All we do is evaluate the line integral over each of the pieces and then add them up. Next we need to talk about line integrals over piecewise smooth curves. In this section we are now going to introduce a new kind of integral. The terms path integral, curve integral, and curvilinear integral are also used; contour integral is used as well, although that is typically reserved for line integrals in the complex plane. This is red curve is the curve in which the line integral is performed. The fact tells us that this line integral should be the same as the second part (i.e. R C yds; C: x= t2;y= t;0 t 2 2. After learning about line integrals in a scalar field, learn about how line integrals work in vector fields. With line integrals we will start with integrating the function \(f\left( {x,y} \right)\), a function of two variables, and the values of \(x\) and \(y\) that we’re going to use will be the points, \(\left( {x,y} \right)\), that lie on a curve \(C\). Let’s start with the curve \(C\) that the points come from. Let’s first see what happens to the line integral if we change the path between these two points. Don’t forget to plug the parametric equations into the function as well. Green's theorem. Theorem: Line Integrals of Vector Valued Functions, \[\textbf{r}(t) = x(t) \hat{\textbf{i}} + y(t) \hat{\textbf{j}} \; \; \; \; a \leq t \leq b \], be a differentiable vector valued function that defines a smooth curve \(C\). Also notice that \({C_3} = - {C_2}\) and so by the fact above these two should give the same answer. Now let’s do the line integral over each of these curves. A breakdown of the steps: This definition is not very useful by itself for finding exact line integrals. We use a \(ds\) here to acknowledge the fact that we are moving along the curve, \(C\), instead of the \(x\)-axis (denoted by \(dx\)) or the \(y\)-axis (denoted by \(dy\)). This will happen on occasion. Finally, the line integral that we were asked to compute is. the line integral ∫ C(F ⋅τ)ds exists. You appear to be on a device with a "narrow" screen width (, \[\int\limits_{C}{{f\left( {x,y} \right)\,ds}} = \int_{{\,a}}^{{\,b}}{{f\left( {h\left( t \right),g\left( t \right)} \right)\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} \,dt}}\], \[\int\limits_{C}{{f\left( {x,y} \right)\,ds}} = \int_{{\,a}}^{{\,b}}{{f\left( {h\left( t \right),g\left( t \right)} \right)\,\,\left\| {\,\vec r'\left( t \right)} \right\|\,dt}}\], \[\int\limits_{C}{{f\left( {x,y} \right)\,ds}} = \int\limits_{{ - C}}{{f\left( {x,y} \right)\,ds}}\], \[\int\limits_{C}{{f\left( {x,y,z} \right)\,ds}} = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} \,dt}}\], \[\int\limits_{C}{{f\left( {x,y,z} \right)\,ds}} = \int_{{\,a}}^{{\,b}}{{f\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\,\,\left\| {\vec r'\left( t \right)} \right\|\,dt}}\], Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, \(\begin{array}{c} \displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \\ \mbox{(Ellipse)}\end{array}\), \(\begin{array}{c} \begin{array}{c}\mbox{Counter-Clockwise} \\x = a\cos \left( t \right)\\ y = b\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array}& \begin{array}{c} \mbox{Clockwise} \\x = a\cos \left( t \right)\\ y = - b\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array} \end{array}\), \(\begin{array}{c}{x^2} + {y^2} = {r^2} \\ \mbox{(Circle)}\end{array}\), \(\begin{array}{c} \begin{array}{c}\mbox{Counter-Clockwise} \\ x = r\cos \left( t \right)\\ y = r\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array} & \begin{array}{c} \mbox{Clockwise} \\ x = r\cos \left( t \right)\\ y = - r\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array} \end{array}\), \(\begin{align*}x & = t\\ y & = f\left( t \right)\end{align*}\), \(\begin{align*}x & = g\left( t \right)\\ y & = t\end{align*}\), \(\begin{array}{l}\mbox{Line Segment From} \\ \left( {{x_0},{y_0},{z_0}} \right) \mbox{ to} \\ \left( {{x_1},{y_1},{z_1}} \right) \end{array}\), \(\begin{array}{c} \vec r\left( t \right) = \left( {1 - t} \right)\left\langle {{x_0},{y_0},{z_0}} \right\rangle + t\left\langle {{x_1},{y_1},{z_1}} \right\rangle \,\,\,,\,\,0 \le t \le 1 \\ \mbox{or} \\ \begin{array}{l} \begin{aligned} x & = \left( {1 - t} \right){x_0} + t\,{x_1}\\ y & = \left( {1 - t} \right){y_0} + t\,{y_1}\\ z & = \left( {1 - t} \right){z_0} + t\,{z_1} \end{aligned} & , \,\,\,\,\,\, 0 \le t \le 1 \end{array} \end{array}\), \({C_1}:y = {x^2},\,\,\, - 1 \le x \le 1\). It required integration by parts. Fortunately, there is an easier way to find the line integral when the curve is given parametrically or as a vector valued function. That parameterization is. Also notice that, as with two-dimensional curves, we have. Note that this time we can’t use the second parameterization that we used in part (b) since we need to move from right to left as the parameter increases and the second parameterization used in the previous part will move in the opposite direction. Using this notation, the line integral becomes. The field is rotated in 3D to illustrate how the scalar field describes a surface. \nonumber \]. The curve is projected onto the plane \(XY\) (in gray), giving us the red curve, which is exactly the curve \(C\) as seen from above in the beginning. Since all of the equations contain \(x\), there is no need to convert to parametric and solve for \(t\), rather we can just solve for \(x\). We will assume that the curve is smooth (defined shortly) and is given by the parametric equations. x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π. Line integral. \nonumber\]. This will be a much easier parameterization to use so we will use this. note that from \( 0 \rightarrow \pi \;\) only \(\; -(a\: \sin(t)) \;\) exists. In mathematics, a line integral is an integral where the function to be integrated is evaluated along a curve. Then C has the parametric equations. Such an integral ∫ C(F⋅τ)ds is called the line integral of the vector field F along the curve C and is denoted as. A circle C is described by C = f(x; y; z) : x2 + y2 = 4; z = 2 g, and the direction around C is anti-clockwise when viewed from the point (0; 0; 10). Below is the definition in symbols. \[d(s)=\sqrt {\left ( \dfrac{dx}{dx} \right )^2+\left ( \dfrac{dy}{dx} \right )^2}dx \nonumber\], Next we convert the function into a function of \(x\) by substituting in \(y\), \[ f(x,y)=\dfrac{x^3}{y} \; \rightarrow \; f(x)=\dfrac{x^3}{\dfrac{x^2}{2}} \; \rightarrow \; f(x)= 2x. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "line integrals", "showtoc:no" ], \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 4.4: Conservative Vector Fields and Independence of Path, an equation of the function \(f(x)\) AKA \(y=\), an equation of the function \(f(x,y)\) AKA \(z=\), the equation of the path in parametric form \(( x(t),y(t) )\), the bounds in terms of \(t=a\) and \(t=b\). for \(0 \le t \le 1\). Let’s take a look at an example of a line integral. You were able to do that integral right? To this point in this section we’ve only looked at line integrals over a two-dimensional curve. There are several ways to compute the line integral $\int_C \mathbf{F}(x,y) \cdot d\mathbf{r}$: Direct parameterization; Fundamental theorem of line integrals To approximate the work done by F as P moves from ϕ(a) to ϕ(b) along C, we first divide I into m equal subintervals of length ∆t= b− a … This final view illustrates the line integral as the familiar integral of a function, whose value is the "signed area" between the X axis (the red curve, now a straight line) and the blue curve (which gives the value of the scalar field at each point). In fact, we will be using the two-dimensional version of this in this section. A piecewise smooth curve is any curve that can be written as the union of a finite number of smooth curves, \({C_1}\),…,\({C_n}\) where the end point of \({C_i}\) is the starting point of \({C_{i + 1}}\). Example 1. The curve \(C \) starts at \(a\) and ends at \(b\). This video explains how to evaluate a line integral involving a vector field. Find the line integral. The function to be integrated can be defined by either a scalar or a vector field, with … x = x (t), y = y (t). Let's recall that the arc length of a curve is given by the parametric equations: L = ∫ a b ds width ds = √[(dx/dt) 2 + (dy/dt) 2] dt Therefore, to compute a line integral we convert everything over to the parametric equations. R. 1 dA = C. x dy. Opposite directions create opposite signs when computing dot products, so traversing the circle in opposite directions will create line integrals … R C xy 4 ds; Cis the right half of the circle x2 + y2 = 16 3. By "normal integral" I take you to mean "integral along the x-axis". In this case the curve is given by, →r (t) =h(t) →i +g(t)→j a ≤ t ≤ b r → ( t) = h ( t) i → + g ( t) j → a ≤ t ≤ b. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. In a later section we will investigate this idea in more detail. We can do line integrals over three-dimensional curves as well. If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that. The curve is called smooth if \(\vec r'\left( t \right)\) is continuous and \(\vec r'\left( t \right) \ne 0\) for all \(t\). \]. This shows how at each point in the curve, a scalar value (the height) can be associated. The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, hence the line integral over a scalar field on C is the same irrespective of orientation. Then we can view A = A(x,y,z) as a vector valued function of the three variables (x,y,z). Let’s also suppose that the initial point on the curve is \(A\) and the final point on the curve is \(B\). Likewise from \( \pi \rightarrow 2\pi \;\) only \( -(-a\: \sin(t)) \) exists. \nonumber\]. We then have the following fact about line integrals with respect to arc length. Notice that work done by a force field on an object moving along a curve depends on the direction that the object goes. Note that this time, unlike the line integral we worked with in Examples 2, 3, and 4 we got the same value for the integral despite the fact that the path is different. \end{align*} \]. (Public Domain; Lucas V. Barbosa). Next, take the rate of change of the arc length (\(ds\)): \[\dfrac{dx}{dt}=1 \;\;\;\dfrac{dy}{dt}=\dfrac{2}{3} \nonumber\], \[ ds=\sqrt{\left (\dfrac{dx}{dt} \right )^2+\left (\dfrac{dy}{dt} \right )^2}dt=\sqrt{1^2+\left (\dfrac{2}{3} \right )^2}dt=\sqrt{13/9} \; dt=\dfrac{\sqrt{13}}{3}dt. There are two parameterizations that we could use here for this curve. Maybe the contribution from a segment is cancelled by a segment an angle π/2 away (I didn't think about it, I just made up a number), maybe it's not cancelled by any segment. \], \[\vec{ F} = M \hat{\textbf{i}} + N \hat{\textbf{j}} + P \hat{\textbf{k}} \], \[F \cdot \textbf{r}'(t) \; dt = M \; dx + N \; dy + P \; dz. The curve \(C\), in blue, is now shown along this surface. The line integral of an electric field along the circumference of a circle of radius r drawn with a point Q at the centre will be _____ A. The line integral is then: Example 1 . The length of the line can be determined by the sum of its arclengths, \[\lim_{n \to \infty }\sum_{i=1}^{n}\Delta_i =\int _a^b d(s)=\int_a^b\sqrt {\left ( \dfrac{dx}{dt} \right )^2+\left ( \dfrac{dy}{dt} \right )^2}dt\], note that the arc length can also be determined using the vector components \( s(t)=x(t)i+y(t)j+z(t)k \), \[ds= \left | \dfrac{ds}{dt} \right|=\sqrt {\left ( \dfrac{dx}{dt} \right )^2+\left ( \dfrac{dy}{dt} \right )^2+\left ( \dfrac{dz}{dt} \right )^2} dt =\left |\dfrac{dr}{dt}\right | dt\], so a line integral is sum of arclength multiplied by the value at that point, \[\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(c_i)\Delta s_i=\int_a^b f(x,y)ds=\int_a^b f(x(t),y(t))\sqrt {\left ( \dfrac{dx}{dt} \right )^2+\left ( \dfrac{dy}{dt} \right )^2}dt\]. Is smooth ( defined shortly ) and \ ( d ( s ) )... Regularly we can simplify the notation for the ellipse and the other counter-clockwise and a curve (... 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