I am trying to evaluate a line integral over the boundary of the area in the x-y plane between the parabola \(\displaystyle y=x^2 \) and the line \(\displaystyle y=1 \), 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.

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?

Actually, there isn't any difference! If you path went from a to c, passing through b, then \(\displaystyle \int_a^c f(x,y)dx+ g(x,y)dy= \int_a^b f(x,y)dx+ g(x,y)dy+ \int_b^c f(x,y)dx+ g(x,y)dy\). 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!

This is what I have done.

I have parametrised the line as \(\displaystyle x=1-2t, y=1\) from t =0 to1 [/tex] and the parabola separately as \(\displaystyle x=t, y=t^2\) from t =-1 to 1.

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.

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.

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)

As I said before, \(\displaystyle \int_a^c = \int_a^b+ \int_b^c\). There is no reason for absolute values. (In fact, in general even \(\displaystyle \left|\int_a^c\right|\ne \left|\int_a^b\right|+ \left|\int_b^c\right|\).)

Could 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.

That's correct. If x= 1- 2t and y= 1, then dx= -2dt and dy= 0. The integral becomes \(\displaystyle \int_0^1 (y^2- x)dx+ (3x+ y)dy= \int_0^1 (1- (1- 2t))(-2dt)+ (3(1- 2t)+ 1)(0)\)\(\displaystyle = \int_0^1 (-4t)dt= \left[-2t^2\right]_0^1= -2\)