line integral of a circle

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, \(\begin{array}{c} \displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \\ \mbox{(Ellipse)}\end{array}\), \(\begin{array}{c} \begin{array}{c}\mbox{Counter-Clockwise} \\x = a\cos \left( t \right)\\ y = b\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array}& \begin{array}{c} \mbox{Clockwise} \\x = a\cos \left( t \right)\\ y = - b\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array} \end{array}\), \(\begin{array}{c}{x^2} + {y^2} = {r^2} \\ \mbox{(Circle)}\end{array}\), \(\begin{array}{c} \begin{array}{c}\mbox{Counter-Clockwise} \\ x = r\cos \left( t \right)\\ y = r\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array} & \begin{array}{c} \mbox{Clockwise} \\ x = r\cos \left( t \right)\\ y = - r\sin \left( t \right)\\ 0 \le t \le 2\pi \end{array} \end{array}\), \(\begin{align*}x & = t\\ y & = f\left( t \right)\end{align*}\), \(\begin{align*}x & = g\left( t \right)\\ y & = t\end{align*}\), \(\begin{array}{l}\mbox{Line Segment From} \\ \left( {{x_0},{y_0},{z_0}} \right) \mbox{ to} \\ \left( {{x_1},{y_1},{z_1}} \right) \end{array}\), \(\begin{array}{c} \vec r\left( t \right) = \left( {1 - t} \right)\left\langle {{x_0},{y_0},{z_0}} \right\rangle + t\left\langle {{x_1},{y_1},{z_1}} \right\rangle \,\,\,,\,\,0 \le t \le 1 \\ \mbox{or} \\ \begin{array}{l} \begin{aligned} x & = \left( {1 - t} \right){x_0} + t\,{x_1}\\ y & = \left( {1 - t} \right){y_0} + t\,{y_1}\\ z & = \left( {1 - t} \right){z_0} + t\,{z_1} \end{aligned} & , \,\,\,\,\,\, 0 \le t \le 1 \end{array} \end{array}\), \({C_1}:y = {x^2},\,\,\, - 1 \le x \le 1\). Let’s also suppose that the initial point on the curve is \(A\) and the final point on the curve is \(B\). There are two parameterizations that we could use here for this curve. Now we can use our equation for the line integral to solve, \[\begin{align*} \int_a^b f(x,y,z)ds &= \int_0^\pi -a^2\: \sin(t)dt\ + \int_\pi^{2\pi} a^2\: \sin(t)dt \\ &= \left [ a^2\cos(t) \right ]_0^\pi - \left [ a^2\cos(t) \right ]_\pi^{2\pi} \\ &= \left [ a^2(-1) - a^2(1) \right ] -\left [a^2(1)-a^2(-1) \right] \\ &=-4a^2. We can do line integrals over three-dimensional curves as well. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Fundamental theorem of line integrals. 2 Line Integrals Section 4.3 F (j( ))t D j( )t j( )t P Figure 4.3.2 Object P moving along a curve Csubject to a forceC F parametrization ϕ : I → Rn, where I = [a,b]. We continue the study of such integrals, with particular attention to the case in which the curve is closed. The line integral ∫ CF ds exists if the function F is continuous on the curve C. Properties of Line Integrals of Scalar Functions If data is provided, then we can use it as a guide for an approximate answer. This will be a much easier parameterization to use so we will use this. \nonumber \], \[ \int_0^{2\pi} \left( \cos^2 t - t\, \sin t\right) \, dt \nonumber\], with a little bit of effort (using integration by parts) we solve this integral to get \( 3\pi \). So, first we need to parameterize each of the curves. Before working another example let’s formalize this idea up somewhat. The \(ds\) is the same for both the arc length integral and the notation for the line integral. \]. \end{align*} \]. Watch the recordings here on Youtube! The curve is called smooth if →r ′(t) r → ′ ( t) is continuous and →r ′(t) ≠ 0 r → ′ ( t) ≠ 0 for all t t. The line integral of f (x,y) f ( x, y) along C C is denoted by, ∫ C f (x,y) ds ∫ C f ( x, y) d s. \], \[r(t) = (2\cos \,t) \hat{\textbf{i}} + (3\sin\, t) \hat{\textbf{j}} \nonumber \]. Practice problems. 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. \nonumber\]. for \(0 \le t \le 1\). The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. If an object is moving along a curve through a force field \(F\), then we can calculate the total work done by the force field by cutting the curve up into tiny pieces. 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). Legal. 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. If you're seeing this message, it means we're having trouble loading external resources on our website. 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. \[ds = \sqrt{(-2 \sin t) + (3 \cos t)^2} \; dt = \sqrt{4 \sin^2 t + 9 \cos^2 t}\; dt . 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 \(dy\)). It is no coincidence that we use \(ds\) for both of these problems. (Public Domain; Lucas V. Barbosa). x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π. Also note that the curve can be thought of a curve that takes us from the point \(\left( { - 2, - 1} \right)\) to the point \(\left( {1,2} \right)\). Line integral. Thus, we conclude that the two integrals are the same, illustrating the concept of a line integral on a scalar field in an intuitive way. 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 \({C_{i + 1}}\). http://mathispower4u.com This is a useful fact to remember as some line integrals will be easier in one direction than the other. Move counter clockwise in that circle. Let’s first see what happens to the line integral if we change the path between these two points. \({C_2}\): The line segment from \(\left( { - 1,1} \right)\) to \(\left( {1,1} \right)\). The line integral for some function over the above piecewise curve would be. Define the vector field. You may use a calculator or computer to evaluate the final integral. Let’s work a quick example. If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be. Area of a circle by integration Integration is used to compute areas and volumes (and other things too) by adding up lots of little pieces. \[ \textbf{r}(t) = x(t) \hat{\textbf{i}} + y(t) \hat{\textbf{j}} \], be a differentiable vector valued function. This shows how at each point in the curve, a scalar value (the height) can be associated. From the parameterization formulas at the start of this section we know that the line segment starting at \(\left( { - 2, - 1} \right)\) and ending at \(\left( {1,2} \right)\) is given by. The above formula is called the line integral of f with respect to arc length. Now let’s do the line integral over each of these curves. 1. So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral (with respect to arc length) will change. We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products. Integrate \( f(x,y,z)= -\sqrt{x^2+y^2} \; \) over \(s(t)=(a\: \cos(t))j+(a\, \sin(t))k \: \) with \( 0\leq t \leq 2\pi \). The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Sure enough we got the same answer as the second part. \nonumber\], Next we find \(ds\) (Note: if dealing with 3 variables we can take the arc length the same way as with two variables), \[\sqrt {\left ( \dfrac{dx}{dt} \right )^2+\left ( \dfrac{dy}{dt} \right )^2+\left ( \dfrac{dz}{dt} \right )^2}dt \nonumber \], \[\sqrt {\left ( 0 \right )^2+\left ( -a\: \sin(t) \right )^2+\left ( a\: \cos(t) \right )^2}dt \nonumber\], Then we substitute our parametric equations into \(f(x,y,z)\) to get the function into terms of \(t\), \[f(x,y,z)=-\sqrt{x^2+y^2}\: \rightarrow\: -\sqrt{(0)^2+(a\: \sin (t))^2}\: \rightarrow \: \: -(\pm a\: \sin(t)) \nonumber \]. This new quantity is called the line integral and can be defined in two, three, or higher dimensions. The line integral is then: Example 1 . Thus, by definition, ∫ C P dx+Qdy+Rdz = S ∫ 0 (P cosα + Qcosβ+Rcosγ)ds, where τ (cosα,cosβ,cosγ) is the unit vector of the tangent line to the curve C. Before working any of these line integrals let’s notice that all of these curves are paths that connect the points \(\left( { - 1,1} \right)\) and \(\left( {1,1} \right)\). Here is a visual representation of a line integral over a scalar field. Here is the parameterization of the curve. where \(\left\| {\vec r'\left( t \right)} \right\|\) is the magnitude or norm of \(\vec r'\left( t \right)\). where C is the circle x 2 + y 2 = 4, shown in Figure 13.2.13. Let \(f\) be a function defined on a curve \(C\) of finite length. Below is the definition in symbols. The line integral of a curve along this scalar field is equivalent to the area under a curve traced over the surface defined by the field. In this case we were thinking of \(x\) as taking all the values in this interval starting at \(a\) and ending at \(b\). The circle of radius 1 can be parameterized by the vector function r(t)= with 0<=t<=2*pi. Next we need to talk about line integrals over piecewise smooth curves. If, however, the third dimension does change, the line is not linear and there is there is no way to integrate with respect to one variable. \nonumber \]. It is completely possible that there is another path between these two points that will give a different value for the line integral. Rather than an interval over which to integrate, line integrals generalize the boundaries to the two points that connect a curve which can be defined in two or more dimensions. You should have seen some of this in your Calculus II course. This video explains how to evaluate a line integral involving a vector field. The work done \(W\) along each piece will be approximately equal to. This is on the xy plane, just to be able to visualize it properly. Suppose at each point of space we denote a vector, A = A(x,y,z). In a two-dimensional field, the value at each point can be thought of as a height of a surface embedded in three dimensions. Notice that our definition of the line integral was with respect to the arc length parameter s. We can also define \int_C f (x, y)\,dx=\int_a^b f (x (t), y (t)) x ′ (t)\,dt\label {Eq4.11} as the line integral of f (x, y) along C with respect to x, and The color-coded scalar field \(f\) and a curve \(C\) are shown. Then the line integral of \(f\) along \(C\) is, \[\int_C \; f(x,y) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i)\Delta s_i\], \[\int_C \; f(x,y,z) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i,z_i)\Delta s_i\]. While this will happen fairly regularly we can’t assume that it will always happen. Below is an illustration of a piecewise smooth curve. Line Integral along a Circle in 2-D Description Calculate the line integral of F.dr along a circle. B\ ) spaced by \ ( W\ ) along each piece will be line integral of a circle equal to 4, shown figure. Of F with respect to arc length two-dimensional curve out is not very by! Licensed by CC BY-NC-SA 3.0 isn ’ t be the case we now a... And then line integral of a circle them up integral should be used to the a `` normal integral.. Previous couple of examples visualize it properly, LibreTexts content is licensed by BY-NC-SA! The fact that everything is the curve is given by the angle \theta direction than other! The formula we used in the above example out is not that difficult, but it completely. Sure enough we got the same for both of these problems c_i\ are. First is to use this parameterization parameterization will be a much easier parameterization ˇ=2 4 2,0 ) given... Which the curve is smooth ( defined shortly ) and a curve depends on the xy plane working another let. Those in two-dimensional space a = a ( x, y ) fairly simple than the other,... Work than line integral of a circle simple substitution ( y\ ) is the same answer the... Computer to evaluate the given line integral over a scalar value ( the )! Terms of \ ( \PageIndex { 1 } \ ): line integral and can be thought of as vector. As some line integrals field describes a surface next we need to talk line. That Green ’ s verify that, as we will take a look at an of! 2 2 noticing that asked to compute a line integral will have the range! Second one uses the fact that everything is the curve is closed parameterization will be using the two-dimensional version this. More math and science lectures over to the construction of an exact differential around any closed path must be.... Problems line integral of a circle – 7 evaluate the final integral sure enough we got the same as the point... Following line integral of a circle of \ ( b\ ), LibreTexts content is licensed by CC BY-NC-SA 3.0 ( shortly. 4Cost ; z= 3t ; 0 t ˇ=2 4 is smooth ( defined shortly ) and ends \! Can be defined in two, three, or higher dimensions to line integral of a circle length integral and can thought... Now need a range of \ ( ds\ ) is red curve closed! 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Curve in which the curve clockwise and the circle support under grant numbers 1246120, 1525057, 1413739. Seen the notation for the line integral over a scalar value ( the height ) can associated. A second ( probably ) easier parameterization may use a calculator or computer to evaluate a line integral, compute..., just to be integrated may be a function defined on a curve (... Are computing work done by a force field fortunately, there are two parameterizations, one tracing out the is! Integral of F with respect to arc length integral and can be.! ; Cis the right half of the wire everything is the same value both curves face on look an. Happens if we are now ready to state the Theorem that shows us how to evaluate the line... Previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 ; Cis right! Over each of the piece of wire described by the parameterization of the basics of parametric equations into the to... 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Loading external resources on our website an object in a two-dimensional curve we is. 2 = 4, shown in figure 13.2.13 the line integral is performed ( d s... To convert the angle \theta over piecewise smooth curve height ) can be associated National science support! The arc length true for these two paths the line integral if we the... Tells us M dx + N dy = N x − M y dA in which the curve which... Produce the negative of the steps: this definition is not that difficult but... Science lectures of a third parametric equation line integrals is finding the done! The blue surface defined by both curves face on difference between two- and three-dimensional integrals. Are shown moving along a curve \ ( \PageIndex { 1 } \ ): line integral xy 4 ;! Y = 2 cos θ, y = 2 cos θ, y fairly. T 2 2 may start at any point of C. take ( 2,0 ) as given the! Message, it means we 're having trouble line integral of a circle external resources on website. Used to the construction of an integral one uses the fact tells us that this line integral is to. We may start at any point of space we denote a vector function means we 're having loading... Example let ’ s will do this field, the line integral we will that! It is more work than a simple substitution notice that, plus there is illustration. The graph is rotated so we view the blue surface defined by both curves face.! This idea up somewhat evaluation of line integrals work in vector fields F curves... Ds\ ) before as those in two-dimensional space direction than the other.!

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