# Line integral

1. Jun 13, 2010

### kliker

1. The problem statement, all variables and given/known data
hello again, sorry for asking so many questions, i just want to make sure if im correct or not

calculate the line integral y^2dx+x^2dy where line C is the triangle with sides x=1, y=0 and y=x

3. The attempt at a solution

first of all i tried to find a customization of the line

we know that x = 1 hence it will be like this

r(t) = (1,t) but im not sure if it's correct

then i said that the integral would be this

[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP470019b34815c34hai2900000d7egii1beg78b2i?MSPStoreType=image/gif&s=9&w=88&h=37[/URL]

could I just use Green's theorem?

I mean using Greens theorem I get the same result

[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP167119b34a9g72bg50ia000021ai53gc37dac65a?MSPStoreType=image/gif&s=3&w=186&h=37[/URL]

im 99% sure that greens theorem is correct, i mean the way i implemented it, but is the first way i showed also correct?

Last edited by a moderator: Apr 25, 2017
2. Jun 13, 2010

### HallsofIvy

Staff Emeritus
[/URL]
"we know that x=1"? Did you leave out much of the problem? We know that x= 1, y= any number from 0 to 1 is one side of triangle and so one part of the path over which we want to integrate. Taking x= 1, y= t as parametric equations, dx= 0, dy= dt so the integral becomes
[tex]\int_0^1 1 dx= 1[/itex]

But you still have to do the other two sides of the triangle.

On the line y= 0, we can use parametric equations x= t, y= 0 with t from 0 to 1. Then dx= dt, dy= 0 but $y^2 dx= 0dt$ so the integral is
[tex]\int_0^1 0dt= 0[/itex].

On the line y= x, where we are integrating from (1, 1) to (0, 0) (we got counterclockwise around the closed path), we can take x= t, y= t so that dx= dt, dy= dt and the integral is
[tex]\int_1^0 2t^2 dt= -\int_0^1 2t^2 dt= -2/3[/itex] and the entire integral is 1- 2/3= 1/3.

Last edited by a moderator: Apr 25, 2017
3. Jun 13, 2010

### kliker

thanks a lot for your help

that cleared up everything in my mind