Contour/Path Integration (concept)?

Question

I don't understand what the need for a contour or path is in an integral in vector calculus/complex analysis. For instance

∫zdz (on γ) where let's say γ is the semicircle with radius one centered on R = 2 where R is the real line in the complex plane. To me this is like integrating two functions, one being the semicircle and the other being f(z) = z. I don't get it, can anyone help me? (I don't need help with the integration procedure itself, just the *why* about contours/paths).

Answer 1

Okay, here's my attempt to give you an intuitive representation of the integral along a contour like the semicircle.



First, I'll have to pass on the complex plane portion, since I'm only a bit familiar with actual applications of complex numbers in electrical engineering with AC current. I'm sure in physics there are other applications. But the complex plane is essentially comparable to the xy plane, so I'll try to explain what the contour integral on a semicircular path on the the xy plane would give you.



If you think about it, the semicircular path is almost like taking the x-axis and bending it in a semicircle. When you integrate a function along the x-axis it simply gives you the area under the function. So you can see that if you integrate along the semicircular path, it would give you the area (or actually "surface area") of that half-pipe surface formed between the xy plane and the function z.



Some made up "real world" applications? Pretend you wanted to make a shark net to protect a harbor from sharks somewhere,and you are trying to figure out how much material you need. The shape of the net is to be a semicicle around the harbor. The countour of the ocean bottom would be like the function z. If you were able to model the ocean floor surface mathematically, the contour integral you calculate of the depth of the ocean along the semicircle would give you a number for the amount of netting material you would need.



Another example - pretend you were going to paint a dam like the Hoover Dam (~semicircular in shape). If you could mathematically model the surface of the gorge which is being dammed up, the contour integral would give you the amount of surface area that would have to be painted, and thus an estimate of how much paint would be needed.



Back to the purely mathematical geometric example - If you have a surface z, and take the contour integral along a semicircle on the xy plane, it simply gives you the surface area of that half-pipe surface bounded by the function z and the xy plane.



Anyways - I hope that gives you some insight into what the contour integral gives you. I'll leave the complex plane application to someone else to explain.



Was your problem simplay a math assignment, or was it in some other subject like physics?



Hope this helps



EDIT:







Oh, I actually remembered (just barely) another actual application of the complex number plane in the real world, but it's still quite abstract. It's used in controls engineering, or more specifically in process controls.



Excuse the lack of details, but I remember this from long ago in a process controls course I took in chemical engineering, and never fully grasped everything, and I don't believe it went sufficiently in depth to fully get into your example.



However, in process controls, you mathematically model a system (mechanical or chemical process) which includes a number of different components, and which has feedback to keep the system at a setpoint - maybe you are trying to control the temperature of a tank in which a chemical reaction is taking place that produces heat, so you control the amount of cooling water flowing through a heat exchanger in the tank. The controller uses the deviation of the tank's temperature from the temperature setpoint as input into its programming to adjusting how much it should open the cooling water valve.



Mathematically modelling the behavior of such a system uses complicated mathematical functions called transfer functions, and Laplace Transforms.



As I recall (again barely recall), we would analyze the system to find out under what conditions the system would become unstable (out of control) or stable (self-controlling/correcting). Solving the mathematical equations would produce roots of the equations which were complex numbers, and we would plot these on the complex plane.



Certain of these roots plotted on the complex plane would represent unstable equilibrium points for the system, and others stable equilibrium points. At the unstable equilibrium points the system would act like an upside-down pendulum – being stable only if it were positioned exactly correctly, but quickly moving away from that equilibrium point with the slightest deviation. The stable equilibrium points would be like the correctly hujng pendulum, with any deviations self-corrected back to the stable equilibrium point. All the other points on the complex plane would represent intermediate points in the system behavior. If the system were artificially positioned at any intermediate point, the system would self-correct away from the unstable points and would move toward the stable points.



You can now imagine, that there is a surface described over the complex plane which we can call z, which measures the stability of the system. The various points on the complex plane between the equilibrium points describe the state of the system "in flux" between the equilibrium points. As the system moved away from the unstable points to the stable points, it would follow a contour on the complex plane.



The semicircular path in your problem could be describing the path the system took in response to a deviation (like a spike in temperature, or a change in setpoint). The path might be a natural response of the system to the deviation (like the control system acting on its own), or a forced path (like the operator of the system manually controlling the cooling water valve).



The contour integral might then be a measure of the cumulative amount of heat removed from the system during the course change, as it reacted to the deviation, or the amount of time the system was “out of spec”.



Again this is very qualitative, but I believe is a real application of the type of problem you are dealing with.



However, I think one would need to be at the Master's or even Phd level of chemical or controls engineering to get into the level of detail that your problem is addressing.



Anyways, I hope this was understandable and gives you more insight into how that type of problem might be applicable to the "real" world. It’s actually an interesting application, but you would have to really get into it to fully appreciate it.

Answer 2

when you have a complex-valued function of a complex variable, such as f(z) = u(z) + iv(z), it's basically like a vector function of two variables F(x,y) = Fx(x,y)i + Fy(x,y)j in vector calculus, where Fx and Fy are the component functions of F and each is a function of two variables, namely x and y. since F is no longer defined on R, but on R^2, you can't just specify an interval (a,b) like in single-variable calculus and integrate the function on (a,b) because in R^2, the line/contour/path over which you're integrating the function f(z) or F(x,y) might be a curve that lies in the R^2 plane. in the direct method of coutour integration and in vector calculus, the first thing you do is parameterize the function with a single real variable t. in doing so, you're basically converting this indefinable curved path to an easily definable real interval (t1,t2) on the t axis. so to answer your question, the PATH which is the semicircle with radius = 2 centered on 2+i0 on the complex plane is the REGION over which the function is integrated, and that, is very different from f(z), which is the actual function you're integrating. to put it all simply, what you're doing is integrating the FUNCTION f(z) on the REGION the semicircle.





Ok, by the word "region" I just meant the path/line/contour or really just the domain over which the function is integrated. now you asked how the semicircle had anything to do with the bounds of integration, well in the case of single variable calculus, for a definite integral you had to write Int{ x = a to b} f(x) dx, where the interval (a,b) is the section of the real x axis where f(x) is integrated. now in a 2D plane, the "interval" is no longer limited to between two values of an axis, but a curved line, like a "curved interval" no longer confined to the x axis but in the x-y plane. so to give your an intuitive idea or motivation, imagine the floor of your bedroom as the x-y plane or the complex plane, there's a curtain in your room, and if you look straight down from above your room the folds of the curtain make a curved line projection onto the floor. that line is the path. i know that most curtains have uniform height, but let's say some parts of your curtain hang lower than other parts and let the real valued fuction f(x,y) or f(z) be the height of the curtain at the point (x,y) or z on the floor plane, then the integral of f over the curved path the curtain makes gives you the area of the sheet of cloth that the curtain is made of if you were to stretch it flat and measure it. this is the example my prof gave me and it really enlightened and i hope it does the same to you.



Now in the case where f(z) is complex valued, unlike the above case where f(z) is the height which is real valued, f is just u(z)+iv(z) where u and v are both real functions. And in the case where the path exists in a 3D space like say a helix, then there's this curtain that exists in the 4th dimention whose area is the line/countour integral of the height function.

Answer 3

You have done contour integrals all along possibly without really knowing it. When you evaluate the integral from 0 to 4 of x^2 with respect to x, what path are you integrating along? The answer is the line segment beginning at (0,0) and ending at (0,4) presumably. Actually in this case you are integrating along any line segment that begins at (a,0) and ends at (a,4) [and direction does matter]. It's easy to see that direction matters. If you integrate x^2 dx from 4 to 0, you get a different answer (the opposite ofcourse).



Now, when you begin studying rhiemann integrals in Calculus I, no mention is made of the contour (path) as it is always taken as a simple line segment. Well that's not quite true, if one limit is infinity it is a ray, and if the integral goes from -infinity to infinity it is a line. Did it ever occur to you that one could integrate along a contour other than a line segment. Well you certainly can.



As for what does it return....use greens theorem. Greens theorem gives a contour integral as a double integral of a particular integrand at least when partial derivitives of the initial integrand with respect to x and also to y are continuous. A double integral can be thought of as the volume under the surface defined by the integrand yielded by greens theorem that lies within the region over which the double integral is being evaluated.



I'm not sure how much this helps. You may be looking for an 'intuitive' definition, but this is not always so easy. For instance, think about lesbague integrals v.s rhiemann integrals.



Pardon my spelling. Good luck.

Answer 4

Sometimes contour integrals give you more "physical" information. Think of a rocket near a planet. Gravity and thrusters both exert a force on the vehicle resulting in accelerations--one toward the planet, one in the direction of the rocket. If you want to know the total acceleration, resulting velocity and position, you'll need a contour integral along the path the vehicle takes.

Answer 5

i anticipate you propose Sin[t]^2 necessary of (sin t^2) dt. necessary from x to 2x^2+a million. Ans a million/4 (2 - 2 x + 4 x^2 + Sin[2 x] - Sin[2 + 4 x^2]) f ' [x] a million/4 (-2 + 8 x + 2 Cos[2 x] - 8 x Cos[2 + 4 x^2])