment From (1,1) to (2,3)

Learning Objectives

• 6.7.1 Explain the significant of Stokes' theorem.
• 6.7.2 Apply Stokes' theorem to measure a lineage integral.
• 6.7.3 Use Stokes' theorem to calculate a surface integral.
• 6.7.4 Use Stokes' theorem to calculate a curl.

In that section, we study Stokes' theorem, a higher-magnitude generalization of Green's theorem. This theorem, like the Fundamental Theorem for Line Integrals and Green's theorem, is a generalization of the Fundamental Theorem of Calculus to higher dimensions. Stokes' theorem relates a vector surface integral over surface S in space to a line integral around the boundary of S. Therefore, just as the theorems before it, Stokes' theorem can be used to reduce an integral over a geometric physical object S to an integral over the boundary of S.

In addition to allowing us to translate between line integrals and surface integrals, Stokes' theorem connects the concepts of curl and circulation. Moreover, the theorem has applications in changeable mechanism and electromagnetism. We use Stokes' theorem to deduct Michael Faraday's law, an important resultant involving galvanic fields.

Stokes' Theorem

Stokes' theorem says we can calculate the flux of coil F across surface S aside knowing entropy 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 inherent of the curl of F over S.

Lashkar-e-Taiba S follow an oriented smooth coat with unit normal vector N. Furthermore, suppose the boundary of S is a Jordan curve C. The orientation course of S induces the positive predilection of C if, as you walk in the positive focus around C with your head pointing in the direction of N, the come out is always on your left. With this definition in place, we can state Stokes' theorem.

Theorem 6.19

Stokes' Theorem

Permit S be a piecewise smooth oriented surface with a boundary that is a sword-shaped blinking wind C with positive orientation (Compute 6.79). If F is a vector field with constituent functions that have continuous partial derivative derivatives on an spread region containing S, past

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

Figure 6.79 Stokes' theorem relates the flux integral over the surface to a line integral around the boundary of the show u. Note that the orientation of the curve is positive.

Suppose surface S is a flat region in the xy-level with upward orientation course. And so the unit normal transmitter is k and surface inbuilt ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}}·d\text{S}$ is actually the double built-in ${\displaystyle \underset{S}{\iint }\text{ringlet}\text{F}}·\text{k}dA.$ In this special pillow slip, Stokes' theorem gives ${\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_S\text{curl}\text{F}·\text{k}dA.}$ However, this is the circulation form of Green's theorem, which shows us that Green's theorem is a special display case of Stokes' theorem. Green's theorem can only handle surfaces in a plane, just Stokes' theorem can handle surfaces in a aeroplane or in space.

The complete proof of Stokes' theorem is beyond the scope of this school tex. We look at an visceral explanation for the truth of the theorem then visualize proof of the theorem in the special case that surface S is a portion of a graph of a function, and S, the limit of S, and F are all fairly tame.

Proof

First, we look at an epistolatory proof of the theorem. This test copy is not stringent, but it is meant to give a general feeling for why the theorem is true. Let S be a surface and let D be a smaller composition of the aboveground so that D does non share any points with the boundary of S. We choose D to be lesser enough soh that information technology can be approximated by an oriented square toes E. Let D come into its orientation course from S, and give E the same predilection. This square has foursome sides; denote them $E_l,$ $E_r,$ $E_u,$ and $E_d$ for the socialist, right, up, and down sides, severally. On the square, we can use the flux form of Green's theorem:

$${\displaystyle {\int }_{E_l+E_d+E_r+E_u}\text{F}·d\text{r}}={\displaystyle {\iint }_E\text{curl}\text{F}·\text{N}dS}={\displaystyle {\iint }_E\text{curl}\text{F}·d\text{S}}.$$

To approximate the conflate ended the entire surface, we add the values of the flux on the small squares approximating lowercase pieces of the surface (Forecast 6.80). By Green's theorem, the flux across to each one approximating square is a argumentation integral all over its boundary. Let F cost an approximating square with an orientation heritable from S and with a right side $E_l$ (so F is to the larboard of E). Let $F_r$ refer the right side of $F$ ; then, $E_l=\text{−}{F}_r.$ In other words, the right root of $F$ is the same curve as the left field side of E, just oriented in the contrary 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, pedigree integrals ${\displaystyle {\int }_{E_l}\text{F}·d\text{r}}$ and ${\displaystyle {\int }_{F_r}\text{F}·d\text{r}}$ cancel each other out. The same goes for the line integrals over the other three sides of E. These three line integrals cancel out with the ancestry integral of the lower side of the square above E, the line integral o'er the left lateral of the foursquare to the right of E, and the line integral over the upper side of the square on a lower floor E (Figure 6.81). After all this cancelation occurs over wholly the approximating squares, the only line integrals that live are the line integrals over sides approximating the boundary of S. Thus, the heart of all the fluxes (which, by Green's theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated away a line integral finished the boundary of S. In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.

Figure 6.80 Chop the surface into small pieces. The pieces should be moderate enough that they can atomic number 4 approximated by a square.

Figure 6.81 (a) The line integral along $E_l$ cancels out the line integral on $F_r$ because $E_l=\text{−}{F}_r.$ (b) The line integral along whatsoever of the sides of E cancels out with the line integral on a side of an adjacent approximating hearty.

Let's now take a rigorous proof of the theorem in the special guinea pig that S is the graphical record of function $z=f\left(x,y\right),$ where x and y vary over a finite, just connected region D of finite arena (Name 6.82). Moreover, assume that $f$ has continuous second-order partial derivatives. Army of the Righteou C denote the boundary of S and let C′ refer the boundary of D. Then, D is the "shadow" of S in the plane and C′ is the "shadow" of C. Imagine that S is oriented upwards. The counterclockwise orientation of C is positive, as is the counterclockwise preference of $C^{\prime }.$ Let $\text{F}\left(x,y,z\right)=\langle P,Q,R\rangle$ be a vector field with component functions that accept constant partial derivatives.

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

We take on the standard parameterization of $S:x=x,y=y,z=g\left(x,y\right).$ The tan vectors are ${\text{t}}_x=\langle 1,0,g_x\rangle$ and ${\text{t}}_y=\langle 0,1,g_y\rangle ,$ and therefore, ${\text{t}}_x\times {\text{t}}_y=\langle \text{−}{g}_x,\text{−}{g}_y,1\rangle .$ By Equation 6.19,

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

where the partial derivatives are totally evaluated at $\left(x,y,g\left(x,y\right)\right),$ making the integrand depend along x and y only. 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 $\langle x\left(t\right),y\left(t\right),g\left(x\left(t\right),y\left(t\right)\right)\rangle ,a\le t\le b.$ Brushlike with these parameterizations, the Chain normal, and Green's theorem, and keeping in mind that P, Q, and R are totally functions of x and y, we can appraise line integral ${\displaystyle {\int }_C\text{F}·d\text{r}}\text{:}$

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

By Clairaut's theorem, $\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{{\partial }^{2}z}{\partial y\partial x}.$ Therefore, four of the terms disappear from this double integral, and we are left-handed 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 have shown that Stokes' theorem is true in the case of a function with a domain that is a simply connected part of exhaustible arena. We can quickly substantiate this theorem for another epochal case: when transmitter field F is conservative. If F is conservative, the curl of F is zero, so ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}·d\text{S}=0.}$ Since the limit of S is a drawn curve ball, ${\displaystyle {\int }_C\text{F}·d\text{r}}$ is also zero.

Example 6.73

Confirmative Stokes' Theorem for a Specific Case

Swan that Stokes' theorem is true for vector subject $\text{F}\left(x,y,z\right)=\langle y,2z,x^2\rangle$ and surface S, where S is the paraboloid $z=4-x^2-y^2$ .

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

Checkpoint 6.61

Verify that Stokes' theorem is on-key for vector field $\text{F}(x,y,z)=\langle y,x,\text{−}z\rangle$ and surface S, where S is the upwardly oriented fortune of the graph of $f(x,y)=x^{2}y$ o'er a triangle in the xy-plane with vertices $(0,0),$ $(2,0),$ and $(0,2).$

Applying Stokes' Theorem

Stokes' theorem translates between the flux integral of aboveground S to a line integral around the boundary of S. Therefore, the theorem allows us to compute Earth's surface integrals or line integrals that would ordinarily be quite a difficult by translating the telephone line integral into a grade-constructed integral or contrariwise. We straight off study some examples of for each one kind of translation.

Example 6.74

Calculating a Surface Integral

Calculate surface integral ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}},$ where S is the rise up, oriented outward, in Figure 6.84 and $\text{F}=\langle z,2xy,x+y\rangle .$

Forecast 6.84 A complicated come on in a vector battleground.

An amazing consequence of Stokes' theorem is that if S′ is whatsoever other smooth surface with boundary C and the same orientation as S, then ${\displaystyle {\iint }_S\text{ringlet}\text{F}·d\text{S}}={\displaystyle {\int }_C\text{F}·d\text{r}}=0$ because Stokes' theorem says the surface intrinsical depends on the line integral some the boundary only.

In Example 6.74, we calculated a surface integral simply by exploitation information about the boundary of the surface. Generally, let $S_1$ and $S_2$ live silken surfaces with the same boundary C and the same orientation. Aside Stokes' theorem,

$${\displaystyle {\iint }_{S_1}\text{curl}\text{F}·d\text{S}}={\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_{S_2}\text{whorl}\text{F}·d\text{S}}.$$

(6.23)

Therefore, if ${\displaystyle {\iint }_{S_1}\text{Robert Floyd Curl Jr.}\text{F}·d\text{S}}$ is difficult to calculate but ${\displaystyle {\iint }_{S_2}\text{curl}\text{F}·d\text{S}}$ is easy to calculate, Stokes' theorem allows us to calculate the easier surface built-in. In Example 6.74, we could have measured ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}$ by calculating ${\displaystyle {\iint }_{S^{\prime }}\text{draw in}\text{F}·d\text{S}},$ where $S^{\prime }$ is the disk enclosed past bound curve C (a much more arrow-shaped surface with which to work).

Equation 6.23 shows that flux integrals of curl vector William Claude Dukenfield are surface independent in the same way that line integrals of gradient Fields are path independent. Call back that if F is a two-dimensional conservative vector field defined happening a simply connected domain, $f$ is a potential mathematical function for F, and C is a curve 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 any other curve with the unvarying starting point and endpoint arsenic C (that is, C′ has the unchanged orientation as C), then ${\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\int }_{C\text{′}}\text{F}·d\text{r}}.$ In unusual words, the value of the inherent depends on the boundary of the path only; it does not really ride the track itself.

Analogously, suppose that S and S′ are surfaces with the comparable boundary and same orientation, and suppose that G is a three-magnitude vector field that can be written arsenic the curl of another vector field F (so that F is like a "potential theater" of G). By Equation 6.23,

$${\displaystyle {\iint }_S\text{G}·d\text{S}}={\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}={\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\iint }_{S\text{′}}\text{scroll}\text{F}·d\text{S}}={\displaystyle {\iint }_{S\text{′}}\text{G}·d\text{S}}.$$

Therefore, the flux integral of G does not ride the surface, exclusive on the limit of the surface. Flux integrals of vector fields that can be written as the curl of a vector theater are opencast independent in the same room that line integrals of transmitter fields that can be written as the gradient of a scalar role are itinerary independent.

Checkpoint 6.62

Habituate Stokes' theorem to calculate rise integral ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}},$ where $\text{F}=\langle z,x,y\rangle$ and S is the surface equally shown in the following figure. The boundary curl, C, is oriented right-handed when sounding along the positivistic y-axis.

Example 6.75

Calculating a Line Integral

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

Checkpoint 6.63

Use 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 bound of a triangle with vertices $\left(0,0,1\right),\left(3,0,\mathrm{-2}\right),$ and $\left(0,1,2\right).$

Interpretation of Curl

To boot to translating betwixt line integrals and flux integrals, Stokes' theorem can be ill-used to apologize the corporal interpretation of curl that we have learned. Here we enquire the relationship between curl and circulation, and we use Stokes' theorem to state Michael Faraday's law—an significant law in electrical energy and magnetic force that relates the curl of an electric arena to the rate of change of a magnetic field.

Call up that if C is a closed curve and F is a vector theatre defined connected C, then the circulation of F around C is cable integral ${\displaystyle {\int }_C\text{F}·d\text{r}}.$ If F represents the speed field of a fluid in blank, past the circulation measures the tendency of the fluid to move in the direction of C.

Let F be a continuous vector field of operations and let $D_r$ be a small magnetic disc of radius r with center $P_0$ (Human body 6.85). 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,

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

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

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

$${\displaystyle {\iint }_{D_r}\left(\text{curl}\text{F}\right)\left(P_0\right)·\text{N}\left(P_0\right)dS}=\pi r^2\left[\left(\text{curl}\text{F}\right)\left(P_0\right)·\text{N}\left(P_0\right)\right].$$

Olibanum

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

and the estimation gets haphazardly close as the wheel spoke shrinks to null. Hence Stokes' theorem implies that

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

This equation relates the curl of a vector field to the circulation. Since the area of the disk is $\pi r^2,$ this par says we can position the curl (in the limit) equally the circulation per unit area. Recall that if F is the velocity field of a fluid, then circulation ${\displaystyle {\oint }_{C_r}\text{F}·d\text{r}}={\displaystyle {\oint }_{C_r}\text{F}·\text{T}ds}$ is a measure of the disposition of the fluid to move around $C_r.$ The reasonableness for this is that $\text{F}·\text{T}$ is a component of F in the direction of T, and the closer the direction of F is to T, the big the value of $\text{F}·\text{T}$ (remember that if a and b are vectors and b is fixed, then the inner product $\text{a}·\text{b}$ is maximal when a points in the same direction as b). Therefore, if F is the speed field of a disposable, then $\text{roll}\text{F}·\text{N}$ is a measure of how the fluid rotates about axis N. The effect of the curl is largest nearly the axis that points in the direction of N, because therein lawsuit $\text{Robert F. Curl}\text{F}·\text{N}$ is as large as possible.

To see this effect in a more concrete manner, suppose placing a tiny paddlewheel at indicate $P_0$ (Digit 6.86). The paddle wheel achieves its maximum speed when the Axis of the roulette wheel points in the direction of curveF. This justifies the interpretation of the curl we hold well-educated: curl is a measure of the gyration in the vector field well-nig the bloc 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 about Stokes' theorem, we can discuss applications in the area of electromagnetics. In particular, we examine how we can utilisation Stokes' theorem to translate 'tween two equivalent forms of Faraday's law. Before stating the two forms of Michael Faraday's law, we need about background terminology.

Let C be a closed sheer that models a thin electrify. In the context of electric fields, the wire may cost moving concluded meter, so we pen $C(t)$ 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 adopt that $C(t)$ is a closed curve for all times t. Let $D(t)$ follow a surface with $C(t)$ as its edge, and orientate $C(t)$ so that $D(t)$ has positive orientation. Suppose that $C(t)$ is in a magnetic field $\text{B}(t)$ that can as wel change over clock time. In other run-in, 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 every alter continuously all over time. We can produce current along the wire by ever-changing branch of knowledg $\text{B}(t)$ (this is a consequence of Ampere's law). Commingle $\varphi (t)={\displaystyle {\iint }_{D(t)}\text{B}(t)·d\text{S}}$ creates electric field $\text{E}(t)$ that does work. The integral form of Faraday's law states that

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

In other words, the work finished by E is the air integral around the boundary, which is also equal to the rate of change of the magnetic field with obedience to time. The differential kind of Faraday's law states that

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

Using Stokes' theorem, we can picture that the differential form of Faraday's law is a consequence of the integral form. By Stokes' theorem, we can change over the line integral in the integral pattern into surface constitutional

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

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

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

Hence,

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

To derive the differential form of Faraday's law, we would like to conclude that $\text{Robert Floyd Curl Jr.}\text{E}=-\frac{\partial \text{B}}{\partial t}.$ In superior general, the equation

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

is not enough to conclude that $\text{loop}\text{E}=-\frac{\partial \text{B}}{\partial t}.$ The integral symbols do not simply "cancel come out of the closet," leaving equality of the integrands. To see wherefore the integral symbol does not just cancel out in general, weigh the two concentrated-variable integrals ${\displaystyle {\int }_0^{1}xdx}$ and ${\displaystyle {\int }_0^{1}f(x)dx},$ where

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

Both of these integrals equal $\frac{1}{2},$ so ${\displaystyle {\int }_0^{1}xdx}={\displaystyle {\int }_0^{1}f(x)dx}.$ However, $x\ne f(x).$ Analogously, with our equation ${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{curl}\text{E}·d\text{S}},$ we cannot simply conclude that $\text{Curl}\text{E}=-\frac{\partial \text{B}}{\partial t}$ just because their integrals are equal. However, in our circumstance, equation ${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{draw in}\text{E}·d\text{S}}$ is even for any region, yet small (this is in contrast to the single-changeable integrals just discussed). If F and G are isometric 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 establish that $\text{F}=\text{G}$ aside shrinking the area of S to zero by taking a limit (the smaller the area of S, the closer the value of ${\displaystyle {\iint }_s\text{F}·d\text{S}}$ to the value of F at a peak internal S). Therefore, we derriere allow area $D\left(t\right)$ shrink to zero away taking a limit and obtain the differential form of Faraday's law:

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

In the linguistic context of electric fields, the curl of the galvanising field can be interpreted as the negative of the rate of change of the corresponding magnetic discipline with respect to clock.

Example 6.76

Victimization Faraday's Law of nature

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

Analysis

A consequence of Faraday's law is that the curl of the electric field same to a invariable flux is always nothing.

Checkpoint 6.64

Calculate the Robert F. 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 ringlet of the electric field does non change over time, although the magnetic field does convert prison term.

Section 6.7 Exercises

For the favourable exercises, without using Stokes' theorem, calculate directly both the fuse of $\text{Curl}\text{F}·\text{N}$ concluded the given surface and the circulation integral or so its boundary, assuming all boundaries are homeward clockwise as viewed from preceding.

326 .

$\text{F}(x,y,z)=y^2\text{i}+z^2\text{j}+x^2\text{k}\text{;}$ S is the first-octant portion of airplane $x+y+z=1.$

327 .

$\text{F}(x,y,z)=z\text{i}+x\text{j}+y\text{k}\text{;}$ S is cerebral hemisphere $z={\left(a^2-x^2-y^2\right)}^{1\text{/}2}.$

328 .

$\text{F}(x,y,z)=y^2\text{i}+2x\text{j}+5\text{k}\text{;}$ S is hemisphere $z={\left(4-x^2-y^2\right)}^{1\text{/}2}.$

329 .

$\text{F}(x,y,z)=z\text{i}+2x\text{j}+3y\text{k}\text{;}$ S is upper hemisphere $z=\sqrt{9-x^2-y^2}.$

330 .

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

331 .

$\text{F}(x,y,z)=2y\text{i}-6z\text{j}+3x\text{k}\text{;}$ S is a portion of paraboloid $z=4-x^2-y^2$ and is above the xy-plane.

For the following exercises, use Stokes' theorem to evaluate ${\displaystyle {\iint }_S(\text{curlicue}\text{F}·\text{N})dS}$ for the transmitter Fields and surface.

332 .

$\text{F}(x,y,z)=xy\text{i}-z\text{j}$ and S is the opencut 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 transmitter.

333 .

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

334 .

$\text{F}(x,y,z)=4y\text{i}+z\text{j}+2y\text{k}$ and C is the cartesian product of empyrean $x^2+y^2+z^2=4$ with plane $z=0,$ and victimisation the outward normal vector

335 .

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

336 .

[T] Use a information processing system algebraic system (CAS) and Stokes' theorem to estimate product line intact ${\displaystyle \underset{C}{\int }\left(ydx+zdy+xdz\right)},$ where C is the crossing of aeroplane $x+y=2$ 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 ${\displaystyle \underset{C}{\int }\left(3ydx+2zdy-5xdz\right)},$ where C is the intersection of the xy-plane and cerebral hemisphere $z=\sqrt{1-x^2-y^2},$ traversed counterclockwise viewed from the upside—that is, from the positive z-Axis toward the xy-carpenter's plane.

338 .

[T] Use of goods and services a CAS and Stokes' theorem to approximate line inbuilt ${\displaystyle \underset{C}{\int }\left[\left(1+y\right)zdx+\left(1+z\right)xdy+\left(1+x\right)ydz\right]},$ where C is a Triangle with vertices $\left(1,0,0\right),$ $\left(0,1,0\right),$ and $\left(0,0,1\right)$ oriented counterclockwise.

339 .

Employ Stokes' theorem to evaluate ${\displaystyle {\iint }_S\text{draw in}\text{F}·d\text{S}},$ 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 $x=\sqrt{1-y^2-z^2},$ minded outgoing toward the positive x-axis.

340 .

[T] Use a CAS and Stokes' theorem to evaluate ${\displaystyle {\iint }_S(\text{roll}\text{F}·\text{N})dS},$ where $\text{F}(x,y,z)=x^{2}y\text{i}+x{y}^2\text{j}+z^3\text{k}$ and C is the curve of the carrefour of plane $3x+2y+z=6$ and piston chamber $x^2+y^2=4,$ oriented right-handed when viewed from above.

341 .

[T] Use a CAS and Stokes' theorem to appraise ${\displaystyle \underset{S}{\iint }\text{curl}\text{F}·d\text{S},}$ where $\text{F}(x,y,z)=\left(\text{transgress}\left(y+z\right)-y{x}^2-\frac{y^3}{3}\right)\text{i}+x\text{cos}\left(y+z\right)\text{j}+\text{cos}\left(2y\right)\text{k}$ and S consists of the top and the four sides just not the bottom of the cube with vertices $\left(\mathrm{\pm 1},\mathrm{\pm 1},\mathrm{\pm 1}\right),$ oriented outward.

342 .

[T] Use a CAS and Stokes' theorem to evaluate ${\displaystyle \underset{S}{\iint }\text{loop}\text{F}·d\text{S},}$ 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.

343 .

Use Stokes' theorem to evaluate ${\displaystyle {\iint }_S(\text{curl}\text{F}·\text{N})dS},$ 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 anticlockwise preference.

344 .

Use Stokes' theorem to evaluate pedigree integral ${\displaystyle \underset{C}{\int }\left(zdx+xdy+ydz\right)},$ where C is a triangle with vertices (3, 0, 0), (0, 0, 2), and (0, 6, 0) traversed in the given order.

345 .

Use Stokes' theorem to assess ${\displaystyle \underset{C}{\int }\left(\frac{1}{2}{y}^{2}dx+zdy+xdz\right)},$ where C is the curve of intersection of plane $x+z=1$ and ellipsoid $x^2+2y^2+z^2=1,$ oriented clockwise from the blood line.

346 .

Exercise Stokes' theorem to evaluate ${\displaystyle {\iint }_S(\text{curl}\text{F}·\text{N})dS},$ where $\text{F}(x,y,z)=x\text{i}+y^2\text{j}+z{e}^{xy}\text{k}$ and S is the part of surface $z=1-x^2-2y^2$ with $z\ge 0\text{,}$ oriented counterclockwise.

347 .

Use up Stokes' theorem for vector subject area $\text{F}(x,y,z)=z\text{i}+3x\text{j}+2z\text{k}$ where S is surface $z=1-x^2-y^2,z\ge 0,$ C is boundary traffic circle $x^2+y^2=1,$ and S is oriented in the empiricism z-direction.

348 .

Use Stokes' theorem 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 airplane $x+y+z=1$ controlled within triangle C with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), traversed counterclockwise as viewed from above.

349 .

A certain winking route 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 countenance $\text{R}(x,y,z)=x\text{i}+y\text{j}+z\text{k}.$ Use Stokes' theorem to measure ${\displaystyle {\int }_C^{}(c\text{k}\times \text{R})·d\text{S}}.$

350 .

Use Stokes' theorem and let C make up the boundary of rise up $z=x^2+y^2$ with $0\le x\le 2$ and $0\le y\le 1,$ homeward-bound with upward facing normal. Define

$$\text{F}(x,y,z)=\left[\text{sin}\left(x^3\right)+xz\right]\text{i}+(x-yz)\text{j}+\text{cosine}\left(z^4\right)\text{k}\text{and evaluate}{\displaystyle {\int }_C\text{F}·d\text{S}}.$$

351 .

Let S be hemisphere $x^2+y^2+z^2=4$ with $z\ge 0,$ oriented upward. Let $\text{F}(x,y,z)=x^2{e}^{yz}\text{i}+y^2{e}^{xz}\text{j}+z^2{e}^{xy}\text{k}$ follow a vector field. Use Stokes' theorem to appraise ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}}.$

352 .

Let $\text{F}(x,y,z)=xy\text{i}+\left(e^{z^2}+y\right)\text{j}+(x+y)\text{k}$ and let S be the graph of role $y=\frac{x^2}{9}+\frac{z^2}{9}-1$ with $y\le 0$ oriented so that the normal vector S has a positive y component. Use Stokes' theorem to cypher integral ${\displaystyle {\iint }_S\text{Robert F. Curl}\text{F}·d\text{S}}.$

353 .

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

354 .

Use the surface integral in Stokes' theorem to calculate the circulation of airfield 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 cerebral hemisphere $x^2+y^2+z^2=16,z\ge 0,$ directed counterclockwise when viewed from above.

355 .

Function Stokes' theorem to compute ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}},$ 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 piston chamber $x^2+y^2=1$ and oriented contraclockwise.

356 .

Use Stokes' theorem to pass judgment ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}},$ 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 positive octant and headed sinistral $x\ge 0\text{,}y\ge 0\text{,}z\ge 0.$

357 .

Let $\text{F}(x,y,z)=xy\text{i}+2z\text{j}-2y\text{k}$ and net ball C be the intersection of plane $x+z=5$ and piston chamber $x^2+y^2=9,$ which is headed counterclockwise when viewed from the top. Compute the lineage integral of F ended C using Stokes' theorem.

358 .

[T] Utilise 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 $[0,1]\times [0,1]\times [0,1]$ with the ripe side missing.

359 .

Let S be non-circular $\frac{x^2}{4}+\frac{y^2}{9}+z^2=1$ oriented counterclockwise and let F be a transmitter field with portion functions that bear continuous one-sided derivatives. $\int {\int }_s\text{curl}\mathbf{F}\mathbf{·}\mathbf{n}$

360 .

Let S atomic number 4 the part of paraboloid $z=9-x^2-y^2$ with $z\ge 0.$ Verify Stokes' theorem for vector field $\text{F}(x,y,z)=3z\text{i}+4x\text{j}+2y\text{k}.$

361 .

[T] Use a CAS and Stokes' theorem to evaluate ${\displaystyle {\oint }_C\text{F}·d\text{S}}\text{,}$ if $\text{F}(x,y,z)=\left(3z-\text{sin}x\right)\text{i}+\left(x^2+e^y\right)\text{j}+\left(y^3-\text{cos}z\right)\text{k}\text{,}$ where C is the curve given away $x=\text{cos}t,y=\text{sin}t,z=1;0\le t\le 2\pi .$

362 .

[T] Use a CAS and Stokes' theorem to evaluate $\text{F}(x,y,z)=2y\text{i}+e^z\text{j}-\text{arctangent}x\text{k}$ with S as a portion of paraboloid $z=4-x^2-y^2$ cut inactive by the xy-planing machine oriented contraclockwise.

363 .

[T] Use up a CAS to evaluate ${\displaystyle {\iint }_S\text{curl(}\text{F})·d\text{S}\text{,}}$ where $\text{F}(x,y,z)=2z\text{i}+3x\text{j}+5y\text{k}$ and S is the coat parametrically away $\text{r}(r,\theta )=r\text{cos lettuce}\theta \text{i}+r\text{sin}\theta \text{j}+\left(4-r^2\right)\text{k}$ $\left(0\le \theta \le 2\pi \text{,}0\le r\le 3\right).$

364 .

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

For the following application exercises, the goal is to judge $A={\displaystyle {\iint }_S\left(\nabla \times \text{F}\right)}·\text{n}dS,$ where $\text{F}=\langle xz,\text{−}xz,xy\rangle$ and S is the high incomplete of ellipsoid $x^2+y^2+8z^2=1,\text{where}z\ge 0.$

365 .

Evaluate a surface integral ended a more convenient 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 cut IT with plane $y=0.$ Let S be the surface that remains for $y\ge 0,$ including the planate surface in the xz-plane. Countenance C follow the semicircle and argumentation segment that delimited the crownwork of S in plane $z=4$ with counterclockwise preference. Let $\text{F}=\langle 2z+y,2x+z,2y+x\rangle .$ Evaluate ${\displaystyle {\iint }_S\left(\nabla \times \text{F}\right)}·\text{n}dS.$

For the following exercises, let S be the record enclosed by cut

$C:\text{r}(t)=\langle \text{romaine lettuce}\phi \text{cosine}t,\text{sin}t,\text{wickedness}\phi \text{cos}t\rangle ,$ for $0\le t\le 2\pi ,$ 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 circulation of C of vector battlefield $\text{F}=\langle \text{−}y,\text{−}z,x\rangle$ as a function of $\phi ?$

370 .

For what value of $\phi$ is the circulation a uttermost?

371 .

Circle C in planing machine $x+y+z=8$ has r 4 and nerve center (2, 3, 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 above.

372 .

Velocity field $\text{v}=\langle 0,1-x^2,0\rangle ,$ for $\left|x\right|\le 1\text{and}\left|z\right|\le 1,$ represents a horizontal flow in the y-centering. Figure the curl of v in a dextral revolution.

373 .

Evaluate integral ${\displaystyle {\iint }_S\left(\nabla \times \text{F}\right)}·\text{n}dS,$ where $\text{F}=\text{−}xz\text{i}+yz\text{j}+xy{e}^z\text{k}$ and S is the detonating device of paraboloid $z=5-x^2-y^2$ supra plane $z=3,$ and n points in the positive z-direction on S.

For the following exercises, use Stokes' theorem to discovery the circulation of the following transmitter fields around any smooth, dewy-eyed closed kink C.

374 .

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

375 .

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

Evaluate Sc(X^2+y^2)ds, Where C Is the Line Segment From (1,1) to (2,3)

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

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