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Green's theorem

In vector calculus, Green's theorem relates a line integral around a simple closed curve C to a double integral over the plane region D (surface in ) bounded by C.

Extending the Fundamental Theorem of Calculus

Recall that the Fundamental Theorem of Calculus says that

\[{\int }_{a}^{b}{F}^{'}(x)dx=F(b)-F(a).\]

As a geometric statement, this equation says that the integral over the region below the graph of \({F}^{'}(x)\) and above the line segment \([a,b]\) depends only on the value of F at the endpoints a and b of that segment. Since the numbers a and b are the boundary of the line segment \([a,b],\) the theorem says we can calculate integral \({\int }_{a}^{b}F\text{'}(x)dx\) based on information about the boundary of line segment \([a,b]\) (). The same idea is true of the Fundamental Theorem for Line Integrals:

\[{\int }_{C}∇f\cdot d\text{r}=f(\text{r}(b))-f(\text{r}(a)).\]

When we have a potential function (an “antiderivative”), we can calculate the line integral based solely on information about the boundary of curve C.

Green’s theorem takes this idea and extends it to calculating double integrals. Green’s theorem says that we can calculate a double integral over region D based solely on information about the boundary of D. Green’s theorem also says we can calculate a line integral over a simple closed curve C based solely on information about the region that C encloses. In particular, Green’s theorem connects a double integral over region D to a line integral around the boundary of D.

Circulation Form of Green’s Theorem

The first form of Green’s theorem that we examine is the circulation form. This form of the theorem relates the vector line integral over a simple, closed plane curve C to a double integral over the region enclosed by C. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.

Notice that Green’s theorem can be used only for a two-dimensional vector field F. If F is a three-dimensional field, then Green’s theorem does not apply. Since

\[{\int }_{C}Pdx+Qdy={\int }_{C}\text{F}\cdot \text{T}ds,\]

this version of Green’s theorem is sometimes referred to as the tangential form of Green’s theorem.

The proof of Green’s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that D is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which \(\text{F}=〈P,Q〉\) is conservative. In this case,

\[{\int }_{C}Pdx+Qdy=0\]

because the circulation is zero in conservative vector fields. By , F satisfies the cross-partial condition, so \({P}_{y}={Q}_{x}.\) Therefore,

\[{∬}_{D}({Q}_{x}-{P}_{y})dA={∬}_{D}0dA=0={\int }_{C}Pdx+Qdy,\]

which confirms Green’s theorem in the case of conservative vector fields.

Condensed: the full section is in OpenStax Calculus Volume 3.

Flux Form of Green’s Theorem

The circulation form of Green’s theorem relates a double integral over region D to line integral \({\int }_{C}\text{F}\cdot \text{T}ds,\) where C is the boundary of D. The flux form of Green’s theorem relates a double integral over region D to the flux across boundary C. The flux of a fluid across a curve can be difficult to calculate using the flux line integral. This form of Green’s theorem allows us to translate a difficult flux integral into a double integral that is often easier to calculate.

Because this form of Green’s theorem contains unit normal vector N, it is sometimes referred to as the normal form of Green’s theorem.

Condensed: the full section is in OpenStax Calculus Volume 3.

Green’s Theorem on General Regions

Green’s theorem, as stated, applies only to regions that are simply connected, that is, Green’s theorem as stated so far cannot handle regions with holes. Here, we extend Green’s theorem so that it does work on regions with finitely many holes ().

Before discussing extensions of Green’s theorem, we need to go over some terminology regarding the boundary of a region. Let D be a region and let C be a component of the boundary of D. We say that C is positively oriented if, as we walk along C in the direction of orientation, region D is always on our left. Therefore, the counterclockwise orientation of the boundary of a disk is a positive orientation, for example. Curve C is negatively oriented if, as we walk along C in the direction of orientation, region D is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation, for example.

Let D be a region with finitely many holes (so that D has finitely many boundary curves), and denote the boundary of D by \(∂D\) (). To extend Green’s theorem so it can handle D, we divide region D into two regions, \({D}_{1}\) and \({D}_{2}\) (with respective boundaries \(∂{D}_{1}\) and \(∂{D}_{2}),\) in such a way that \(D={D}_{1}\cup {D}_{2}\) and neither \({D}_{1}\) nor \({D}_{2}\) has any holes ().

Assume the boundary of D is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region \({D}_{1}\) and \({D}_{2}\) is positively oriented. If F is a vector field defined on D, then Green’s theorem says that

\[\begin{array}{ll}{\int }_{∂D}\text{F}\cdot d\text{r} & ={\int }_{∂{D}_{1}}\text{F}\cdot d\text{r}+{\int }_{∂{D}_{2}}\text{F}\cdot d\text{r} \\ & ={∬}_{{D}_{1}}{Q}_{x}-{P}_{y}dA+{∬}_{{D}_{2}}{Q}_{x}-{P}_{y}dA \\ & ={∬}_{D}({Q}_{x}-{P}_{y})dA.\end{array}\]

Therefore, Green’s theorem still works on a region with holes.

To see how this works in practice, consider annulus D in and suppose that \(\text{F}=〈P,Q〉\) is a vector field defined on this annulus. Region D has a hole, so it is not simply connected. Orient the outer circle of the annulus counterclockwise and the inner circle clockwise () so that, when we divide the region into \({D}_{1}\) and \({D}_{2},\) we are able to keep the region on our left as we walk along a path that traverses the boundary. Let \({D}_{1}\) be the upper half of the annulus and \({D}_{2}\) be the lower half. Neither of these regions has holes, so we have divided D into two simply connected regions.

\[{\int }_{∂D}\text{F}\cdot d\text{r}={∬}_{D}({Q}_{x}-{P}_{y})dA.\]

Condensed: the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • Green’s theorem relates the integral over a connected region to an integral over the boundary of the region. Green’s theorem is a version of the Fundamental Theorem of Calculus in one higher dimension.
  • Green’s Theorem comes in two forms: a circulation form and a flux form. In the circulation form, the integrand is \(\text{F}\cdot \text{T}.\) In the flux form, the integrand is \(\text{F}\cdot \text{N}.\)
  • Green’s theorem can be used to transform a difficult line integral into an easier double integral, or to transform a difficult double integral into an easier line integral.
  • A vector field is source free if it has a stream function. The flux of a source-free vector field across a closed curve is zero, just as the circulation of a conservative vector field across a closed curve is zero.

Key Equations

Green’s theorem, circulation form\({\int }_{C}Pdx+Qdy={∬}_{D}{Q}_{x}-{P}_{y}dA,\) where C is the boundary of D
Green’s theorem, flux form\({\int }_{C}\text{F}\cdot \text{N}ds={∬}_{D}{P}_{x}+{Q}_{y}dA\)
Green’s theorem, extended version\({\int }_{∂D}\text{F}\cdot d\text{r}={∬}_{D}{Q}_{x}-{P}_{y}dA\)

Green’s Theorem

For the following exercises, evaluate the line integrals by applying Green’s theorem.

For the following exercises, use Green’s theorem.

For the following exercises, use Green’s theorem to find the area.

For the following exercises, use Green’s theorem to calculate the work done by force F on a particle that is moving counterclockwise around closed path C.

Green's Theorem

This section relies on the calculation of circulation around a closed curve from and as well as the idea of circulation density that was described in .

Introduction

We know from that a vector field is path-independent if and only if the circulation around every closed curve in its domain is \(0\). It is probably not surprising that, given the multitude of names we have for path-independent vector fields, they are important vector fields that arise frequently. However, not every vector field is path-independent, and many times we will want to calculate the circulation around a closed curve in a vector field that is not path-independent. This section explores a connection between line integrals and double integrals that you may find surprising. It will be the first of three major theorems that connect types of integrals that seem very different on the surface.

Exploration

We will consider the vector field \(\vF = \langle 2y,3x^2 y\rangle\), which is defined on the entire \(xy\)-plane. Suppose that we want to calculate the circulation of \(\vF\) around the circle \(C\) of radius \(2\), centered at \((0,0)\), and oriented counterclockwise.

Verify that \(\vF\) is not path-independent by calculating the circulation of \(\vF\) around the circle \(C\) (Use ). The SageMath cell below is set up to assist you with this, but you will need to supply a parametrization of \(C\) on line 4.

Recall from that if \(\vF = \langle F_1(x,y),F_2(x,y)\rangle\) is a vector field, then the circulation density (in 2D) is given by \[\begin{aligned}\end{aligned}\] What is the circulation density of \(\vF = \langle 2y,3x^2 y\rangle\)?

Sketch the curve \(C\) and shade the region it bounds. Describe the region bounded by \(C\) in both rectangular and polar coordinates.

Calculate the double integral of \(\displaystyle \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\) over the region inside the circle \(C\). This integral is not the most fun to do by hand, so a SageMath cell has been provided to assist you.

What do you notice about your results to parts (a) and (d)? Do you think this will happen in general? Write a sentence or two about what you think is happening here.

A natural question after completing the Preview Activity is if there is something special about the vector field \(\vF\) or the curve \(C\) that led to the results you obtained, and investigating this will be our principal task in this section.

Circulation

In , you integrated the circulation density of a smooth vector field over a disk and found the result was equal to the circulation of the vector field along the region's circular boundary. This relationship is the theme of this section. To see why this would make sense, consider the region \(R\) bounded by the curve \(C\) We have placed a square grid inside \(R\) to suggest the idea of breaking \(R\) up into many smaller regions, most of which are square. This idea should make you think of the methods we have already seen of breaking up a region into smaller and smaller regions for Riemann sums.

The critical idea here is that if we integrate the circulation density over each of the small regions and add those up, this is the same as integrating the circulation density over the entirety of the region \(R\) because of the fundamental properties of integrals. Also, integrating circulation per unit area over a two-dimensional region should be related to the total circulation on that region. Now look at and think of these two square regions as being two of the square regions inside \(R\) in .

We orient the boundary \(C_i\) of square region \(R_i\) in the manner suggested by the circular arrows. This means that the vertical boundary in common between \(R_1\) and \(R_2\) is oriented up when we calculate the line integral \(\oint_{C_1}\vF\cdot d\vr\) and oriented down when we evaluate \(\oint_{C_2}\vF\cdot d\vr\). Thus, this line segment does not contribute to the sum \(\oint_{C_1}\vF\cdot d\vr + \oint_{C_2}\vF\cdot d\vr\). Therefore, \[\begin{aligned}\end{aligned}\] is equal to the line integral of \(\vF\) along the boundary of the large rectangle.

Returning to , if we find the circulation along the boundary of each of the smaller regions, the line integrals along the boundaries that lie inside the region \(R\) will all offset. Thus the value should equal \(\oint_C\vF\cdot d\vr\). If we make our grid fine enough, all of the smaller regions into which \(R\) is divided will be very close to rectangular.

Condensed: the full section is in Boelkins et al., Active Calculus Multivariable.

Green's Theorem

So far in this section, we have restricted ourselves to relatively nice closed curves when thinking about circulation. While the main theorem of this section will not allow us to consider arbitrary closed curves, it does cover more varied curves than we have discussed so far. A simple closed curve is a closed curve that does not cross itself, and these are the curves to which our next theorem applies.

The restriction that the curve in prohibits curves such as the one below, which crosses itself.

At first glance, it may seem that is of purely intellectual interest. However, we have already encountered situations where parameterizing a curve can be complicated. This is particularly true when the curve has corners that require us to give separate parameterizations for several pieces of the curve, such as with the rectangular curve pictured in . However, as with this rectangle, is often the case that a curve that is difficult to parametrize bounds a region that is not too complex to describe using rectangular or polar coordinates. For instance, the rectangular region in can be described as \(1\leq x\leq 4\) and \(2\leq y\leq 4\), which is a simpler description than needing to parameterize each of the four sides of the rectangle separately. Additionally, the integrand of the double integral in involves partial derivatives that can sometimes result in an integrand that is easy to work with. The purpose of is, at its core, to allow you to exchange one type of integration problem (a line integral) for another type of integration problem (a double integral). This will be a recurring theme as this chapter continues.

Activity

For each of the curves described below, find the circulation of the given vector field around the curve. Do this both by calculating the line integral directly as well as by calculating the double integral from .

The curve \(C_1\) is the circle of radius \(3\) centered at the point \((2,1)\) (oriented counterclockwise) and the vector field is \(\vF = \langle y^2, 5x+2xy\rangle\).

The curve \(C_2\) is the triangle with vertices \((0,0)\), \((3,0)\), and \((3,3)\) (oriented counterclockwise) and the vector field is \(\vG = \langle y^2, 3xy\rangle\).

Condensed: the full section is in Boelkins et al., Active Calculus Multivariable.

What happens when vector fields are not smooth?

Notice that the assumptions in require that the region \(R\) be bounded by a simple closed curve \(C\) and that the vector field have continuous partial derivatives on \(R\) and \(C\). In , we will explore what happens when the region \(R\) cannot be bounded by a simple closed curve, as sometimes multiple applications of can be used in those circumstances. Now, however, we will take a look at what happens in some cases where the vector field \(\vF\) is not smooth.

Activity

Consider the vector field \(\vF = \displaystyle\frac{-y}{{x^2+y^2}}\vi + \frac{x}{{x^2+y^2}}\vj\). Notice that \(\vF\) is smooth everywhere in the plane other than at the point \((0,0)\). This vector field is plotted in , but we have not plotted the vectors close to the origin as their magnitudes get so large that they make it hard to interpret the figure. Here applies to any simple closed curve \(C\) that neither passes through \((0,0)\) nor bounds a region containing \((0,0)\).

Find the circulation density of \(\vF\) (i.e., the integrand of the double integral in ).

Suppose that \(C\) is the unit circle centered at the origin. Without doing any calculations, what can you say about \(\oint_C \vF\cdot d \vr\)? What does this tell you about if \(\vF\) is path-independent?

What would you get if you integrated the circulation density of \(\vF\) over the region bounded by \(C\)?

Do the previous two parts contradict ? Explain your reasoning.

Is the vector field \(\vG = \displaystyle\frac{x}{x^2+y^2}\vi +\frac{y}{x^2+y^2}\vj\), which is shown in , path-independent? Why or why not?

Suppose that \(C\) is the unit circle centered at the origin. Find \(\oint_C \vG\cdot d \vr\). Can you do this using ?

We can now see that is a powerful tool. However it cannot be used for all line integrals. In particular, the restriction that the vector field be smooth on the entire region bounded by a simple closed curve \(C\) creates limitations. This can occur even in cases where holes in the vector field's domain or points where the vector field is not continuously differentiable lie away from \(C\).

Green's theorem

In vector calculus, Green's theorem relates a line integral around a simple closed curve C to a double integral over the plane region D (surface in \(\R^2\)) bounded by C. It is the two-dimensional special case of Stokes' theorem (surface in \(\R^3\)). In one dimension, it is equivalent to the fundamental theorem of calculus. In two dimensions, it is equivalent to the divergence theorem.

It is named after mathematical physicist George Green.

Theorem

Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then

\[\oint_C (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) dA\]

where the path of integration along C is counterclockwise.

Application

Green's theorem in the plane relates line integrals around a simple closed curve to double integrals over the regions it encloses. It has two equivalent forms: the circulation form, which says the tangential line integral of a vector field \(\mathbf{F}=(P,Q)\) around a positively oriented simple closed curve \(C\) equals the double integral of the scalar curl \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\) over the region \(D\) bounded by \(C\); and the flux (divergence) form, which says that the normal line integral of \(\mathbf{F}\) around \(C\) equals the double integral of the divergence \(\nabla \cdot \mathbf{F}\) over \(D\). In applications, the circulation form is used for two-dimensional circulation and rotational flow calculations, while the flux form measures the net outflow across a closed boundary. Green's theorem also yields practical boundary-integral formulas for the area and centroid of a plane region.

Proof when D is a simple region

The following is a proof of half of the theorem for the simplified area \(D\), a type I region where \(C_1\) and \(C_3\) are curves connected by vertical lines (possibly of zero length). A similar proof exists for the other half of the theorem when \(D\) is a type II region where \(C_2\) and \(C_4\) are curves connected by horizontal lines (again, possibly of zero length). Putting these two parts together, the theorem is thus proven for regions of type III (defined as regions which are both type I and type II). The general case can then be deduced from this special case by decomposing \(D\) into a set of type III regions.

If it can be shown that

and

are true, then Green's theorem follows immediately for the region \(D\). We can prove (1) easily for regions of type I, and (2) for regions of type II. Green's theorem then follows for regions of type III.

Assume region \(D\) is a type I region and can thus be characterized, as pictured on the right, by \[D = \{(x,y)\mid a\le x\le b, g_1(x) \le y \le g_2(x)\}\] where \(g_1\) and \(g_2\) are continuous functions on \([a,b].\) Compute the double integral in (1):

Now compute the line integral in (1). \(C\) can be rewritten as the union of four curves: \(C_1,C_2,C_3,C_4.\)

With \(C_1\), use the parametric equations: \(x=x,y=g_1(x),a\leq x\leq b\). Then \[\int_{C_1} L(x,y)\, dx = \int_a^b L(x,g_1(x))\, dx.\]

Condensed: the full section is in Wikipedia.

Proof for rectifiable Jordan curves

We are going to prove the following

Theorem, Let \(\Gamma\) be a rectifiable, positively oriented Jordan curve in \(\R^2\) and let \(R\) denote its inner region. Suppose that \(A, B:\overline{R} \to \R\) are continuous functions with the property that \(A\) has second partial derivative at every point of \(R\), \(B\) has first partial derivative at every point of \(R\) and that the functions \(D_1 B, D_2 A : R \to \R\) are Riemann-integrable over \(R\). Then \[\int_{\Gamma}(A\,dx + B\,dy) = \int_R \left(D_1 B(x,y)-D_2 A(x,y)\right)\,d(x,y).\]

We need the following lemmas whose proofs can be found in:

Lemma 1 (Decomposition Lemma), Assume \(\Gamma\) is a rectifiable, positively oriented Jordan curve in the plane and let \(R\) be its inner region. For every positive real \(\delta\), let \(\mathcal{F}(\delta)\) denote the collection of squares in the plane bounded by the lines \(x=m\delta, y=m\delta\), where \(m\) runs through the set of integers. Then, for this \(\delta\), there exists a decomposition of \(\overline{R}\) into a finite number of non-overlapping subregions in such a manner that

  1. Each one of the subregions contained in \(R\), say \(R_1, R_2,\ldots, R_k\), is a square from \(\mathcal{F}(\delta)\).
  2. Each one of the remaining subregions, say \(R_{k+1},\ldots,R_s\), has as boundary a rectifiable Jordan curve formed by a finite number of arcs of \(\Gamma\) and parts of the sides of some square from \(\mathcal{F}(\delta)\).
  3. Each one of the border regions \(R_{k+1},\ldots,R_s\) can be enclosed in a square of edge-length \(2\delta\).
  4. If \(\Gamma_i\) is the positively oriented boundary curve of \(R_i\), then \(\Gamma=\Gamma_1+\Gamma_2+\cdots+\Gamma_s.\)
  5. The number \(s-k\) of border regions is no greater than \(4\!\left(\frac{\Lambda}{\delta} + 1\right)\), where \(\Lambda\) is the length of \(\Gamma\).

Lemma 2, Let \(\Gamma\) be a rectifiable curve in the plane and let \(\Delta_\Gamma(h)\) be the set of points in the plane whose distance from (the range of) \(\Gamma\) is at most \(h\). The outer Jordan content of this set satisfies \(\overline{c}\,\,\Delta_\Gamma(h)\le2h\Lambda +\pi h^2\).

Lemma 3, Let \(\Gamma\) be a rectifiable closed curve in \(\R^2\) and let \(f:\text{range of }\Gamma \to \R\) be a continuous function. Then \[\left\vert\int_\Gamma f(x,y)\,dy\right\vert \le\frac{1}{2} \Lambda\Omega_f,\] and \[\left\vert\int_\Gamma f(x,y)\,dx\right\vert \le\frac{1}{2} \Lambda\Omega_f,\] where \(\Omega_f\) is the oscillation of \(f\) on the range of \(\Gamma\).

Now we are in position to prove the theorem:

Proof of Theorem. Let \(\varepsilon\) be an arbitrary positive real number. By continuity of \(A\), \(B\) and compactness of \(\overline{R}\), given \(\varepsilon>0\), there exists \(0<\delta<1\) such that whenever two points of \(\overline{R}\) are less than \(2\sqrt{2}\,\delta\) apart, their images under \(A, B\) are less than \(\varepsilon\) apart. For this \(\delta\), consider the decomposition given by the previous Lemma. We have \[\int_\Gamma A\,dx+B\,dy=\sum_{i=1}^k \int_{\Gamma_i} A\,dx+B\,dy\quad +\sum_{i=k+1}^s \int_{\Gamma_i}A\,dx+B\,dy.\]

Put \(\varphi := D_1 B - D_2 A\).

For each \(i\in\{1,\ldots,k\}\), the curve \(\Gamma_i\) is a positively oriented square, for which Green's formula holds. Hence \[\sum_{i=1}^k \int_{\Gamma_i}A\,dx + B\,dy =\sum_{i=1}^k \int_{R_i} \varphi = \int_{\bigcup_{i=1}^k R_i}\,\varphi.\]

Condensed: the full section is in Wikipedia.

Validity under different hypotheses

The hypothesis of the last theorem are not the only ones under which Green's formula is true. Another common set of conditions is the following:

The functions \(A, B:\overline{R} \to \R\) are still assumed to be continuous. However, we now require them to be Fréchet-differentiable at every point of \(R\). This implies the existence of all directional derivatives, in particular \(D_{e_i}A=:D_i A, D_{e_i}B=:D_i B, \,i=1,2\), where, as usual, \((e_1,e_2)\) is the canonical ordered basis of \(\R^2\). In addition, we require the function \(D_1 B-D_2 A\) to be Riemann-integrable over \(R\).

As a corollary of this, we get the Cauchy Integral Theorem for rectifiable Jordan curves:

Theorem (Cauchy), If \(\Gamma\) is a rectifiable Jordan curve in \(\Complex\) and if \(f : \text{closure of inner region of } \Gamma \to \Complex\) is a continuous mapping holomorphic throughout the inner region of \(\Gamma\), then \[\int_\Gamma f=0,\] the integral being a complex contour integral.

Proof

We regard the complex plane as \(\R^2\). Now, define \(u,v:\overline{R} \to \R\) to be such that \(f(x + iy) = u(x, y) + iv(x, y).\) These functions are clearly continuous. It is well known that \(u\) and \(v\) are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations: \(D_1 v + D_2 u = D_1 u - D_2 v = \text{zero function}\).

Now, analyzing the sums used to define the complex contour integral in question, it is easy to realize that \[\int_\Gamma f=\int_\Gamma u\,dx-v\,dy\quad+i\int_\Gamma v\,dx+u\,dy,\] the integrals on the RHS being usual line integrals. These remarks allow us to apply Green's Theorem to each one of these line integrals, finishing the proof.

Multiply-connected regions

Theorem. Let \(\Gamma_0,\Gamma_1,\ldots,\Gamma_n\) be positively oriented rectifiable Jordan curves in \(\R^{2}\) satisfying \[\begin{aligned} \Gamma_i \subset R_0,&&\text{if } 1\le i\le n\\ \Gamma_i \subset \R^2 \setminus \overline{R}_j,&&\text{if }1\le i,j \le n\text{ and }i\ne j, \end{aligned}\] where \(R_i\) is the inner region of \(\Gamma_i\). Let \[D = R_0 \setminus (\overline{R}_1 \cup \overline{R}_2 \cup \cdots \cup \overline{R}_n).\]

Suppose \(p: \overline{D} \to \R\) and \(q: \overline{D} \to \R\) are continuous functions whose restriction to \(D\) is Fréchet-differentiable. If the function \[(x,y)\longmapsto\frac{\partial q}{\partial e_1}(x,y)-\frac{\partial p}{\partial e_2}(x,y)\] is Riemann-integrable over \(D\), then \[\begin{aligned} & \int_{\Gamma_0} p(x,y)\,dx+q(x,y)\,dy-\sum_{i=1}^n \int_{\Gamma_i} p(x,y)\,dx + q(x,y)\,dy \\[5pt] = {} & \int_D\left\{\frac{\partial q}{\partial e_1}(x,y)-\frac{\partial p}{\partial e_2}(x,y)\right\} \, d(x,y). \end{aligned}\]

Relationship to Stokes' theorem

Green's theorem is a special case of the Kelvin-Stokes theorem, when applied to a region in the \(xy\)-plane.

We can augment the two-dimensional field into a three-dimensional field with a z component that is always 0. Write F for the vector-valued function \(\mathbf{F}=(L,M,0)\). Start with the left side of Green's theorem: \[\oint_C (L\, dx + M\, dy) = \oint_C (L, M, 0) \cdot (dx, dy, dz) = \oint_C \mathbf{F} \cdot d\mathbf{r}.\]

The Kelvin-Stokes theorem: \[\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS.\]

The surface \(S\) is just the region in the plane \(D\), with the unit normal \(\mathbf{\hat n}\) defined (by convention) to have a positive z component in order to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes \[\nabla \times \mathbf{F} \cdot \mathbf{\hat n} = \left[ \left(\frac{\partial 0}{\partial y} - \frac{\partial M}{\partial z}\right) \mathbf{i} + \left(\frac{\partial L}{\partial z} - \frac{\partial 0}{\partial x}\right) \mathbf{j} + \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \mathbf{k} \right] \cdot \mathbf{k} = \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right).\]

Thus we get the right side of Green's theorem \[\iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \, dA.\]

Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives: \[\oint_C L \,dx + M \,dy = \oint_{\partial D} \! \omega = \int_D d\omega = \int_D \frac{\partial L}{\partial y} \,dy \wedge \,dx + \frac{\partial M}{\partial x} \,dx \wedge \,dy = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \,dx \,dy.\]

Relationship to the divergence theorem

Considering only two-dimensional vector fields, Green's theorem is equivalent to the two-dimensional version of the divergence theorem:

\(\iint_D\left(\nabla\cdot\mathbf{F}\right)dA=\oint_C \mathbf{F} \cdot \mathbf{\hat n} \, ds,\)

where \(\nabla\cdot\mathbf{F}\) is the divergence on the two-dimensional vector field \(\mathbf{F}\), and \(\mathbf{\hat n}\) is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal \(\mathbf{\hat n}\) in the right side of the equation. Since in Green's theorem \(d\mathbf{r} = (dx, dy)\) is a vector pointing tangential along the curve, and the curve C is the positively oriented (i.e. anticlockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right of this; one choice would be \((dy, -dx)\). The length of this vector is \(\sqrt{dx^2 + dy^2} = ds.\) So \((dy, -dx) = \mathbf{\hat n}\,ds.\)

Start with the left side of Green's theorem: \[\oint_C (L\, dx + M\, dy) = \oint_C (M, -L) \cdot (dy, -dx) = \oint_C (M, -L) \cdot \mathbf{\hat n}\,ds.\] Applying the two-dimensional divergence theorem with \(\mathbf{F} = (M, -L)\), we get the right side of Green's theorem: \[\oint_C (M, -L) \cdot \mathbf{\hat n}\,ds = \iint_D\left(\nabla \cdot (M, -L) \right) \, dA = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \, dA.\]

Area calculation

Green's theorem can be used to compute area by line integral. The area of a planar region \(D\) is given by\[A = \iint_D dA.\]

Choose \(L\) and \(M\) such that \(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1\). The area is given by\[A = \oint_{C} (L\, dx + M\, dy).\]

Possible formulas for the area of \(D\) include\[A=\oint_C x\, dy = -\oint_C y\, dx = \frac{1}{2} \oint_C (-y\, dx + x\, dy).\]

History

William Thomson (Lord Kelvin) named this theorem after George Green, who stated a similar result in an 1828 paper titled An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, which was almost forgotten until Thomson rediscovered it in 1845. In 1846, Augustin-Louis Cauchy published a paper stating Green's theorem as the penultimate sentence. A proof of the theorem was finally provided in 1851 by Bernhard Riemann in his doctoral dissertation.

Zvino iwe Hapana calculator inogadzirisa iyi imwe, asi mapepa ayo anogona kuongororwa. Tsananguro:

Kudzidza (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Calculate the line integral

    \[{\int }_{C}{x}^{2}ydx+(y-3)dy,\]

    where C is a rectangle with vertices \((1,1),\) \((4,1),\) \((4,5),\) and \((1,5)\) oriented counterclockwise.

    Kuratidza mhinduro

    Let \(\text{F}(x,y)=〈P(x,y),Q(x,y)〉=〈{x}^{2}y,y-3〉.\) Then, \({Q}_{x}=0\) and \({P}_{y}={x}^{2}.\) Therefore, \({Q}_{x}-{P}_{y}=\text{-}{x}^{2}.\)

    Let D be the rectangular region enclosed by C (). By Green’s theorem,

    \[\begin{array}{ll}{\int }_{C}{x}^{2}ydx+(y-3)dy & ={∬}_{D}({Q}_{x}-{P}_{y})dA \\ & =\int {\int }_{D}\text{-}{x}^{2}dA={\int }_{1}^{5}{\int }_{1}^{4}\text{-}{x}^{2}dxdy \\ & ={\int }_{1}^{5}-21dy=-84.\end{array}\]

  2. Calculate the work done on a particle by force field

    \[\text{F}(x,y)=〈y+\text{sin}\ x,{e}^{y}-x〉\]

    as the particle traverses circle \({x}^{2}+{y}^{2}=4\) exactly once in the counterclockwise direction, starting and ending at point \((2,0).\)

    Kuratidza mhinduro

    Let C denote the circle and let D be the disk enclosed by C. The work done on the particle is

    \[W={\int }_{C}(y+\text{sin}\ x)dx+({e}^{y}-x)dy.\]

    As with , this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green’s theorem ().

    Let \(\text{F}(x,y)=〈P(x,y),Q(x,y)〉=〈y+\text{sin}\ x,{e}^{y}-x〉.\) Then, \({Q}_{x}=-1\) and \({P}_{y}=1.\) Therefore, \({Q}_{x}-{P}_{y}=-2.\)

    By Green’s theorem,

    \[\begin{array}{ll}W & ={\int }_{C}(y+\text{sin}(x))dx+({e}^{y}-x)dy \\ & ={∬}_{D}({Q}_{x}-{P}_{y})dA={∬}_{D}-2dA \\ & =-2(\text{area}(D))=-2\pi ({2}^{2})=-8\pi .\end{array}\]

  3. Use Green’s theorem to calculate line integral

    \[{\int }_{C}\text{sin}({x}^{2})dx+(3x-y)dy,\]

    where C is a right triangle with vertices \((-1,2),\) \((4,2),\) and \((4,5)\) oriented counterclockwise.

    Kuratidza mhinduro

    \(\frac{45}{2}\)

  4. Calculate the area enclosed by ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\) ().

    Kuratidza mhinduro

    Let C denote the ellipse and let D be the region enclosed by C. Recall that ellipse C can be parameterized by

    \[x=a\ \text{cos}\ t,y=b\ \text{sin}\ t,0\le t\le 2\pi .\]

    Calculating the area of D is equivalent to computing double integral \({∬}_{D}dA.\) To calculate this integral without Green’s theorem, we would need to divide D into two regions: the region above the x-axis and the region below. The area of the ellipse is

    \[{\int }_{\text{-}a}^{a}{\int }_{0}^{\sqrt{{b}^{2}-{(bx\text{/}a)}^{2}}}dydx+{\int }_{\text{-}a}^{a}{\int }_{\text{-}\sqrt{{b}^{2}-{(bx\text{/}a)}^{2}}}^{0}\ dydx.\]

    These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green’s theorem to transform \({∬}_{D}dA\) into a line integral around the boundary C.

    Consider vector field

    \[\text{F}(x,y)=〈P,Q〉=〈-\frac{y}{2},\frac{x}{2}〉.\]

    Then, \({Q}_{x}=\frac{1}{2}\) and \({P}_{y}=-\frac{1}{2},\) and therefore \({Q}_{x}-{P}_{y}=1.\) Notice that F was chosen to have the property that \({Q}_{x}-{P}_{y}=1.\) Since this is the case, Green’s theorem transforms the line integral of F over C into the double integral of 1 over D.

    By Green’s theorem,

    \[\begin{array}{ll} \\ \\ \\ {∬}_{D}dA & ={∬}_{D}({Q}_{x}-{P}_{y})dA \\ & ={\int }_{C}\text{F}\cdot d\text{r}=\frac{1}{2}{\int }_{C}\text{-}ydx+xdy \\ & =\frac{1}{2}{\int }_{0}^{2\pi }\text{-}b\ \text{sin}\ t(\text{-}a\ \text{sin}\ t)+a(\text{cos}\ t)b\ \text{cos}\ tdt \\ & =\frac{1}{2}{\int }_{0}^{2\pi }ab\ {\text{cos}}^{2}t+ab\ {\text{sin}}^{2}tdt=\frac{1}{2}{\int }_{0}^{2\pi }abdt=\pi ab.\end{array}\]

    Therefore, the area of the ellipse is \(\pi ab.\)

  5. Find the area of the region enclosed by the curve with parameterization \(\text{r}(t)=〈\text{sin}\ t\ \text{cos}\ t,\text{sin}\ t〉,0\le t\le \pi .\)

    Kuratidza mhinduro

    \(\frac{2}{3}\)

  6. Let C be a circle of radius r centered at the origin () and let \(\text{F}(x,y)=〈x,y〉.\) Calculate the flux across C.

    Kuratidza mhinduro

    Let D be the disk enclosed by C. The flux across C is \({\int }_{C}\text{F}\cdot \text{N}ds.\) We could evaluate this integral using tools we have learned, but Green’s theorem makes the calculation much more simple. Let \(P(x,y)=x\) and \(Q(x,y)=y\) so that \(\text{F}=〈P,Q〉.\) Note that \({P}_{x}=1={Q}_{y},\) and therefore \({P}_{x}+{Q}_{y}=2.\) By Green’s theorem,

    \[{\int }_{C}\text{F}\cdot \text{N}ds=\int {\int }_{D}2dA=2\int {\int }_{D}dA.\]

    Since \(\int {\int }_{D}dA\) is the area of the circle, \(\int {\int }_{D}dA=\pi {r}^{2}.\) Therefore, the flux across C is \(2\pi {r}^{2}.\)

  7. Let S be the triangle with vertices \((0,0),\) \((1,0),\) and \((0,3)\) oriented clockwise (). Calculate the flux of \(\text{F}(x,y)=〈P(x,y),Q(x,y)〉=〈{x}^{2}+{e}^{y},x+y〉\) across S.

    Kuratidza mhinduro

    To calculate the flux without Green’s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green’s theorem to translate the flux line integral into a single double integral is much more simple.

    Let D be the region enclosed by S. Note that \({P}_{x}=2x\) and \({Q}_{y}=1;\) therefore, \({P}_{x}+{Q}_{y}=2x+1.\) Green’s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because \({\int }_{C}\text{F}\cdot \text{N}ds=\text{-}{\int }_{\text{-}S}\text{F}\cdot \text{N}ds\) and \(\text{-}S\) is oriented counterclockwise. By Green’s theorem, the flux is

    \[\begin{array}{ll} \\ \\ \\ {\int }_{C}\text{F}\cdot \text{N}ds & ={\int }_{\text{-}S}\text{F}\cdot \text{N}ds \\ & =\text{-}{∬}_{D}({P}_{x}+{Q}_{y})dA \\ & =\text{-}{∬}_{D}(2x+1)dA.\end{array}\]

    Notice that the top edge of the triangle is the line \(y=-3x+3.\) Therefore, in the iterated double integral, the y-values run from \(y=0\) to \(y=-3x+3,\) and we have

    \[\begin{array}{ll}\text{-}{∬}_{D}(2x+1)dA & =\text{-}{\int }_{0}^{1}{\int }_{0}^{-3x+3}(2x+1)dydx \\ & =\text{-}{\int }_{0}^{1}(2x+1)(-3x+3)dx=\text{-}{\int }_{0}^{1}(-6{x}^{2}+3x+3)dx \\ & =\text{-}{[-2{x}^{3}+\frac{3{x}^{2}}{2}+3x]}_{0}^{1}=-\frac{5}{2}.\end{array}\]

  8. Calculate the flux of \(\text{F}(x,y)=〈{x}^{3},{y}^{3}〉\) across a unit circle oriented counterclockwise.

    Kuratidza mhinduro

    \(\frac{3\pi }{2}\)

  9. Water flows from a spring located at the origin. The velocity of the water is modeled by vector field \(\text{v}(x,y)=〈5x+y,x+3y〉\) m/sec. Find the amount of water per second that flows across the rectangle with vertices \((-1,-2),(1,-2),(1,3),\ \text{and}\ (-1,3),\) oriented counterclockwise ().

    Kuratidza mhinduro

    Let C represent the given rectangle and let D be the rectangular region enclosed by C. To find the amount of water flowing across C, we calculate flux \({\int }_{C}\text{v}⋅\text{N}\text{ds}.\) Let \(P(x,y)=5x+y\) and \(Q(x,y)=x+3y\) so that \(\text{v}=(P,Q).\) Then, \({P}_{x}=5\) and \({Q}_{y}=3.\) By Green’s theorem,

    \[\begin{array}{ll}{\int }_{C}\text{v}⋅\text{N}\text{ds} & ={∬}_{D}({P}_{x}+{Q}_{y})dA \\ & ={∬}_{D}8dA \\ & =8(\text{area of}\ D)=80.\end{array}\]

    Therefore, the water flux is 80 m2/sec.

  10. Verify that rotation vector field \(\text{F}(x,y)=〈y,\text{-}x〉\) is source free, and find a stream function for F.

    Kuratidza mhinduro

    Note that the domain of F is all of \({ℝ}^{2},\) which is simply connected. Therefore, to show that F is source free, we can show any of items 1 through 4 from the previous list to be true. In this example, we show that item 4 is true. Let \(P(x,y)=y\) and \(Q(x,y)=\text{-}x.\) Then \({P}_{x}+{Q}_{y}=0+0=0.\) Thus, F is source free.

    To find a stream function for F, proceed in the same manner as finding a potential function for a conservative field. Let g be a stream function for F. Then \({g}_{y}=y,\) which implies that

    \[g(x,y)=\frac{{y}^{2}}{2}+h(x).\]

    Since \(\text{-}{g}_{x}=Q=\text{-}x,\) we have \(h\text{'}(x)=x.\) Therefore,

    \[h(x)=\frac{{x}^{2}}{2}+C.\]

    Letting \(C=0\) gives stream function

    \[g(x,y)=\frac{{x}^{2}}{2}+\frac{{y}^{2}}{2}.\]

    To confirm that g is a stream function for F, note that \({g}_{y}=y=P\) and \(\text{-}{g}_{x}=\text{-}x=Q.\)

    Notice that source-free rotation vector field \(\text{F}(x,y)=〈y,\text{-}x〉\) is perpendicular to conservative radial vector field \(∇g=〈x,y〉\) ().

  11. Find a stream function for vector field \(\text{F}(x,y)=〈x\ \text{sin}\ y,\text{cos}\ y〉.\)

    Kuratidza mhinduro

    \(g(x,y)=\text{-}x\ \text{cos}\ y\)

  12. For vector field \(\text{F}(x,y)=〈{e}^{x}\text{sin}\ y,{e}^{x}\text{cos}\ y〉,\) verify that the field is both conservative and source free, find a potential function for F, and verify that the potential function is harmonic.

    Kuratidza mhinduro

    Let \(P(x,y)={e}^{x}\text{sin}\ y\) and \(Q(x,y)={e}^{x}\text{cos}\ y.\) Notice that the domain of F is all of two-space, which is simply connected. Therefore, we can check the cross-partials of F to determine whether F is conservative. Note that \({P}_{y}={e}^{x}\text{cos}\ y={Q}_{x},\) so F is conservative. Since \({P}_{x}={e}^{x}\text{sin}\ y\) and \({Q}_{y}=-{e}^{x}\text{sin}\ y,{P}_{x}+{Q}_{y}=0\) and the field is source free.

    To find a potential function for F, let \(f\) be a potential function. Then, \(∇f=\text{F},\) so \({f}_{x}={e}^{x}\text{sin}\ y.\) Integrating this equation with respect to x gives \(f(x,y)={e}^{x}\text{sin}\ y+h(y).\) Since \({f}_{y}={e}^{x}\text{cos}\ y,\) differentiating \(f\) with respect to y gives \({e}^{x}\text{cos}\ y={e}^{x}\text{cos}\ y+h\text{'}(y).\) Therefore, we can take \(h(y)=0,\) and \(f(x,y)={e}^{x}\text{sin}\ y\) is a potential function for \(f.\)

    To verify that \(f\) is a harmonic function, note that \({f}_{xx}=\frac{∂}{∂x}({e}^{x}\text{sin}\ y)={e}^{x}\text{sin}\ y\) and

    \({f}_{yy}=\frac{∂}{∂x}({e}^{x}\text{cos}\ y)=\text{-}{e}^{x}\text{sin}\ y.\) Therefore, \({f}_{xx}+{f}_{yy}=0,\) and \(f\) satisfies Laplace’s equation.

  13. Is the function \(f(x,y)={e}^{x+5y}\) harmonic?

    Kuratidza mhinduro

    No

  14. Calculate integral

    \[{∮}_{∂D}(\text{sin}\ \text{x}-\frac{{y}^{3}}{3})dx+(\frac{{x}^{3}}{3}+\text{sin}\ \text{y})dy,\]

    where D is the annulus given by the polar inequalities \(1\le \text{r}\le 2,\) \(0\le \theta \le 2\pi .\)

    Kuratidza mhinduro

    Although D is not simply connected, we can use the extended form of Green’s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:

    \[\begin{array}{ll}{\int }_{∂D}(\text{sin}\ x-\frac{{y}^{3}}{3})dx+(\frac{{x}^{3}}{3}+\text{sin}\ y)dy & ={∬}_{D}({Q}_{x}-{P}_{y})dA \\ & ={∬}_{D}({x}^{2}+{y}^{2})dA \\ & ={\int }_{0}^{2\pi }{\int }_{1}^{2}{r}^{3}drd\theta ={\int }_{0}^{2\pi }\frac{15}{4}d\theta \\ & =\frac{15\pi }{2}.\end{array}\]

  15. Let \(\text{F}=〈P,Q〉=〈\frac{y}{{x}^{2}+{y}^{2}},-\frac{x}{{x}^{2}+{y}^{2}}〉\) and let C be any simple closed curve in a plane oriented counterclockwise. What are the possible values of \({\int }_{C}\text{F}\cdot d\text{r}?\)

    Kuratidza mhinduro

    We use the extended form of Green’s theorem to show that \({\int }_{C}\text{F}\cdot d\text{r}\) is either 0 or \(-2\pi\), that is, no matter how crazy curve C is, the line integral of F along C can have only one of two possible values. We consider two cases: the case when C encompasses the origin and the case when C does not encompass the origin.

    In this case, the region enclosed by C is simply connected because the only hole in the domain of F is at the origin. We showed in our discussion of cross-partials that F satisfies the cross-partial condition. If we restrict the domain of F just to C and the region it encloses, then F with this restricted domain is now defined on a simply connected domain. Since F satisfies the cross-partial property on its restricted domain, the field F is conservative on this simply connected region and hence the circulation \({\int }_{C}\text{F}\cdot d\text{r}\) is zero.

    In this case, the region enclosed by C is not simply connected because this region contains a hole at the origin. Let \({C}_{1}\) be a circle of radius a centered at the origin so that \({C}_{1}\) is entirely inside the region enclosed by C (). Give \({C}_{1}\) a clockwise orientation.

    Let D be the region between \({C}_{1}\) and C, and C is orientated counterclockwise. By the extended version of Green’s theorem,

    \[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r}+{\int }_{{C}_{1}}\text{F}\cdot d\text{r} & ={∬}_{D}{Q}_{x}-{P}_{y}dA \\ & ={∬}_{D}-\frac{{y}^{2}-{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}+\frac{{y}^{2}-{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}dA \\ & =0,\end{array}\]

    and therefore

    \[{\int }_{C}\text{F}\cdot d\text{r}=-{\int }_{{C}_{1}}\text{F}\cdot d\text{r}.\]

    Since \({C}_{1}\) is a specific curve, we can evaluate \({\int }_{{C}_{1}}\text{F}\cdot d\text{r}.\) Let

    \[x=a\ \text{cos}\ t,y=-a\ \text{sin}\ t,0\le t\le 2\pi\]

    be a parameterization of \({C}_{1}.\) Then,

    \[\begin{array}{ll}{\int }_{{C}_{1}}\text{F}\cdot d\text{r} & ={\int }_{0}^{2\pi }\text{F}(\text{r}(t))\cdot \text{r}\text{'}(t)dt \\ & ={\int }_{0}^{2\pi }〈-\frac{\text{sin}(t)}{a},-\frac{\text{cos}(t)}{a}〉\cdot 〈\text{-}a\ \text{sin}(t),\text{-}a\ \text{cos}(t)〉dt \\ & ={\int }_{0}^{2\pi }{\text{sin}}^{2}(t)+{\text{cos}}^{2}(t)dt={\int }_{0}^{2\pi }dt=2\pi .\end{array}\]

    Therefore, \({\int }_{C}\text{F}\cdot d\text{r}=-2\pi .\)

  16. Calculate integral \({\int }_{∂D}\text{F}\cdot d\text{r},\) where D is the annulus given by the polar inequalities \(2\le r\le 5,0\le \theta \le 2\pi ,\) and \(\text{F}(x,y)=〈{x}^{3},5x+{e}^{y}\text{sin}\ y〉.\)

    Kuratidza mhinduro

    \(105\pi\)

  17. \({\int }_{C}^{}2xydx+(x+y)dy,\) where C is the path from (0, 0) to (1, 1) along the graph of \(y={x}^{3}\) and from (1, 1) to (0, 0) along the graph of \(y=x\) oriented in the counterclockwise direction

  18. \({\int }_{C}^{}2xydx+(x+y)dy,\) where C is the boundary of the region lying between the graphs of \(y=0\) and \(y=4-{x}^{2}\) oriented in the counterclockwise direction

    Kuratidza mhinduro

    \({\int }_{C}^{}2xydx+(x+y)dy=\frac{32}{3}\)

  19. \({\int }_{C}^{}2\ \text{arctan}(\frac{y}{x})dx+\text{ln}({x}^{2}+{y}^{2})dy,\) where C is defined by \(x=4+2\ \text{cos}\ \theta ,y=4\ \text{sin}\ \theta\) oriented in the counterclockwise direction

  20. \({\int }_{C}^{}\text{sin}\ x\ \text{cos}\ ydx+(xy+\text{cos}\ x\ \text{sin}\ y)dy\text{,}\) where C is the boundary of the region lying between the graphs of \(y=x\) and \(y=\sqrt{x}\) oriented in the counterclockwise direction

    Kuratidza mhinduro

    \({\int }_{C}^{}\text{sin}\ x\ \text{cos}\ ydx+(xy+\text{cos}\ x\ \text{sin}\ y)dy=\frac{1}{12}\)

  21. \({\int }_{C}^{}xydx+(x+y)dy,\) where C is the boundary of the region lying between the graphs of \({x}^{2}+{y}^{2}=1\) and \({x}^{2}+{y}^{2}=9\) oriented in the counterclockwise direction

  22. \({\int }_{C}(\text{-}ydx+xdy),\) where C consists of line segment C1 from \((-1,0)\) to (1, 0), followed by the semicircular arc C2 from (1, 0) back to (–1, 0)

    Kuratidza mhinduro

    \({\int }_{C}(\text{-}ydx+xdy)=\pi\)

  23. Let C be the curve consisting of line segments from (0, 0) to (1, 1) to (0, 1) and back to (0, 0). Find the value of \({\int }_{C}xydx+\sqrt{{y}^{2}+1}dy.\)

  24. Evaluate line integral \({\int }_{C}x{e}^{-2x}dx+({x}^{4}+2{x}^{2}{y}^{2})dy,\) where C is the boundary of the region between circles \({x}^{2}+{y}^{2}=1\) and \({x}^{2}+{y}^{2}=4,\) and is a positively oriented curve.

    Kuratidza mhinduro

    \({\int }_{C}x{e}^{-2x}dx+({x}^{4}+2{x}^{2}{y}^{2})dy=0\)

  25. Find the counterclockwise circulation of field \(\text{F}(x,y)=xy\text{i}+{y}^{2}\text{j}\) around and over the boundary of the region enclosed by curves \(y={x}^{2}\) and \(y=x\) in the first quadrant and oriented in the counterclockwise direction.

  26. Evaluate \({\int }_{C}{y}^{3}dx-{x}^{3}{y}^{2}dy,\) where C is the positively oriented circle of radius 2 centered at the origin.

    Kuratidza mhinduro

    \({\int }_{C}{y}^{3}dx-{x}^{3}ydy=-20\pi\)

  27. Evaluate \({\int }_{C}{y}^{3}dx-{x}^{3}dy,\) where C includes the two circles of radius 2 and radius 1 centered at the origin, both with positive orientation.

  28. Calculate \({\int }_{C}\text{-}{x}^{2}ydx+x{y}^{2}dy,\) where C is a circle of radius 2 centered at the origin and oriented in the counterclockwise direction.

    Kuratidza mhinduro

    \({\int }_{C}\text{-}{x}^{2}ydx+x{y}^{2}dy=8\pi\)

  29. Calculate integral \({\int }_{C}2[y+x\ \text{sin}(y)]dx+[{x}^{2}\text{cos}(y)-3{y}^{2}]dy\) along triangle C with vertices (0, 0), (1, 0) and (1, 1), oriented counterclockwise, using Green’s theorem.

  30. Evaluate integral \({\int }_{C}({x}^{2}+{y}^{2})dx+2xydy,\) where C is the curve that follows parabola \(y={x}^{2}\ \text{from}\ (0,0)\text{ to }(2,4),\) then the line from (2, 4) to (2, 0), and finally the line from (2, 0) to (0, 0).

    Kuratidza mhinduro

    \({\int }_{C}({x}^{2}+{y}^{2})dx+2xydy=0\)

  31. Evaluate line integral \({\int }_{C}(y-\text{sin}(y)\text{cos}(y))dx+2x\ {\text{sin}}^{2}(y)dy,\) where C is oriented in a counterclockwise path around the region bounded by \(x=-1,x=2,y=4-{x}^{2},\) and \(y=x-2.\)

  32. Find the area between ellipse \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1\) and circle \({x}^{2}+{y}^{2}=25.\)

    Kuratidza mhinduro

    \(A=19\pi\)

  33. Find the area of the region enclosed by parametric equation

    \[p(\theta )=(\text{cos}(\theta )-{\text{cos}}^{2}(\theta ))\text{i}+(\text{sin}(\theta )-\text{cos}(\theta )\text{sin}(\theta ))\text{j}\ \text{for}\ 0\le \theta \le 2\pi .\]

  34. Find the area of the region bounded by hypocycloid \(\text{r}(t)={\text{cos}}^{3}(t)\text{i}+{\text{sin}}^{3}(t)\text{j}.\) The curve is parameterized by \(t\in [0,2\pi ].\)

    Kuratidza mhinduro

    \(A=\frac{3\pi }{8}\)

  35. Find the area of a pentagon with vertices \((0,4),(4,1),(3,0),(-1,-1),\) and \((-2,2).\)

  36. Use Green’s theorem to evaluate \({\int }_{{C}^{+}}({y}^{2}+{x}^{3})dx+{x}^{4}dy,\) where \({C}^{+}\) is the perimeter of square \([0,1]\ \times \ [0,1]\) oriented counterclockwise.

    Kuratidza mhinduro

    \({\int }_{C+}({y}^{2}+{x}^{3})dx+{x}^{4}dy=0\)

  37. Use Green’s theorem to prove the area of a disk with radius \(a\) is \(A=\pi {a}^{2}.\)

  38. Use Green’s theorem to find the area of one loop of a four-leaf rose \(r=3\ \text{sin}\ 2\theta .\) (Hint: \(xdy-ydx={r}^{2}d\theta ).\)

    Kuratidza mhinduro

    \(A=\frac{9\pi }{8}\)

  39. Use Green’s theorem to find the area under one arch of the cycloid given by parametric plane \(x=t-\text{sin}\ t,y=1-\text{cos}\ t,t\ge 0.\)

  40. Use Green’s theorem to find the area of the region enclosed by curve

    \[\text{r}(t)={t}^{2}\text{i}+(\frac{{t}^{3}}{3}-t)\text{j}\text{,}\ -\sqrt{3}\le t\le \sqrt{3}.\]
    Kuratidza mhinduro

    \(A=\frac{8\sqrt{3}}{5}\)

Tenga yako yemahara

Mubhadharo wakasununguka unowedzera mazita ezvidzidzo, zvinyorwa zvezvidzidzo zvawakati wazvigadzirira, zvidzidzo zvawave uchizvigadzirisa, uye mudzidzisi waungabvunza nezvechidzidzo ichi. Maths inowanikwa kune vese, vasiri vadzidzi kana vadzidzi. Kuwana mazita ezvidzidzo zvinodiwa kunogona kuitwa pawebsite ye Maths.

Kutanga Kupinda

Mazita anoshandiswa pano

Tinya chero chiratidzo kuti uone chirevo chakazara, vhidhiyo, uye kuti chii chinoreva chero chiratidzo.

Mabvunzo vanhu vanobvunza

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.

Zvimwe zvenyaya iyi zvakashandurwa kubva 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). Condensed uye re-akaratidza pano; zvikanganiso ndedu. Nekuti chigadzirwa ichi chine mvumo yekushandiswa pasina kutengeswa, ichi chidzidzo chine mukana wekuwana mari, hachina kutengesa, uye chinopihwa pasi pechibvumirano ichi.

More in Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems