The Fundamental Theorem of Line Integrals |
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DefinitionIf $C$ is a smooth curve given by the vector function $\vec r(t)$, $a\leq t\leq b$ and $f$ is a differentiable function whose gradient vector $\vec\nabla f$ is continuous on $C$. Then-
$$\int_C \vec\nabla f \cdot \mathrm{d}\vec r=f(\vec r(b))-f(\vec r(a))$$^{[1]} This is also known as "Gradient Theorem"^{ }^{[3]} 1 |
MotivationThis is a beautiful theorem which allows us to calculate the value of line integral without even doing the integral under some specific conditions. By this theorem line integral value along any curve over a vector field(not every) depends only on the endpoints of the curve viz. the line of integration. This indirectly implies that we can encode the information of a vector field (vectors associated with each point in space) into a scalar field(values associated with points in space) which is indeed spectacular^{[3]}. Example of such fields are gravitational field, electromagnetic field. 2 |
Bird's Eye ViewIn the the beginning of integral calculus one may find Fundamental Theorem of Calculus written as : If $f^\prime(x)$ is derivative of $f(x)$ continuous on $[a,b]$ then - The extension of that theorem to multiple variables is the current theorem if we think $\vec\nabla f$ as some sort of derivative of the multivariable function $f$.^{[1]} Although $f(x,y,z)$ is a scalar function $\vec\nabla f(x,y,z)$ is a vector valued function, For good illustration in 3-dimension function of two variables are used. 3 |
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Animation 1: Gradient of a scalar function $f(x,y)=xy$ is a vector function $\vec F=y\hat i+x\hat j$. |
So, we can restate the theorem as- Line integral in a vector field $\vec F$ can be evaluated as the difference of $f$-value at the final and initial points of the path of integration $\vec r(t)$. Remarks: We can calculate the value of the line integral in a vector field ($\vec F$) by this theorem if and only if, that vector field can be represented as the gradient of some scalar function ($f$). Such kind of vector fields are called conservative vector fields. 3 |
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Animation 2: Line integral is independent of path in the conservative field. |
Context of the DefinitionWe are going to see the proof of the theorem first, The parametric equation of the curve along which we want to do the line integral is smoothly defined as $\vec r(t)= x(t)\hat i+y(t)\hat j+z(t)\hat k,\, (a\leq t\leq b)$. and, $$\vec F(x,y,z)=\vec\nabla f(x,y,z)=\dfrac{\partial f}{\partial x}\hat i+\dfrac{\partial f}{\partial y}\hat j+\dfrac{\partial f}{\partial z}\hat k$$ \begin{align} The proof is based on the fact that the vector field $\vec F$ can be represented as a gradient($\vec\nabla$) of some scalar function $f$. 4 |
Test the field for conservativenssIf two points are fixed in the vector field and line integral along any path between the two points gives the same answer, the field has conservative nature. Example: 1. integral along the path $(0,0)\rightarrow (0,1)\rightarrow (1,1)$ 2. integral along the path $(0,0)\rightarrow (1,0)\rightarrow (1,1)$ The two answer is same indicating a conservative field. Furthermore we can deduce another theorem : If $C$ is a simple close curve and $\oint_C\vec F \cdot d\vec r = 0$ anywhere in the field $\vec F$ then it is a conservative field and there is a scalar potential $f$ associated with it. 4 |
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Animation 3: Line Integration of the previous example |
The most consistent way to prove a field is conservative is to see if the curl^{[5]} of the field is $0$. In the previous example observing the plot of the vector field ( in animation 3 ) you can also guess this answer. Find the scalar potential function if the field is conservative.An example will be best for it rather than saying too much. Example : Show that $\vec F =(2xy+z^3)\hat i+ x^2\hat j +3xz^2\hat k$ is conservative and find its scalar potential $f$.^{[6]} $$ \vec\nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ \dfrac{\partial }{{\partial x}}}&{ \dfrac{\partial }{{\partial y}}}&{ \dfrac{\partial }{{\partial z}}}\\2xy+z^3 & x^2 & 3xz^2 \end{array}} \right|=0$$ so $\vec F$ is conservative. now we have to find $f$ such that, $\vec\nabla f =\vec F$ or ,$$\dfrac{\partial f}{\partial x}\hat i+\dfrac{\partial f}{\partial y}\hat j+\dfrac{\partial f}{\partial y}\hat k=(2xy+z^3)\hat i+ x^2\hat j +3xz^2\hat k$$ Integrating (1),(2),(3) respectively , if we choose $f(y,z) = 0, g(x,z) = xz^3, h(x,y) = x^2y$ we get , $f=x^2y + xz^3 + k$, where $k$ is any arbitrary constant. 6 |
ApplicationsWe have already seen that if we are doing a line integral in a vector field, we can use this theorem to reduce a significant amount of work by introducing the scalar(potential) function associated with it. Then every integral in that field become merely finding values at the endpoints and subtract them. You have heard about gravitational potential, it is the scalar function related to the gravitational field (a force-vector field). You have used this theorem maybe without knowing it. Like if you are asked what amount of work is needed to lift a $15$ kg load of bricks to a roof of $10$ meter height against the gravitational field, you instantly calculate it using the $W=mgh$ formula, $15\times 9.8\times 10=1470\, Joule$ even without the information in which path you are going to do it i.e. using a pully or along the stairway. Generally, if you want to calculate the work done in a force field you need to do a line integral along the path from the initial point to the endpoint. But as gravitational field is a conservative field you are easily doing it using the difference of potential between the two points(note that $mgh$ is an approximated formula). Similar applications are found in electromagnetism to calculate work done on charges. 11 |
HistoryDuring the decade 1660–1670 Isaac Barrow, James Gregory, and Isaac Newton independently developed the fundamental theorem of calculus. Further development of mathematics in need of describing certain physical theories in a concise way lead to the formulation of Fundamental Theorem for Line Integrals. 11 |
Pause and PonderThis theorem is a really a "calculation saver"; from the next time when you are going to do a really involving line integral in a vector field, check if it is conservative viz. the $(\vec\nabla \times)$ is $0$, then the rest will 'boring' (obviously not the potential finding part !). The conservation formulas are the result of this potential formulation of the vector fields and that's why these fields are called conservative. 12 |
References
Further Reading1. For more detailed study:
2. A large collection of problem and solutions: Murray R. Spiegel. In Vector Analysis and an introduction to Tensor Analysis, Schaum’s Outline, chapter 5. Vector Integration. Line Integrals. McGraw-Hill,1959. 3. Theorems of vector calculus : https://www.math.ucla.edu/~josephbreen/The_Theorems_of_Vector_Calculus.pdf
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