Vector fields, line integrals, and Green s Theorem Line integrals The problem: Suppose ou have a surface = f(, ) defined over a region D. Restrict the domain of the function to the values of and which lie on the boundar of D. On this domain, the values of f(, ) define a curve in space. It does not make sense to ask what the volume is under this curve. However, imagine a fence put up along the boundar, whose height at each point is given b the value of f(, ). It does make sense to ask What s the surface area of that fence? It also leads to questions such as Is there a relationship between that area and the volume under the surface? f(,) f(,) 6 6 4 4 f(,) on boundar f(,) on boundar 6 6 4 4 Figure : The value of that area is one wa to define the value of an integral of f(, ) along the boundar of D. Integrals of this tpe are called line integrals.
Notation There are several was to denote a line integral. The general version usual occurs outside of freshman alc tets, works for an dimension, and uses the following conventions: Ω is a region in R n. The boundar of the region is denoted Ω. The vector is n dimensional: =<,,... n >. f() is an n + dimensional function of n variables defined on Ω. The integral of f over Ω is denoted Ω f() d (in one variable, the area under f(), in two variables, the volume under f(, ) = f(, ), in three variables, the hpervolume under f(,, ) = f(,, ), and so on, for positive valued functions). The line integral of f over the boundar of Ω, Ω is denoted f() ds Ω We ll prett much stick with the two variable / D case (since that s the one we can visualie). In this case, we have the notation D is a region in R (the plane). The boundar of the region is denoted = D. (The stands for curve ) The vector is dimensional: =<, >. f(, ) is an dimensional function of variables defined on D. The integral of f over D is denoted D f(, )da The line integral of f over the boundar of D, denoted D or, is denoted f(, )ds The curve doesn t have to be the boundar of an enclosed region; we can evaluate the line integral along a non closed curve.
Evaluating Line Integrals First, note that that ds in the integral up there is what is was before an infinitesimall small chunk of arc length. (Think about it for a second and make sure that it makes sense that ou ll get the area of the fence b multipling the function value with a chunk of length along the curve, then summing up.) Recall from past eperience that if we can parameterie the curve b r(t) = (t)i + (t)j a t b then So Step b step ds = d + d = f(, )ds = d + d = b Parameterie the curve if not alread done. a (d ) + ( ) d f((t), (t)) ( (t)) + ( (t)) Write the function f(, ) as a function of t b composing f((t), (t)). (d ) ( ) d Write the epression + = ( (t)) + ( (t)). Set up the integral f(, )ds = b b making the above substitutions. Integrate and evaluate. a f((t), (t)) ( (t)) + ( (t))
Eample: Find 4 ds where is the right half of the circle + = 6 traversed counterclockwise from (, 4) to (, 4). Quickl sketch the curve and parameterie it. Recall that the best wa to parameterie a circle or ellipse is b using sine and cosine. onvert the function f(, ) = 4 to f((t), (t)) b substituting our parameteriation. ompute d and d and use them to construct ds = (d ) ( ) d + Assemble the whole thing into an integral and integrate.
The plot of the surface f(, ) = 4 and one interpretation of what ou just found is given below. This fence area is called the lateral surface area. f(,) f(,) 4 4 f(,) on boundar f(,) on boundar 4 4 Figure : Question: Does the direction of travel along the path matter? What would happen if ou had traversed the half circle clockwise from (, 4) to (, 4)?
More on curves The curve in the previous eample is an eample of a smooth curve, which intuitivel is just what it sounds like - no sharp changes in direction. Formall, a plane curve [space curve] given b r(t) = (t)i + (t)j [r(t) = (t)i + (t)j + (t)k] is smooth if d and d [and d] are continuous on [a, b], and not simultaneousl ero on (a, b). A curve is piecewise smooth if the interval [a, b] can be partitioned into a finite number of subintervals, on which is smooth on each subinterval. If a curve is piecewise smooth, we ll have to break up the integral, since each piece will have a different parameteriation. f ds = f ds + f ds +... + f ds n where =... n and each of the i is smooth. As we saw in the first eample, parameteriing a curve gives it an orientation: a specific direction of travel along a curve. Reversing the orientation reverses the sign of the value of the integral, so direction does matter. For eample, both parameteriations below give the segment of the line = + with endpoints (, ) and (, ), but have opposite orientations: r (t) = ti + (t + )j, t r (t) = ( t)i + ( t)j, t
Eample: Write a piecewise smooth parameteriation of the curve shown below. Note also that the orientation is shown on the curve, and be sure our parameteriation reflects that.
Line integrals in D and another phsical interpretation There s no particular reason mathematicall wh we can t etend this to curves in space: then f(, )ds = r(t) = (t)i + (t)j + (t)k (d ) ds = + b a ( ) d + a t b ( ) d f((t), (t), (t)) ( (t)) + ( (t)) + ( (t)) But describing the result as an analog of surface area is trick to visualie. Here is another possible phsical interpretation, that certainl works for curves in D, but also etends nicel up into dimensional curves: densit and mass. Imagine a wire in D or D whose shape is described b the smooth curve. If f(, ) or f(,, ) is a function that gives the densit of the wire, then the line integral f ds gives the mass.