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I=integral(xdy+ydx) where y=sqrt(x) O(0,0) A(1,1)

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John D'Errico
John D'Errico 2016-5-29
What are O(0,0) and A(1,1)?
John D'Errico
John D'Errico 2016-5-29
编辑：John D'Errico 2016-5-29
Are you perhaps asking how to compute the arc length of some curve, between the points(0,0) and (1,1)?
Is this a question of a symbolic solution, so exact, or a numerical one?
Is this your homework? (I am somehow sure of that.) If so, then what have you tried? Answers is not here to do your homework.
The coordonates for the parabolic
no is not my homework, is a mathematic problem that i want to solve. Yes what i want to do is the arc length of the curve between that two points while I an integral on curve gamma let's say of xdy+ydx
i just finished an exercise with a double integral like 3<=x.^2+y.^2<=5 while x>=0 and y>=0. the function was: exp(-x.^2-y.^2)dxdy and now im doing this problem...I finished my 2nd semester and i have done my matlab courses and seminars with 9 xD
im looking to improve my self cuz i see a lot of potential in this program but idk from where to begin...what i have done at school is not enough. this is what i did http://www.apar.pub.ro/informatica_aplicata_2/laborator/ all 5 labs

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 采纳的回答

Roger Stafford
Roger Stafford 2016-5-29
编辑：Roger Stafford 2016-5-29

2 个投票

The integral(xdy+ydx) is equivalent to integral(1*d(x*y)). If the integral lower limit is x*y = 0*0 and the upper limit x*y = 1*1, then the integral must have a value of 1. It has nothing to do with being on the curve y = sqrt(x) except for the two endpoints.

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Roger Stafford
Roger Stafford 2016-5-30
If you do them correctly, you will find that ϒ1, ϒ2, and ϒ3 will all be the same if the two endpoints are (0,0) and (1,1). In other words, ∫xdy+ydx depends only on the two endpoint values of x*y.
so i don't need the equation to solve it?
Roger Stafford
Roger Stafford 2016-5-30
Well, I have the impression that your instructor intends for you to solve those three integrals independently and then notice that your answer in each case is the same. For example, with ϒ1 you can calculate
y = sqrt(x)
dy = (1/2)/sqrt(x)*dx
x*dy+y*dx = x*(1/2)/sqrt(x)*dx + sqrt(x)*dx = 3/2*sqrt(x)*dx
∫ x*dy+y*dx = ∫ 3/2*sqrt(x)*dx = 3/2*x^(3/2)/(3/2) = x^(3/2)
Hence the definite integral is 1^(3/2)-0^(3/2) = 1. This is the same as x*y at (1,1) minus x*y at (0,0), namely 1.
The same kind of computation can be done with ϒ2 and ϒ3 (I assume they intended for the ϒ3 case to be a straight line from point O to point A.)
So, in answer to your question, I would say that actually you do need the equations in order to demonstrate to your instructor’s satisfaction that ∫ x*dy+y*dx does depend only on the difference of x*y at the two endpoints of your curve. (I’ve done one-third of your home work for you. I won’t tell if you don’t.)

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更多回答（2 个）

John BG
John BG 2016-5-30
编辑：John BG 2016-5-31

1 个投票

Hi
1.- from http://mathworld.wolfram.com/LineIntegral.html
.
.
2.- eq [2] you want to integrate a vector function F along a path or line.
F = [F1 , F2, F3] = [y , x, 0]
.
3.- eq [5] is possible because the curl of F is 0, just solve the following (from https://en.wikipedia.org/wiki/Curl_(mathematics) ) manually:
4.- So, the potential function you need to solve the integral is
phi = -[x*y , x*y, k]
5.- So, the integral of the field [y,x,0] along the arcul/arc/path/line (call it whatever you like it) y=x^.5 from point O [0 0] to point A [1 1] is the difference of potential
phi(O)-phi(A) = -phi(A)
and you get the same result whether you follow the previous arcul
[x x^.5]
or following
[x x^2]
If you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John
jgb2012@sky.com
Jesús
Jesús 2022-9-2

0 个投票

(X+y) dx + xdy =0 MATLAB como hacerlo

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