[QUOTE=Debsta;511887]I am trying to evaluate a line integral over the boundary of the area in the x-y plane between the parabola and the line , in an counterclockwise direction.
The integral is
Int((y^2-x)dx+(3x+y)dy) (still learning latex ...sorry)
The LaTex for this is [ math ]\int (y^2- x)dx+ (2x+y) dy[ /math ]- without the spaces, of course.
Actually, there isn't any difference! If you path went from a to c, passing through b, then . Whether you actually do them separately or not is a matter of convenience- and here, where you have two very different formulas for two parts, yes, it is convenient to do them separately!Can someone please tell me: Do I have to split up the enclosed boundary into the straight line part and the parabola part, and do them separately?
Okay, that will work. As t goes from 0 to 1, the point (x,y) goes from (1, 1) to (-1, 1) on the line y= 1.This is what I have done.
I have parametrised the line as from t =0 to1 [/tex] and the parabola separately as from t =-1 to 1.
As t goes from -1 to 1, for the second parameterization, (x, y) goes from (-1, 1) to (1, 1). Yes, that goes around the path counterclockwise.
As I said before, . There is no reason for absolute values. (In fact, in general even .)I then evaluated the line integrals separately and got -2 and 4.4
Not sure if I need to now simply add them as they are or add absolute values (I'm thinking the first one)
That's correct. If x= 1- 2t and y= 1, then dx= -2dt and dy= 0. The integral becomesCould someone kind out there please check my first value of -2. The problem I encountered is : If dy/dt=0 then dy=0dt and does that mean the second part of the integral (ie the bit with 3x+y) just becomes 0?
Any help appreciated. Thanks.