Green's Theorem |
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TheoremIf $\vec F \left ( x,y \right ) = (P \left( x,y \right ), Q \left( x,y \right )) $ is a smooth vector field and S is a region which has boundary C, then $$\oint_{C} \vec P dx + \vec Q dy = \int \int_{S} (\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial x}) dydx$$ For a vector field $F:S\subseteq\mathbb{R}^3\to\mathbb{R}^3$ we have an extension of the Green's theorem known as the Stokes Theorem,
$$\oint_{C} \vec F. \vec dr = \int \int_{S} curl \vec F dS$$
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MotivationImagine rowing a boat along a closed irregular path, which forms a sort of boundary around a particular region of the water body.
Green's Theorem, one of the four main theorems that shape vector calculus, can help you answer these questions. The following segments are a brief introduction to the theorem.
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Bird's Eye ViewAn elementary idea underlying both single and multivariable calculus is that the sum of all tiny changes in a quantity is equal to the net change in that quantity. Recall the Fundamental Theorem of Calculus, which applies to single variable functions - $\int_{a}^{b} f \left ( x \right ) dx = F \left ( b \right ) - F\left ( a \right )$ The integral on the left-hand side, which describes a one-dimensional curve, is reduced to two zero-dimensional points on the right-hand side. What was defined by a curve is now dependent on its endpoints. Green's Theorem proposes a two dimensional analogue of the Fundamental Theorem of Calculus. It equates a two-dimensional region to a one-dimensional curve; the property of a region is examined by reducing it to the property of its boundary. 3 |
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Animation 1: Green's Theorem is a higher dimensional form the Fundamental Theorem of Calculus |
Context of the DefinitionGreen's theorem states that $ \oint_{C} \vec P dx + \vec Q dy = \int \int_{D} (\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial x}) dydx $ Let's begin from the left-hand side of the equality. $\oint \vec F \cdot \vec dr$$\oint \vec F \cdot \vec dr$ represents the line integral of curve C on the xy-plane. C is oriented counter-clockwise, and lies on the vector field $\vec F = P\hat i + Q\hat j$ of some function $f(x,y)$. Let's suppose you are moving along the curve. You use the tiny vector $\vec dr$ to determine the direction you are facing at any particular point. Upon determining your orientation at a point on the curve, you might ask how the vector field is oriented relative to you. Let's go back to the example described in the Motivation section, of rowing a boat in a river along a boundary. The boundary is represented by the curve C and the fluid flow represents the vector field. The relative orientation between you and the vector field indicates whether, at a point, the fluid is flowing in your direction or against you. This is the purpose of the dot product $\vec F \cdot \vec dr$. If the fluid flows with you, the dot product is positive, and conversely, if it is flowing against you, $\vec F \cdot \vec dr$ is negative. The dot product is defined at a point, however. In order to compute the relative orientation along the entire curve, we integrate the dot product between its endpoints. This is what the line integral $\oint \vec F \cdot \vec dr$ signifies. 6 |
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Animation: The line integral |
We're now familiar with what the line integral signifies. An important property of the line integral is the additive property, by which $\oint_{C} \vec F \cdot \vec dr = \oint_{C_{1}} \vec F \cdot \vec dr + \oint_{C_{2}} \vec F \cdot \vec dr$ when C is spilt into two curves which are both oriented the same way. Notice why this works - as is seen from the animation, the common parts of the smaller boundaries get cancelled out because we traverse them in opposite directions, leaving only the outer boundary (which is equivalent to the original boundary itself). We may exploit this property to divide the curve C into smaller and smaller boundaries $C_{1}, C_{2}, C_{3},...C_{n}$, such that the region within the boundary C, denoted D is divided into infinite boxes with infinitesimal area. The sum of the line integrals of these parts will equal the line integral of C taken as a whole. 8 |
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Animation 3: Splitting the boundary of the curve |
Dividing the boundary into several, much smaller sections allows us to use another equivalent formula to measure the circulation around the boundary - the curl. Consider a sub-section, of the region D, of area $D_{i}$ having the boundary $C_{i}$ . One way to measure the circulation around it is to take its line integral $\oint_{C_{i}} \vec F \cdot \vec dr$. However, since the area is infinitesimal, we may calculate its circulation by computing the 2D-curl of a point within the region (recall how this is similar to the approach we took to arrive at the formal definition of 2D curl using limits). Hence, $\oint_{C_{i}} \vec F \cdot \vec dr = (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}). \mid D_{i} \mid$ Accounting for each of the small boxes, we have $\sum_{i = 1}^{n}\oint_{C_{i}} \vec F \cdot \vec dr = \sum_{i = 1}^{n} (\frac{\partial Q}{\partial x}(x_i, y_i) - \frac{\partial P}{\partial y}(x_i, y_i)). \mid D_{i} \mid$ which is equivalent to $\oint_{C} \vec F \cdot \vec dr = \sum_{i = 1}^{n} (\frac{\partial Q}{\partial x}(x_i, y_i) - \frac{\partial P}{\partial y}(x_i, y_i)). \mid D_{i} \mid$ There is another popular method used to approximate the area of a region which has been divided into infinitesimally small pieces - the double integral! The left hand side of the equation transforms to $\int \int_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$ Consequently, our equation takes the form $\oint \vec F \cdot \vec dr = \int \int_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$ which is what Green's Theorem proposes!
This entire process is well summarised in the following statement - $\oint$ Macroscopic Curl = $\int \int$ (sum of all Microscopic Curl)
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Animation 5: The line integral of the macroscopic curl is equivalent to the double integral of all the microscopic curls |
Formal Proof for Green's TheoremLet $D = \left\{ \left ( x,y \right ) \mid a \leq x \leq b, f\left(x\right) \leq y \leq g\left(x\right) \right\} \textrm{ with } f\left(x\right), g\left(x\right), \text{continuous on} a \leq x \leq b$, and curve $ C = C_{1} + C_{2} \textrm{ where }$ $C_{1}: y = f\left(x\right) \forall a \leq x \leq b $, $C_{2}: y = g\left(x\right) \forall b \leq x \leq a $ Intergrating $\frac{\partial P}{\partial y}$ with repect to y between $ y = f\left(x\right) \textrm{ and }y= g\left(x\right)$, we have $\int_{f(x)}^{g(x)} \frac{\partial P}{\partial y} dy = P(x, g(x)) - P(x, f(x))$ which when integrated over the interval $(a, b)$ gives $\int_{a}^{b}\int_{f(x)}^{g(x)} \frac{\partial P}{\partial y} dy dx = \int_{a}^{b}(P(x, g(x)) - P(x, f(x))) dx$ $= \int_{a}^{b}P(x, g(x))dx - \int_{a}^{b}P(x, f(x)) dx$ $=- \int_{b}^{a}P(x, g(x))dx - \int_{a}^{b}P(x, f(x)) dx$ $ = - \int_{C_{2}} P dx - \int_{C_{1}} P dx$ $ = - \oint_{C} P dx$ Thus, $\oint_{C} P dx = \int \int_{D} (-\frac{\partial P}{\partial y}) dx dy$ 12 |
Similarly, for $D = \left\{ \left ( x,y \right ) \mid m\left(y\right) \leq x \leq n\left(y\right), c leq y \leq d, \right\} \textrm{ with } m\left(y\right), n\left(y\right), \text{continuous on} c \leq y \leq d$ and curve $ C' = C'_{1} + C'_{2} \textrm{ where }$ $C'_{1}: x = m\left(y\right) \forall d \leq x \leq c $, $C'_{2}: x = n\left(y\right) \forall c \leq x \leq d $, Intergrating $\frac{\partial Q}{\partial x}$ with repect to x between $ x = m\left(x\right) \textrm{ and }x= n\left(x\right)$, we have $\int_{m(y)}^{n(y)} \frac{\partial Q}{\partial x} dx = Q(n(y), y) - Q(m(y), y)$ which when integrated over the interval $(a, b)$ gives $\oint_{C'}Q dy $, Thus, $\oint_{C'} Q dy = \int \int_{D} \frac{\partial Q}{\partial x} dy dx$
$\oint_{C} \vec F dr = $ $\oint_{C} \vec P dx + \vec Q dy = \int \int_{D} \frac{\partial Q}{\partial y} - \frac{\partial P}{\partial x} dydx$
The divergence form of Green's TheoremUpon rotating the vector field $\vec F = (P, Q)$ by 90 degrees, we get $F' = (-Q, P)$ such that curl(F) = div(F'). then, $\int_{C} \vec F' \cdot \hat n ds = \int \int_{R} div \vec F' dxdy$
where the left hand side represents the flux integral through the region R bounded by C. $\hat n$ gives a unit normal vector orthogonal to C. 13 |
Example:
Using Green's Theorem, evaluate $\oint_{C} (x + y)dx - (x - y) dy$, where C is the ellipse of form $\frac{x^2}{a^2} + {y^2}{b^2} = 1 $ Solution: Green's theorem states that $\oint P dx + Q dy = \int \int_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dydx$ In this case, $P = x+y$ $Q =- (x-y)$ $\frac{\partial Q}{\partial x} = -1$, and $\frac{\partial P}{\partial y} = 1$ Therefore, $I = \oint_{C} (x+y)dx - (x-y)dy = \int \int_{D} (-1-1)dxdy = -2\int \int_{D} dA$ The double integral of $dA$ is nothing but the area of the ellipse, which is equal to $\pi ab$ Therefore, $I = -2\int \int_{D} dA$ $= -2 \pi ab$ 14 |
ApplicationsGreen's Theorem has several applications, some of which are given below -
Green's Theorem and Area ComputationThe planimeter is a mechanical device used to calculate the area of an arbitrary region by calculating the line integral of the region's boundary. It is used in fields ranging from medicine, to calculate the surface area of tumours, to engineering, to calculate the area of physical spaces. Green's Theorem can be thought of as a theoretical planimeter.
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Consider taking a look at how a planimeter works here. Calculation of area using Green's TheoremWe know that the area A of a region D bounded by curve C is given by $A = \int \int_{D} dA$ If we obtained $\int \int_{D} dA$ as our right-hand side of Green's Theorem, we would have to assume, according to the theorem that curl(F) = 1, or $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ In order to satisfy the above equation, our choices for P and Q, where $\vec F = (P, Q)$ are limited to $ P = 0, Q = x$, or $P = -y, Q = 0$ or $P = \frac{-y}{2}, Q = \frac{x}{2}$ Green's Theorem then allows us to compute the area of a region using the following line integrals $A = \oint_{C} x dy = - \oint_{C} y dx = \frac{1}{2} \oint_{C} x dy - y dx$
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Use Green's Theorem to find the area of a circle of radius a, bounded by curve C, parametrised as $x = a cos t$ $y = a sin t$ for $0 \leq t \leq 2\pi$ Solution: Using the formula $A = \frac{1}{2} \oint_{C} x dy - y dx$, $A = \frac{1}{2}( \int_{0}^{2\pi} a cos t (a cos t) dt - \int_{0}^{2\pi} a sin t (-a sin t) dt) $ $ = \frac{1}{2} \int_{0}^{2\pi} a^{2}cos^{2}dt + a^{2}sin^{2}dt $ $ = \frac{1}{2} \int_{0}^{2\pi} a^{2}dt$ $ = \pi a^{2}$
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HistoryGeorge Green was the son of a baker-tuned-miller. It is believed that he received only a year's worth of formal education. Despite this, he had a keen interest in mathematics and regularly read papers in mathematical journals. He taught himself French and took it upon himself to master English mathematics. He was a paying member of the private Nottingham Subscription Library, which gave him access to various foreign journals. In 1828, Green self-published his essay, An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism. At the time, his essay attracted only 51 subscribers. It was only later on, ten years after his death, that Lord Kelvin (then William Thomson) discovered his essay and brought to light Green's exceptional work. What stood out is George Green's approach to a problem in physics by focusing on its underlying mathematical structure, rather than its physical significance. In this essay, Green introduced what we know today as Green's Theorem and Green's Function.
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Pause and Ponder
Hint: Consider a vector field $\vec f (V, P) = (P, 0)$. Let the P-V diagram be the closed curve parametrised by time t, in the V-P plane. The line integral of the curve is associated with the area within the region bounded by the curve. In other words, the area within the region must be indicative of the work done by the gas.
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Further Reading
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References
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