# How to extend the notion of pressure to lower dimensional spaces?

Consider the pressure at each point in three dimensional space being given by the scalar function $P(\vec{x})$. By definition, the total force perpendicular to a given surface $S$ in this space due to the ambient pressure is given by the surface integral $\vec{F} = -\hat{n}\int_SP(\vec{x})d\Sigma$ where $\hat{n}$ is the unit normal to the surface. This is fine and all, but how would one extend this notion to lower dimensions? In two dimensional space, for instance, would the expression become the line integral $\vec{F} = -\hat{n}\int_CP(\vec{x})ds$? In one space dimension would the expression be $F = -\int_a^bP(x)dx$ or would it reduce to $F = -P(x)$?

Edit: I mention the latter expression even though it is not an integral simply because I noticed that the units check out nicely. For instance, according to the definition in three dimensions pressure has units of $N/m^2 = J/m^3$. According to the suspected line integral definition in two dimensions, pressure would have units of $N/m = J/m^2$. Finally, according to the expression $F = -P(x)$ pressure would have units of $N = J/m$ in one space dimension. Any help in sorting this out would be appreciated.

There are two ways to interpret your question. One is to ignore the physical mechanism that produces pressure, and to define it as some force with which a fluid presses in a normal direction on either side of a surface drawn in the fluid. It will press with the same force if the fluid on one side of the surface is replaced by a solid object. The force on a three-dimensional object is $$\vec{F}=-\int\int p(\vec{x})d\vec{n}(\vec{x})$$ where $\vec{n}$ is a vector normal to the surface with the magnitude of an elemental area. (Im not sure why your normal is outside the integral) and here pressure is force/unit area.
Now consider an object in three-dimensional space where there is no dependence on one of the directions, like a tall flagpole, and assume that the situation is the same at all heights $z$ above the ground. The same formula can be used, with pressure now a function of $x,y$ only, $\vec{n}$ a vector in $x,y$, normal to the curve that defines the surface, with the magnitude of of an elemental length, and $\vec{F}$ a force per unit height. Pressure is still a force per unit area.
Next imagine a situation independent of two of the dimensions, like water in a hosepipe, so that pressure depends only on one coordinate, say $x$. Take a section of the pipe $x_a<x<x_b$. The force on that length of water is $$F=A(p(x_a)-p(x_b)).$$ where $A$ is the area of the hosepipe, $F$ is just a force and $p$ is still a force/unit area.