# Does $g(x,y,z)$ (the equation of the surface) need positive z or

1. Dec 9, 2013

### ainster31

Does $g(x,y,z)$ (the equation of the surface) need positive z or negative z when doing a surface integral?

The author specifies that $g$ either has $z$ or $-z$ depending on the orientation of the surface but doesn't specify which belongs to which orientation.

2. Dec 9, 2013

### Office_Shredder

Staff Emeritus
If you use z-f(x,y) then the gradient of g will be (df/dx,df/dy,1) and the unit normal will always be pointing upwards, if you use f(x,y)-z similarly you can see that the unit normal is always pointing downwards (the z-component of the gradient will be -1). Whether you want the unit normal pointing upwards or downwards depends on the context of the problem. For example if you are applying Stoke's theorem to calculate a line integral around the boundary of a surface, then whether you want it pointing up or down depends on which direction the integral is around the boundary.

3. Dec 9, 2013

### HallsofIvy

Staff Emeritus
At every point on any differentiable surface there are an infinite number of "normal" vectors, two of which are "unit normal" vectors. There exist a unit vector in each of the two directions, each of which is -1 times the other. That is why you need to specify a direction. How you do that depends upon how you are "given" the surface. "up" or "down" ("positive z" and "negative z") may make sense if there is a coordinate system associated to the surface. "Inner" and "outer" will always make sense for a sphere or a paraboloid but obviously wouldn't for a plane.