Could you explain the maths of the line integral please?
You mean a path integral? You can imagine this for real functions of 2 variables like this: Place a curved vertical "fence" along a path. For each point on the path, plot the function value on this fence, (or, which is the same, intersect the fence with the surface which is a plot of your 2-d function, yielding a function graph plot on the fence). The integral will be the area under the graph on this fence, much like for ordinary integral. Too simplistic for many cases, but you can get the idea.
Calculus 3 - Line Integrals?
1. C: y = x² between (0,0) and (4,16) ----> dy = 2x dx ∫c (−y sinx dx + cosx dy) = ∫c (−y sinx) dx + ∫c cosx dy = ∫ [0 to 4] (−x² sinx) dx + ∫ [0 to 4] cosx * 2x dx = (2 − 8 sin(4) + 14 cos(4)) + (−2 + 8 sin(4) + 2 cos(4)) = 16 cos(4) ≈ −10.45829793381779 2. C is the line from the origin to the point (4, 3, 6) x = 4t ---> dx = 4 dt y = 3t ---> dy = 3 dt z = 6t ---> dz = 6 dt 0 ≤ t ≤ 1 ∫c (x³ dx + y² dy + z dz) = ∫c x³ dx + ∫c y² dy + ∫c z dz = ∫ [0 to 1] (4t)³ (4 dt) + ∫ [0 to 1] (3t)² (3 dt) + ∫ [0 to 1] (6t) (6 dt) = ∫ [0 to 1] 256t³ dt + ∫ [0 to 1] 27t² dt + ∫ [0 to 1] 36t dt = 64 + 9 + 18 = 91
What are the most interesting applications/problems of integral calculus?
Some elementary applications include calculation ofAreasVolumesLengthsMomentsHydrostatic pressureMultiple integralsIndefinit integralsSome more advanced applications include calculation ofLine integralsSurface integralsDivergencesCurlSolutions of differential equationsSome theoretical applications includeMeasure theroyRieman-Stieltjes integralsLebegue IntegralsProbability theorySuffice it to say each of the above topics can be found in some detail in Wikipedia. There are many more applications than I list here. I have been studying Integral calculus for more than 60 years and I feel like I am still a novice.
What is the use of line integrals? And please don't say finding the mass of a wire.
Line integrals are actually really useful when we start applying them to physics. Say you're pushing a block up a path. How much work is being done? If your path is straight, this is a pretty simple application-of-formulae problem. But if your path is twisty-turny, line integrals are a quick and easy way to compute the work done. Likewise, how much energy would it take to move a satellite from one altitude to another? We're pretty good at estimating the path that it would take, so using a line integral to determine the amount of work is an appropriate way to estimate. In summary, line integrals may appear to be abstract concepts when you're applying them to paths and fields out of your multivariate calculus textbook, but they have immediate applications in several areas of physics.
How can line integral problems be solved?
This is sort of a broad question, but it really depends on the problem. For future references, you can look at this website Calculus III - Line IntegralsFirst of all, we need to look at what a line integral. They are of multiple types, so I’ll treat them one by one.The first one I will explain is a line integral with respect to arc length. Well, this integral, just like the Riemann integral represents area. What sort of area? Well imagine you have a string(that’s your curve). And somewhere, let’s say above it, you have the surface of a function f(x,y)(sort of a bowl). What you do is project each point from the curve to the bowl. Now you have a curtain. Well, on this curtain you have small arc elements, dS. It starts to smell a bit of Riemann. You now have a nice 2D surface. How do you translate this into a Riemann integral? Well, you take the curtain which is sort of bunched up and you stretch it until it becomes as smooth as possible. Now you have a basic Riemann integral that you can solve by calculating the area under the curve.The second line integral is the integral of a vector field. These integrals are giving you the work done by a particle travelling across that line, through that vector field. Well, the most straightforward way to calculate these ones is by parametrizing the curve(Line Integrals of Vector Fields) and then apply the formula that I have provided through that website(I still have to learn how to type in Latex, sorry). Now, you sometimes might have some ugly curves or vector fields? What do you do? Well, here comes into play mister Stokes(or his 2 friends, Riemann and Green). What does Stokes say? He says that if you want to find the circulation of a vector field across a loop, you can calculate the integral of the rotor of the vector field, on a surface that sits on the given loop that you have. Notice that I said loop: this means only closed curves(also no self-intersections). Well, this is pretty straightforward to calculate so I’ll just explain some intuition behind this theorem. Imagine that you have a closed loop and you chop up into small pieces the space enclosed by it. Well, now think about the microscopic circulation inside each small square formed by the cuts. What Stokes says is that the sum of those microscopic circulations is equal to the circulation across the loop itself. Of course that you can choose any surface of your liking, it will be the same answer.Hope that this helped!
How do I use line integrals for this calculus problem?
ok so a line integral. you want to parametrize the curve, right? so radius is 4 cm, x = 4 cos t, y = 4 sin t ds = sqrt( dx^2 + dy ^2) = sqrt[ (-4 sin t dt)^2 + (4 cos t dt)^2 ] = 4 dt The length of the whole loop is 4 pi dm = dm/ds ds = (2 gm / 4 pi) ds = (2 gm / pi) dt Trying to think of how to set up a center of mass integral Basically you want the average mass-position 1 / (2 grams) integral from 0 to pi dm (x(t) i + y(t) j) is that it? 1 / (2 grams) * integral from 0 to pi of (4 cos t i + 4 sin t j) * (2 grams/pi) * dt I hope I got it right please see if this helps!
Please answer my calculus question below which is about scalar line integrals with arc length as parameter?
Evaluate the following line integrals. ∫xyds; C is the unit circle r(s) =
Evaluate Line Integral -shortest path between 3 points?
If the problem is stated exactly as your first sentence gives it, then I agree with you that you need integrals along two straight line segments. P.S. have just read the additional words you later added under "additional details," and I still think you can treat just the two line segments.