[b][size=150]Line integral of a function[/size][/b][br][br][u]Line integral of a function in 2D space[/u][br][br]Suppose [math]f(x,y)[/math] is a function defined on a domain in [math]\mathbb{R}^2[/math] and [math]C[/math] is a smooth curve in the domain with arc length parametrization [math]\vec{r}(s)=\langle x(s), y(s) \rangle[/math], where [math]0\leq s \leq L[/math] ([math]L[/math] is the arc length of [math]C[/math]). The curve is divided into [math]n[/math] parts such that [math]\Delta s_k[/math] is the arc length of the [math]k^{\text{th}}[/math] subdivision and [math]s_k^*[/math] is the The [b]line integral of [math]f[/math] over [math]C[/math][/b] is [br][br][math]\int_C f \ ds=\int_0^L f(x(s),y(s)) \ ds = \lim_{n\to\infty}\sum_{k=1}^n f(x(s_k^*),y(s_k^*)) \Delta s_k[/math][br][br]The line integral is the surface area of the "sheet" under the graph of [math]z=f(x,y)[/math] over the smooth curve [math]C[/math], as shown in the applet below.[br][br]

Suppose we are given a parametrization of the smooth curve [math]C: \vec{r}(t)=\langle x(t),y(t) \rangle[/math] for [math]a\leq t \leq b[/math], which is not necessary an arc length parametrization. We already know that for the arc length function [math]s(t)[/math] of the curve [math]C[/math], [math]\frac{ds}{dt}=|\vec{r'}(t)|[/math] i.e. [math]ds=|\vec{r'}(t)|dt[/math]. Therefore, we have[br][br][math]\int_C f \ ds=\int_a^b f(x(t),y(t))|\vec{r'}(t)| \ dt[/math][br][br]where [math]|\vec{r'}(t)|=\sqrt{(x'(t))^2+(y'(t))^2}[/math].[br][br][br][br][u]Example[/u]: Find the surface area of the cylinder [math]x^2+y^2=1[/math] between xy plane and the parabolic cylinder [math]z=1-x^2[/math].[br][br][u]Answer[/u]:[br][br]The required surface area is the surface area of the "sheet" under the parabolic cylinder over the circle [math]C: \ x^2+y^2=1[/math]. We can parametrize the circle by [math]\vec{r}(t)=\langle \cos(t), \sin(t) \rangle[/math] for [math]0\leq t \leq 2\pi[/math]. Then[br][br][math]\vec{r'}(t)=\langle -\sin(t), \cos(t) \rangle[/math][br][math]\implies |\vec{r'}(t)|=\sqrt{(-\sin(t))^2+\cos^2(t)}=1[/math][br][br]Therefore, the required surface area can be computed as follows:[br][br][math]\int_C (1-x^2) \ ds = \int_0^{2\pi}(1-\cos^2(t)) \cdot 1 \ dt[/math][br][math]=\int_0^{2\pi}\left(1-\frac{\cos(2t)+1}2\right) \ dt=\int_0^{2\pi}\frac{1-\cos(2t)}2 \ dt=\left[\frac t2-\frac 14\sin(2t)\right]_0^{2\pi}=\pi[/math][br][br]The applet below is the illustration of the surface.[br]

[u]Line integral of a function in 3D space[/u][br][br]Suppose [math]f(x,y,z)[/math] is a function defined on a domain in [math]\mathbb{R}^3[/math] containing a smooth curve [math]C: \vec{r}(t)=\langle x(t), y(t), z(t) \rangle[/math], for [math]a\leq t \leq b[/math]. Then the [b]line integral of [math]f[/math] over [math]C[/math][/b] is [br][br][math]\int_C f \ ds=\int_a^b f(x(t),y(t),z(t)) |\vec{r'}(t)| \ dt[/math][br][br]where [math]|\vec{r'}(t)|=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}[/math].[br][br][br][u]Example[/u]: Evaluate [math]\int_L (xy+2z) \ ds[/math], where [math]L[/math] is the line segment from [math](1,0,0)[/math] to [math](0,1,1)[/math].[br][br][u]Answer[/u]:[br][br]We first need to parametrize the line segment [math]L[/math]: Let [math]\vec{r}(t)=\langle x(t),y(t),z(t)\rangle=(1-t)\langle 1,0,0 \rangle +t \langle 0,1,1\rangle = \langle 1-t, t, t \rangle[/math], for [math]0\leq t \leq 1[/math]. Then [math]|\vec{r'}(t)|=|\langle -1,1,1 \rangle|=\sqrt{3}[/math]. Therefore, we have[br][br][math]\int_L (xy+2z) \ ds=\int_0^1 [(1-t)t+2t]\sqrt{3} \ dt=\int_0^1\sqrt{3}(3t-t^2) \ dt = \sqrt{3}\left(\frac 32-\frac 13\right)=\frac{7\sqrt{3}}6[/math].[br][br][u]Question[/u]: Will the value of the line integral change if the line integral is over the line segment from [math](0,1,1)[/math] to [math](1,0,0)[/math] ?[br]

[b][size=150]Line integral of a vector field[/size][/b][br][br]Let [math]\vec{F}(x,y)=\langle f(x,y), g(x,y) \rangle[/math] be a vector field in [math]\mathbb{R}^2[/math] defined on a domain containing a smooth oriented curve [math]C[/math] parametrized by arc length [math]\vec{r}(s)=\langle x(s),y(s) \rangle[/math]. Let [math]\vec{T}[/math] be the unit tangent vector at each point of [math]C[/math]. The [b]line integral of [math]\vec{F}[/math][/b] is[br][br][math]\int_C \vec{F}\cdot\vec{T} \ ds[/math][br][br]Consider any given parametrization [math]\vec{r}(t)[/math] of [math]C[/math] (not necessary an arc length parametrization) for [math]a\leq t \leq b[/math]. Since [math]\vec{T} \ ds=\underbrace{\frac{\vec{r'}(t)}{|\vec{r'}(t)|}}_{\vec{T}(t)} \underbrace{|\vec{r'}(t)| \ dt}_{ds}=\vec{r'}(t) \ dt[/math], we have the following[br][br][math]\int_C \vec{F}\cdot\vec{T} \ ds=\int_a^b \langle f(x(t),y(t)), g(x(t),y(t)) \rangle \cdot \langle x'(t),y'(t)\rangle \ dt=\int_a^b[f(x(t),y(t))x'(t)+g(x(t),y(t))y'(t)] \ dt[/math][br][br]Other notations for the above line integral: [math]\int_C \vec{F}\cdot\vec{T} \ ds=\int_C f \ dx+ g \ dy = \int_C \vec{F}\cdot d\vec{r}[/math][br][br][br][u]Note[/u]: In some textbooks, [math]\int_C f \ dx[/math] is called the [b]line integral of [math]f[/math] with respect to [math]x[/math][/b] and [br][br][math]\int_C f \ dx=\int_a^b f(x(t),y(t))x'(t) \ dt[/math][br][br]Similarly, [math]\int_C g \ dy[/math] is called the [b]line integral of [math]g[/math] with respect to [math]y[/math][/b] and [br][br][math]\int_C g \ dy=\int_a^b f(x(t),y(t))y'(t) \ dt[/math][br][br]and hence [math]\int_C f \ dx+ g \ dy=\int_a^b[f(x(t),y(t))x'(t)+g(x(t),y(t))y'(t)] \ dt[/math].[br][br][br][u]Example[/u]: Evaluate [math]\int_C yx^2 \ dx+2\sin(\pi y) \ dy[/math], where [math]C[/math] is the line segment from [math](0,2)[/math] to [math](1,4)[/math].[br][br][u]Answer[/u]:[br][br]First, we parametrize [math]C[/math]: [math]\vec{r}(t)=\langle x(t),y(t) \rangle=(1-t)\langle 0,2 \rangle+t\langle 1,4 \rangle=\langle t,2+2t \rangle[/math] for [math]0\leq t \leq 1[/math]. Then [math]\vec{r'}(t)=\langle x'(t),y'(t) \rangle = \langle 1, 2 \rangle[/math].[br][br][math]\int_C yx^2 \ dx+2\sin(\pi y) \ dy=\int_0^1 [(2+2t)(t)^2(1)+2\sin(\pi(2+2t))(2)] \ dt[/math][br][math]=\left[\left(\frac 23 t^3+\frac 12 t^4\right)-\frac 2{\pi}\cos(2\pi+2\pi t)\right]_0^1=\frac 76[/math].[br][br][br][br]We can define the line integral of a 3D vector field over a 3D smooth oriented curve in a similar way: Suppose [math]\vec{G}(x,y,z)=\langle u(x,y,z), v(x,y,z), w(x,y,z) \rangle[/math] be a 3D vector field and [math]C[/math] is a smooth oriented curve parametrized by [math]\vec{r}(t)=\langle x(t),y(t),z(t) \rangle[/math] for [math]a\leq t \leq b[/math]. Then the [b]line integral of [math]G[/math] over [math]C[/math] is[br][br][math]\int_C \vec{G}\cdot d\vec{r}=\int_C u \ dx+v \ dy + w \ dz=\int_a^b[u(x(t),y(t),z(t))x'(t)+v(x(t),y(t),z(t))y'(t)+w(x(t),y(t),z(t))z'(t)] \ dt[/math]

[b][size=150]Properties of line integrals[/size][/b][br][br][u]Reversing the orientation of a curve[/u][br][br]Let [math]C[/math] be a smooth curve in 2D (or 3D) space with a parametrization [math]\vec{r}(t)=\langle x(t), y(t) \rangle[/math] for [math]a\leq t \leq b[/math], which determines the orientation of the curve (the direction that is traced out as [math]t[/math] increases). We define [math]-C[/math] to be the curve with the same points as [math]C[/math], but the orientation is reversed. In fact, we can parametrize [math]-C[/math] as follows:[br][br][math]-C: \vec{r}_{-C}(t)=\vec{r}(a+b-t)[/math] for [math]a\leq t \leq b[/math].[br][br]As you can see, as [math]t[/math] increases from [math]a[/math] to [math]b[/math], the curve is traced from [math]\vec{r}(b)[/math] to [math]\vec{r}(a)[/math] i.e. in reverse direction.[br][br]For any function [math]f(x,y)[/math], its line integral over [math]-C[/math] is[br][br][math]\int_{-C} f(x,y) \ ds=\int_a^b f(\vec{r}_{-C}(t))|\vec{r'}_{-C}(t)| \ dt = \int_a^b f(\vec{r}(a+b-t))|-\vec{r'}(a+b-t)| \ dt[/math][br][br][math]=\int_a^b f(\vec{r}(u))|\vec{r'}(u)| \ du[/math] (use the substitution [math]u=a+b-t[/math])[br][math]=\int_{C} f(x,y) \ ds[/math][br][br]Therefore, reversing the direction of the curve does not change the value of the line integral of a function i.e. [math]\int_{-C} f(x,y) \ ds=\int_{C} f(x,y) \ ds[/math].[br][br][br]Now we consider any 2D vector field [math]\vec{F}(x,y)=\langle f(x,y),g(x,y) \rangle[/math]. its line integral over [math]-C[/math] is[br][br][math]\int_{-C} f \ dx+ g \ dy=\int_a^b [f(\vec{r}(a+b-t))(-x'(a+b-t))+g(\vec{r}(a+b-t))(-y'(a+b-t))] \ dt[/math][br][math]=-\int_a^b [f(\vec{r}(u))x'(u)+g(\vec{r}(u))y'(u)] \ du[/math] (use the substitution [math]u=a+b-t[/math])[br][math]=-\int_C f \ dx + g \ dy[/math][br][br]Therefore, we have [math]\int_{-C} f \ dx+ g \ dy=-\int_C f \ dx + g \ dy[/math].[br][br][br][u]Line integrals over piecewise smooth curves[/u][br][br]We can define a line integral of a vector field (or function) over a curve [math]C[/math] that is [b]piecewise smooth[/b] i.e. [math]C[/math] is a curve formed by connecting a finite number of smooth curves [math]C_1, C_2 , \ldots, C_n[/math] together. Suppose [math]\vec{F}[/math] is any 2D vector field. Then we have[br][br][math]\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}+ \cdots+\int_{C_n} \vec{F}\cdot d\vec{r}[/math][br][br][br][u]Example[/u]: Evaluate [math]\int_C x^2y \ dx + x \ dy[/math], where [math]C[/math] is the loop along the triangle from [math](0,0)[/math] to [math](1,0)[/math] to [math](1,2)[/math] and back to [math](0,0)[/math].[br][br][u]Answer[/u]:[br][br]Consider the following parametrization of [math]C[/math]:[br][br][math]C_1: \vec{r}_1(t)=\langle t,0 \rangle[/math] for [math]0\leq t \leq 1[/math] ([math](0,0)[/math] to [math](1,0)[/math])[br][math]C_2: \vec{r}_2(t)=(1-t)\langle 1,0\rangle+t\langle 1,2\rangle=\langle 1,2t \rangle[/math] for [math]0\leq t \leq 1[/math] ([math](1,0)[/math] to [math](1,2)[/math])[br][math]C_3: \vec{r}_3(t)=(1-t)\langle 1,2\rangle+t\langle 0,0\rangle=\langle 1-t,2-2t \rangle[/math] for [math]0\leq t \leq 1[/math] ([math](1,2)[/math] to [math](0,0)[/math])[br][br]Then we have[br][br][math]\int_C x^2y \ dx + x \ dy = \int_{C_1} x^2y \ dx + x \ dy + \int_{C_2} x^2y \ dx + x \ dy + \int_{C_3} x^2y \ dx + x \ dy[/math][br][math]=\int_0^1 (0\cdot 1+ t\cdot 0) \ dt + \int_0^1 (2t\cdot 0+1\cdot 2) \ dt + \int_0^1 ((1-t)^2(2-2t)\cdot (-1)+(1-t)\cdot (-2)) \ dt[/math][br][math]=0+2-\frac 32=\frac 12[/math][br][br]