Nonstandard integral

1. Feb 17, 2008

tronter

1. Find the amount of work done by the constant force field $dx + 3dy - dz$ as it moves a particle along the intersection of the planes $x+y+z = 1$ and $x-2y = -2$ from where it intersects the $y,z$ plane ($x = 0$) to where it intersects the $z,x$ plane ($y = 0$).

So the intersection is $$z = 0$$ and $$z = 3$$. Then what would I do?

2. Feb 17, 2008

HallsofIvy

Staff Emeritus
I have absolutely no idea what you mean by "the intersection is z= 0 and z= 3" or how you got that! The intersection of two planes is a line, not two points. You are to integrate along that line.

If x+ y+ z= 1, then x= 1- y- zo. If x- 2y= -2, then x= 2y- 2 so we have 2y- 2= 1- y- z.
z= 1- 3y. In other words, the intersection of x+ y+ z= 1 and x- 2y= -2 is given by x= 2t- 2, y= t, z= 1- 3t. That crosses the yz plane when x= 2t- 2= 0 or t= 1. It crosses the xz plane when y= t= 0.

Integrate dx+ 3dy- dz over that line from t= 1 to t= 0.

(I also have no idea why you consider that integral "nonstandard". It's about as "standard" as there is!)

3. Feb 18, 2008