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Green’s Theorem
Apply the circulation form of Green’s theorem.
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:
\[{\int }_{C}\text{∇}f\cdot d\text{r}=f(\text{r}(b))-f(\text{r}(a)).\]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 \(C\) shown in . 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 \(C\).
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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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.
விடை தெரியப்படுத்து
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,
\[\begin{array}{ll}{\int }_{C}{x}^{2}ydx+(y-3)dy & ={∬}_{D}({Q}_{x}-{P}_{y})dA \\ & =\int {\int }_{D}\text{-}{x}^{2}dA={\int }_{1}^{5}{\int }_{1}^{4}\text{-}{x}^{2}dxdy \\ & ={\int }_{1}^{5}-21dy=-84.\end{array}\] -
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 \((2,0).\)
விடை தெரியப்படுத்து
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,
\[\begin{array}{ll}W & ={\int }_{C}(y+\text{sin}(x))dx+({e}^{y}-x)dy \\ & ={∬}_{D}({Q}_{x}-{P}_{y})dA={∬}_{D}-2dA \\ & =-2(\text{area}(D))=-2\pi ({2}^{2})=-8\pi .\end{array}\] -
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.
விடை தெரியப்படுத்து
\(\frac{45}{2}\)
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Calculate the area enclosed by ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\) ().
விடை தெரியப்படுத்து
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 \(\pi ab.\)
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Find the area of the region enclosed by the curve with parameterization \(\text{r}(t)=〈\text{sin}\ t\ \text{cos}\ t,\text{sin}\ t〉,0\le t\le \pi .\)
விடை தெரியப்படுத்து
\(\frac{2}{3}\)
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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.
விடை தெரியப்படுத்து
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 \(2\pi {r}^{2}.\)
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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.
விடை தெரியப்படுத்து
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
\[\begin{array}{ll}\text{-}{∬}_{D}(2x+1)dA & =\text{-}{\int }_{0}^{1}{\int }_{0}^{-3x+3}(2x+1)dydx \\ & =\text{-}{\int }_{0}^{1}(2x+1)(-3x+3)dx=\text{-}{\int }_{0}^{1}(-6{x}^{2}+3x+3)dx \\ & =\text{-}{[-2{x}^{3}+\frac{3{x}^{2}}{2}+3x]}_{0}^{1}=-\frac{5}{2}.\end{array}\] -
Calculate the flux of \(\text{F}(x,y)=〈{x}^{3},{y}^{3}〉\) across a unit circle oriented counterclockwise.
விடை தெரியப்படுத்து
\(\frac{3\pi }{2}\)
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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 ().
விடை தெரியப்படுத்து
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.
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Verify that rotation vector field \(\text{F}(x,y)=〈y,\text{-}x〉\) is source free, and find a stream function for F.
விடை தெரியப்படுத்து
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 \(\text{∇}g=〈x,y〉\) ().
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Find a stream function for vector field \(\text{F}(x,y)=〈x\ \text{sin}\ y,\text{cos}\ y〉.\)
விடை தெரியப்படுத்து
\(g(x,y)=\text{-}x\ \text{cos}\ y\)
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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.
விடை தெரியப்படுத்து
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, \(\text{∇}f=\text{F},\) 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.
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Is the function \(f(x,y)={e}^{x+5y}\) harmonic?
விடை தெரியப்படுத்து
No
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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 \(1\le \text{r}\le 2,\) \(0\le \theta \le 2\pi .\)
விடை தெரியப்படுத்து
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:
\[\begin{array}{ll}{\int }_{∂D}(\text{sin}\ x-\frac{{y}^{3}}{3})dx+(\frac{{x}^{3}}{3}+\text{sin}\ y)dy & ={∬}_{D}({Q}_{x}-{P}_{y})dA \\ & ={∬}_{D}({x}^{2}+{y}^{2})dA \\ & ={\int }_{0}^{2\pi }{\int }_{1}^{2}{r}^{3}drd\theta ={\int }_{0}^{2\pi }\frac{15}{4}d\theta \\ & =\frac{15\pi }{2}.\end{array}\] -
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 \({\int }_{C}\text{F}\cdot d\text{r}?\)
விடை தெரியப்படுத்து
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, \({\int }_{C}\text{F}\cdot d\text{r}=-2\pi .\)
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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 \(\text{F}(x,y)=〈{x}^{3},5x+{e}^{y}\text{sin}\ y〉.\)
விடை தெரியப்படுத்து
\(105\pi\)
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\({\int }_{C}^{}2xydx+(x+y)dy,\) 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
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\({\int }_{C}^{}2xydx+(x+y)dy,\) 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
விடை தெரியப்படுத்து
\({\int }_{C}^{}2xydx+(x+y)dy=\frac{32}{3}\)
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\({\int }_{C}^{}2\ \text{arctan}(\frac{y}{x})dx+\text{ln}({x}^{2}+{y}^{2})dy,\) where C is defined by \(x=4+2\ \text{cos}\ \theta ,y=4\ \text{sin}\ \theta\) oriented in the counterclockwise direction
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\({\int }_{C}^{}\text{sin}\ x\ \text{cos}\ ydx+(xy+\text{cos}\ x\ \text{sin}\ y)dy\text{,}\) where C is the boundary of the region lying between the graphs of \(y=x\) and \(y=\sqrt{x}\) oriented in the counterclockwise direction
விடை தெரியப்படுத்து
\({\int }_{C}^{}\text{sin}\ x\ \text{cos}\ ydx+(xy+\text{cos}\ x\ \text{sin}\ y)dy=\frac{1}{12}\)
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\({\int }_{C}^{}xydx+(x+y)dy,\) 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
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\({\int }_{C}(\text{-}ydx+xdy),\) 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)
விடை தெரியப்படுத்து
\({\int }_{C}(\text{-}ydx+xdy)=\pi\)
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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 \({\int }_{C}xydx+\sqrt{{y}^{2}+1}dy.\)
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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.
விடை தெரியப்படுத்து
\({\int }_{C}x{e}^{-2x}dx+({x}^{4}+2{x}^{2}{y}^{2})dy=0\)
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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.
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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.
விடை தெரியப்படுத்து
\({\int }_{C}{y}^{3}dx-{x}^{3}ydy=-20\pi\)
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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.
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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.
விடை தெரியப்படுத்து
\({\int }_{C}\text{-}{x}^{2}ydx+x{y}^{2}dy=8\pi\)
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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.
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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).
விடை தெரியப்படுத்து
\({\int }_{C}({x}^{2}+{y}^{2})dx+2xydy=0\)
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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 \(y=x-2.\)
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Find the area between ellipse \(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1\) and circle \({x}^{2}+{y}^{2}=25.\)
விடை தெரியப்படுத்து
\(A=19\pi\)
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Find the area of the region enclosed by parametric equation
\[p(\theta )=(\text{cos}(\theta )-{\text{cos}}^{2}(\theta ))\text{i}+(\text{sin}(\theta )-\text{cos}(\theta )\text{sin}(\theta ))\text{j}\ \text{for}\ 0\le \theta \le 2\pi .\] -
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 \(t\in [0,2\pi ].\)
விடை தெரியப்படுத்து
\(A=\frac{3\pi }{8}\)
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Find the area of a pentagon with vertices \((0,4),(4,1),(3,0),(-1,-1),\) and \((-2,2).\)
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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.
விடை தெரியப்படுத்து
\({\int }_{C+}({y}^{2}+{x}^{3})dx+{x}^{4}dy=0\)
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Use Green’s theorem to prove the area of a disk with radius \(a\) is \(A=\pi {a}^{2}.\)
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Use Green’s theorem to find the area of one loop of a four-leaf rose \(r=3\ \text{sin}\ 2\theta .\) (Hint: \(xdy-ydx={r}^{2}d\theta ).\)
விடை தெரியப்படுத்து
\(A=\frac{9\pi }{8}\)
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Use Green’s theorem to find the area under one arch of the cycloid given by parametric plane \(x=t-\text{sin}\ t,y=1-\text{cos}\ t,t\ge 0.\)
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Use Green’s theorem to find the area of the region enclosed by curve
\[\text{r}(t)={t}^{2}\text{i}+(\frac{{t}^{3}}{3}-t)\text{j}\text{,}\ -\sqrt{3}\le t\le \sqrt{3}.\]விடை தெரியப்படுத்து
\(A=\frac{8\sqrt{3}}{5}\)
Symbols used here
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
The usual name for an angle.
In either; in both; in A but not B.
Chance of A; chance of A given that B happened.
Inequalities that allow equality; < and > exclude it.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Green’s Theorem
- Apply the circulation form of Green’s theorem.
- Apply the flux form of Green’s theorem.
- Calculate circulation and flux on more general regions.
- The flux
- If
- There is a
- Explain why the total distance through which the wheel rolls the small motion just described is
- Show that
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
உங்களை முயற்சிக்கவும்
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
மேலும் Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems