[b][size=150]Test for conservative vector fields[/size][/b][br][br][u]Definition[/u]: Suppose a curve [math]C[/math] in [math]\mathbb{R}^2[/math] (or [math]\mathbb{R}^3[/math]) is parametrized by [math]\vec{r}(t)[/math] for [math]a\leq t \leq b[/math]. Then [math]C[/math] is a [b]simple curve[/b] if [math]\vec{r}(t_1)\ne\vec{r}(t_2)[/math] for any [math]t_1[/math] and [math]t_2[/math] with [math]a<t_1<t_2<b[/math] i.e. [math]C[/math] never intersects itself between its endpoints. [br][br][math]C[/math] is [b]closed[/b] if [math]\vec{r}(a)=\vec{r}(b)[/math] i.e. the initial and terminal points of [math]C[/math] are equal.[br]

[u]Definition[/u]: A open region [math]R[/math] in [math]\mathbb{R}^2[/math] (or [math]\mathbb{R}^3[/math]) is [b]connected[/b] if it is possible to connect any two points of [math]R[/math] by a continuous curve lying in [math]R[/math]. [br][br][math]R[/math] is [b]simply connected[/b] if every closed simple curve in [math]R[/math] can be deformed and contracted to a point in [math]R[/math].[br]

Recall that a vector field [math]\vec{F}[/math] is said to be [b]conservative[/b] on a region in [math]\mathbb{R}^2[/math] (or [math]\mathbb{R}^3[/math] if there exists a function [math]\phi[/math] such that [math]\vec{F}=\nabla \phi[/math]. And [math]\phi[/math] is called a [b]potential function[/b] of [math]\vec{F}[/math].[br][br]The following is the main theorem for determining whether a 2D vector field is conservative or not:[br][br][u]Theorem[/u]: Let [math]\vec{F}(x,y)=\langle f(x,y), g(x,y) \rangle[/math] be a 2D vector field defined on a connected and simply connected region [math]R[/math], where [math]f[/math] and [math]g[/math] have continuous first partial derivatives on [math]R[/math]. Then [math]\vec{F}[/math] is a conservative vector field on [math]R[/math] if and only if [math]\frac{\partial f}{\partial y}=\frac{\partial g}{\partial x}[/math].[br][br][u]Proof[/u]: Omitted.[br][br][br][u]Example[/u]: Suppose [math]\vec{F}(x,y)=\langle y+x, y-x \rangle[/math] on [math]\mathbb{R}^2[/math]. Determine whether [math]\vec{F}[/math] is conservative on [math]\mathbb{R}^2[/math].[br][br][u]Answer[/u]:[br][br][math]\mathbb{R}^2[/math] is obviously a connected and simply connected region.[br][br][math]\frac{\partial (y+x)}{\partial y}=1[/math] and [math]\frac{\partial (y-x)}{\partial x}=-1[/math].[br][br]They are not equal. Hence [math]\vec{F}[/math] is not conservative on [math]\mathbb{R}^2[/math].[br][br][br][u]Example[/u]: Suppose [math]\vec{F}(x,y)=\langle 2xy^3, 1+3x^2y^2 \rangle[/math] on [math]\mathbb{R}^2[/math]. Show that [math]\vec{F}[/math] is conservative on [math]\mathbb{R}^2[/math] and find its potential function [math]\phi[/math].[br][br][u]Answer[/u]:[br][br][math]\mathbb{R}^2[/math] is obviously a connected and simply connected region.[br][br][math]\frac{\partial (2xy^3)}{\partial y}=6xy^2[/math] and [math]\frac{\partial (1+3x^2y^2)}{\partial x}=6xy^2[/math].[br][br]They are equal. Hence [math]\vec{F}[/math] is conservative on [math]\mathbb{R}^2[/math].[br][br]Let [math]\phi[/math] be its potential function. Then we have[br][br][math]\nabla \phi = \left\langle \frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y}\right\rangle=\langle 2xy^3, 1+3x^2y^2 \rangle[/math][br][br]Hence [math]\frac{\partial\phi}{\partial x}=2xy^3[/math], which implies that[br][br][math]\phi(x,y)=\int 2xy^3 \ dx=x^2y^3 + k(y)[/math], where [math]k(y)[/math] is a function of [math]y[/math].[br][br]Since [math]\frac{\partial\phi}{\partial y}=1+3x^2y^2[/math], using the above result, we have[br][br][math]\frac{\partial\phi}{\partial y}=3x^2y^2+k'(y)=1+3x^2y^2[/math][br][math]\implies k'(y)=1 \implies k(y)=y+C[/math], where [math]C[/math] is an arbitrary constant.[br][br]Hence [math]\phi(x,y)=x^2y^3+y+C[/math].[br][br]

[b][size=150]Fundamental theorem for line integrals[/size][/b][br][br]The following theorem is a generalization of Fundamental theorem of calculus (FTC):[br][br][u]Theorem[/u] [b](Fundamental theorem for line integrals)[/b] Let [math]\vec{F}[/math] be a continuous vector field on an open connected region [math]R[/math] in [math]\mathbb{R}^2[/math] (or [math]\mathbb{R}^3[/math]). is conservative on i.e. there exists a potential function [math]\phi[/math] such that [math]\vec{F}=\nabla \phi[/math] if and only if [br][br][math]\int_C \vec{F}\cdot d\vec{r}=\phi(B)-\phi(A)[/math][br][br]for all points [math]A[/math] and [math]B[/math] in [math]R[/math] and all smooth oriented curves [math]C[/math] from [math]A[/math] to [math]B[/math].[br][br][br][u]Proof[/u]: Suppose [math]\vec{F}[/math] is conservative and [math]\phi[/math] is its potential function. Then [math]\vec{F}(x,y)=\left\langle \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}\right\rangle[/math]. Therefore, we have[br][br][math]\int_C\vec{F}\cdot d\vec{r}=\int_a^b\left(\frac{\partial\phi}{\partial x}\frac{dx}{dt}+\frac{\partial\phi}{\partial y}\frac{dy}{dt}\right) \ dt[/math][br][br]where [math]\vec{r}(t)=\langle x(t),y(t) \rangle[/math] for [math]a\leq t \leq b[/math] is a parametrization of the curve [math]C[/math], [math]\vec{r}(a)=\langle x(a),y(a)\rangle = A[/math], and [math]\vec{r}(b)=\langle x(b),y(b)\rangle = B[/math][br][br]By chain rule, [math]\frac{d}{dt}(\phi(x(t),y(t)))=\frac{\partial\phi}{\partial x}\frac{dx}{dt}+\frac{\partial\phi}{\partial y}\frac{dy}{dt}[/math]. Hence, we have[br][br][math]\int_C\vec{F}\cdot d\vec{r}=\int_a^b \frac{d}{dt}(\phi(x(t),y(t))) \ dt = \phi(x(b),y(b))-\phi(x(a),y(a))=\phi(B)-\phi(A)[/math][br][br]([u]Note[/u]: The second equality is due to FTC.)[br][br][br][u]Remark[/u]: [br][list=1][*]This theorem is analogous to FTC - [math]\int_a^b F'(x) \ dx=F(b)-F(a)[/math].[/*][*]The theorem implies that if [math]\vec{F}[/math] is conservative, [math]\int_C\vec{F}\cdot d\vec{r}[/math] is [b]path independent[/b] i.e. only the endpoints of [math]C[/math] matter.[/*][*]If [math]C[/math] is a simple closed smooth oriented curve i.e. [math]A=B[/math] and [math]\vec{F}[/math] is conservative, then [math]\int_C \vec{F}\cdot d\vec{r}=0[/math][/*][/list][br][br][br][br][u]Example[/u]: Let [math]\vec{F}(x,y)=\langle y, x \rangle[/math]. We already showed that [math]\vec{F}[/math] is conservative on [math]\mathbb{R}^2[/math]. Verify that its line integral over a curve [math]C[/math] from [math](0,0)[/math] to [math](1,1)[/math] by computing the line integral over [math]C_i[/math] ([math]i=1,2,3[/math]), where[br][br][math]C_1: \vec{r}_1(t)=\langle t,t \rangle[/math] for [math]0\leq t \leq 1[/math][br][math]C_2: \vec{r}_2(t)=\langle t,t^2 \rangle[/math] for [math]0\leq t \leq 1[/math][br][math]C_3: \vec{r}_3(t)=\langle t,t^3 \rangle[/math] for [math]0\leq t \leq 1[/math][br][br]Moreover, compute the value of the line integral using fundamental theorem for line integrals.[br][br][u]Answer[/u]:[br][br]Recall that [math]\vec{F}[/math] is conservative and its potential function [math]\phi(x,y)=xy[/math].[br][br][math]\int_{C_1}\vec{F}\cdot d\vec{r}=\int_0^1(t\cdot 1+t\cdot 1) \ dt =1[/math][br][math]\int_{C_2}\vec{F}\cdot d\vec{r}=\int_0^1(t^2\cdot 1+t\cdot (2t)) \ dt =1[/math][br][math]\int_{C_3}\vec{F}\cdot d\vec{r}=\int_0^1(t^3\cdot 1+t\cdot (3t^2)) \ dt =1[/math][br][br]By fundamental theorem for line integrals, for any smooth oriented curve [math]C[/math] from [math](0,0)[/math] to [math](1,1)[/math], we have[br][br][math]\int_C\vec{F}\cdot d\vec{r}=\phi(1,1)-\phi(0,0)=1\cdot 1 - 0\cdot 0 =1 [/math].[br]

The following theorem states the connections between conservative vector fields, line integrals over closed curves, and path independence:[br][br][u]Theorem[/u]: Let [math]\vec{F}[/math] be a 2D vector field defined on a connected region [math]R[/math]. The following statements are equivalent:[br][br][list=1][*][math]\vec{F}[/math] is conservative on [math]R[/math].[/*][br][*][math]\int_C\vec{F}\cdot d\vec{r}=0[/math] for any closed curve [math]C[/math] in [math]D[/math].[/*][br][*][math]\int_C\vec{F}\cdot d\vec{r}[/math] is path independent i.e. the line integral has the same value for any two points [math]A[/math] and [math]B[/math] and any curve from [math]A[/math] to [math]B[/math][/*][/list][br][br][u]Proof[/u]:[br][br][math](1)\implies (2)[/math]: By fundamental theorem for line integrals (see the remark).[br][br][math](2)\implies (3)[/math]: Suppose [math]C_1[/math] and [math]C_2[/math] are two curves from [math]A[/math] to [math]B[/math]. Consider the closed curve [math]C[/math] by connecting [math]C_1[/math] and [math]-C_2[/math] i.e. the curve starts at [math]A[/math], goes along [math]C_1[/math] to [math]B[/math], and then goes back to [math]A[/math] along [math]-C_2[/math]. Therefore, we have[br][br][math]0=\int_C\vec{F}\cdot d\vec{r}=\int_{C_1}\vec{F}\cdot d\vec{r}+\int_{-C_2}\vec{F}\cdot d\vec{r}=\int_{C_1}\vec{F}\cdot d\vec{r}-\int_{C_2}\vec{F}\cdot d\vec{r}[/math][br][math]\implies \int_{C_1}\vec{F}\cdot d\vec{r}=\int_{C_2}\vec{F}\cdot d\vec{r}[/math][br]Hence, [math]\int_C\vec{F}\cdot d\vec{r}[/math] is path independent.[br][br][math](3)\implies (1)[/math]: The rigorous proof is omitted. Here is the idea:[br][br]Fix a point [math](a,b)[/math] in [math]R[/math]. Define [math]\phi(x,y)[/math] to be the value of the line integral of [math]\vec{F}[/math] over a curve from [math](a,b)[/math] to [math](x,y)[/math]. As we assume the line integral is path independent, [math]\phi(x,y)[/math] is well-defined. It can be shown that [math]\phi[/math] is a potential function of [math]\vec{F}[/math] i.e. [math]\vec{F}=\nabla \phi[/math].[br][br][br][br][u]Remark[/u]: The above results still hold for 3D vector fields.

[b][size=150]Applications of line integrals[/size][/b][br][br][u]Work done in a force field[/u][br][br]Let [math]\vec{F}[/math] be a continuous force field in a region [math]D[/math] in [math]\mathbb{R}^2[/math] (or [math]\mathbb{R}^3[/math]). Suppose an object is moved by this force field along a curve [math]C[/math] in [math]D[/math] which is parametrized by [math]\vec{r}(t)[/math]. The [b]work done [math]W[/math] [/b] by [math]\vec{F}[/math] to move this object along the curve [math]C[/math] is [br][br][math]W=\int_C\vec{F}\cdot \vec{T} \ ds = \int_C \vec{F}\cdot d\vec{r}[/math][br][br]([u]Note[/u]: Recall that if a constant force [math]\vec{F}[/math] is applied to an object so that it moves along a straight line with displacement [math]\vec{d}[/math], the work done by the force is [math]\vec{F}\cdot \vec{d}[/math]. The above definition is a natural generalization to the case when the force is a variable force field and the object is moved along a curve instead of straight line.)[br][br][u]Example[/u]: How much work is required to move an object in the force field [math]\vec{F}(x,y,z)=\langle yz,xy,xz \rangle[/math] along the path [math]\vec{r}(t)=\langle t^2,t,t^4\rangle[/math] for [math]0\leq t \leq 1[/math]?[br][br][u]Answer[/u]:[br][br][math]\vec{r'}(t)=\langle 2t,1,4t^3\rangle[/math]. Then we have[br][br][math]W=\int_C\vec{F}\cdot d\vec{r}=\int_0^1\langle t\cdot t^4, t^2\cdot t, t^2\cdot t^4\rangle \cdot \langle 2t,1,4t^3\rangle \ dt[/math][br][math]=\int_0^1(2t^6+t^3+4t^9) \ dt[/math][br][math]=\left[\frac{2t^7}7+\frac{t^4}4+\frac{2t^9}5\right]_0^1=\frac{131}{140}[/math][br][br][br]If the force field [math]\vec{F}[/math] is a [b]conservative force field[/b] i.e. there exists a potential function [math]\phi[/math] such that [math]\vec{F}=\nabla \phi[/math], the work done by [math]\vec{F}[/math] on an object moving from point [math]A[/math] to point [math]B[/math] does not depend on the path taken by the object because of the above theorem. In fact, the work done is the difference between the potential function at point [math]A[/math] and point [math]B[/math]. The potential function [math]\phi[/math] is usually called the [b]potential energy[/b] of the force.[br][br][br][u]Example[/u]: Gravitational force between point masses obeys inverse square laws - the force acts along the line joining the centers of two masses and they vary as [math]\frac 1{r^2}[/math], where [math]r[/math] is the distance between the centers. The gravitational force of attraction is given by the vector field[br][br][math]\vec{F}(x,y,z)=\frac{k}{(x^2+y^2+z^2)^{\frac 32}}\langle x,y,z \rangle[/math][br][br]where [math]k<0[/math] is a physical constant. Find the work done in moving a mass from point [math](1,1,1)[/math] to [math](a,a,a)[/math] along the line segment joining them, where [math]a>1[/math]. Moreover, find the work done as [math]a\to \infty[/math].[br][br][br][u]Answer[/u]: We already knew that an inverse square field is a conservative field and its potential function (energy) is [br][br][math]\phi(x,y,z)=-\frac k{(x^2+y^2+z^2)^{\frac 12}}[/math][br][br]Therefore, the work done [math]W_a[/math] along any path [math]C[/math] from [math](1,1,1)[/math] to [math](a,a,a)[/math] is[br][br][math]W_a=\int_C \vec{F}\cdot d\vec{r}=\phi(a,a,a)-\phi(1,1,1)=-\frac k{a\sqrt{3}}+\frac k{\sqrt{3}}=\frac{k}{\sqrt{3}}\left(1-\frac 1a\right)[/math][br][br]As [math]a\to \infty[/math], [math]W_a \to \frac k{\sqrt{3}}[/math].[br][br]

[u]Flux and circulation of a 2D vector field[/u][br][br]Let [math]C[/math] be a simple curve in [math]\mathbb{R}^2[/math] and let [math]\vec{F}=\langle f, g\rangle[/math] be a 2D vector field, which models the velocity of a fluid flow across [math]C[/math]. We define the [b]flux[/b] of the fluid across [math]C[/math] to be the rate at which the fluid is crossing [math]C[/math] toward the right as we traverse [math]C[/math] in the positive direction (the "outward" direction), which can be defined as follows:[br][br][math]\text{Flux across }C=\int_C\vec{F}\cdot \vec{N} \ ds[/math][br][br]where [math]\vec{N}[/math] is the [b]outward unit normal vector[/b] along the curve [math]C[/math]. Suppose the curve [math]C[/math] is parametrized by [math]\vec{r}(t)=\langle x(t),y(t)\rangle [/math] for [math]a\leq t\leq b[/math]. Its tangent vector is [math]\vec{r'}(t)=\langle x'(t),y'(t)\rangle[/math] and hence [math]\vec{n}(t)=\langle y'(t),-x'(t)\rangle[/math] is orthogonal to [math]\vec{r'}(t)[/math] and pointing outward i.e. [math]\vec{n}(t)[/math] is an outward normal vector along [math]C[/math]. Hence, we have [math]\vec{N}(t)=\frac{\vec{n}(t)}{|\vec{n}(t)|}[/math] and the flux can be expressed as follows:[br][br][math]\int_C\vec{F}\cdot \vec{N} \ ds=\int_C\vec{F}\frac{\vec{n}(t)}{|\vec{n}(t)|} \ ds=\int_a^b \vec{F}(x(t),y(t))\cdot \frac{\vec{n}(t)}{|\vec{n}(t)|}|\vec{r'}(t)| \ dt[/math][br][math]=\int_a^b \vec{F}(x(t),y(t))\cdot \frac{\vec{n}(t)}{|\vec{n}(t)|}|\vec{n}(t)| \ dt=\int_a^b\vec{F}(x(t),y(t))\cdot \vec{n}(t) \ dt[/math][br][math]=\int_a^b (f(x(t),y(t))y'(t)-g(x(t),y(t))x'(t)) \ dt[/math][br][br][br][u]Example[/u]: Evaluate the flux of [math]\vec{F}(x,y)=\langle x,y \rangle[/math] across a unit circle oriented counterclockwise.[br][br][u]Answer[/u]:[br][br]Let [math]\vec{r}(t)=\langle \cos(t),\sin(t)\rangle[/math] for [math]0\leq t\leq 2\pi[/math] be the parametrization of [math]C[/math], the unit circle oriented counterclockwise. Then [math]\vec{n}(t)=\langle \cos(t),\sin(t)\rangle[/math]. Therefore, the flux is[br][br][math]\int_C\vec{F}\cdot \vec{N} \ ds=\int_0^{2\pi}\langle \cos(t),\sin(t)\rangle \cdot \langle \cos(t),\sin(t)\rangle \ dt[/math][br][math]=\int_0^{2\pi} (\cos^2(t)+\sin^2(t)) \ dt=\int_0^{2\pi} 1 \ dt=2\pi[/math][br][br](See the applet below for the visualization of the flux.)[br][br][br][br]The line integral of [math]\vec{F}[/math] along an oriented closed curve [math]C[/math], [br][br][math]\oint_C\vec{F}\cdot \vec{T} \ ds=\oint_C\vec{F}\cdot d\vec{r}[/math][br][br]is called the [b]circulation[/b] of [math]\vec{F}[/math] along [math]C[/math], which measures the tendency of the fluid to move in the direction of [math]C[/math].[br][br][u]Example[/u]: Find the circulation on the unit circle oriented counterclockwise for the following vector fields:[br](a) The radial flow [math]\vec{F}(x,y)=\langle x,y\rangle[/math][br](b) The rotation flow [math]\vec{F}(x,y)=\langle -y,x\rangle[/math][br][br][u]Answer[/u]:[br][br]Let [math]\vec{r}(t)=\langle \cos(t),\sin(t)\rangle[/math] for [math]0\leq t\leq 2\pi[/math] be the parametrization of [math]C[/math], the unit circle oriented counterclockwise. Then [math]\vec{r'}(t)=\langle -\sin(t),\cos(t)\rangle[/math].[br][br](a) [math]\oint_C\vec{F}\cdot d\vec{r}=\int_0^{2\pi} \langle \cos(t),\sin(t)\rangle \cdot \langle -\sin(t),\cos(t)\rangle \ dt=\int_0^{2\pi} 0 \ dt=0[/math][br][br](b) [math]\oint_C\vec{F}\cdot d\vec{r}=\int_0^{2\pi} \langle -\sin(t),\cos(t)\rangle \cdot \langle -\sin(t),\cos(t)\rangle \ dt=\int_0^{2\pi} (\sin^2(t)+\cos^2(t)) \ dt=\int_0^{2\pi} 1 \ dt=2\pi[/math][br][br](See the applet below for the visualization of circulation.)[br][br][br]