It Smells Vaguely of Physics

In physics, the force per unit surface area (pressure) exerted on a body submerged in fluid is directly proportional to the depth of the object within the fluid. The deeper, the higher the pressure. The pressure force is exerted normally to the surface and toward the interior of the object. The units of pressure are $\frac{N}{m^{2}}$ . The net result of this pressure is a "buoyant force" exerted on the object in opposition to gravity ("upward"). The magnitude of the buoyant force is directly proportional to the volume of the submerged object (a common assumption being that the surrounding fluid is of uniform density).

I thought it would be fun to see if a version of this would still hold if the dimensionality was decreased by one. In 2D space, the "pressure" on a circle of radius $R$ centered on the origin is represented by $(P - R\,sin\theta = P - y)$ , where P is the reference pressure at $y=0$ . $\frac{d(Pressure)}{dy}$ is thus a negative constant, similarly to the 3D case. In 2D, instead of the "pressure" having units of $\frac{N}{m^{2}}$ , it now has units of $\frac{N}{m}$ . The magnitude of the force on an infinitesimal piece of arc is thus the "pressure" multiplied by the differential arc length $R\,d{\theta}$ . In the 3D case (for comparison), the differential force is equal to the pressure multiplied by the differential surface area. Lastly, the direction of the force vector is described by $(-cos\theta \hat{\imath} - sin\theta \hat{\jmath})$ , indicating that the force is normal to the surface and is directed toward the interior of the circle.

Now to evaluate the integral for the net force:

$\large F = \int_0^{2 \pi} (P - R\,sin\theta) (R\, d{\theta}) (-cos\theta \hat{\imath} - sin\theta \hat{\jmath})$ $=\large R \int_0^{2 \pi} (-Pcos\theta \hat{\imath} - Psin\theta \hat{\jmath} + R\,sin\theta\, cos\theta \hat{\imath} + R\,sin^{2}\theta\hat{\jmath} )d\theta$

The $cos\theta$ , $sin\theta$ , and $sin\theta cos\theta$ terms evaluate to zero over $(0,2\pi)$ , leaving:

$\large F = R \int_0^{2 \pi} (R\,sin^{2}\theta\,\hat{\jmath} )d\theta = \pi R^{2}\hat{\jmath}$

So we again have a buoyant force which is purely vertical (in opposition to the pressure gradient). Except now, the buoyant force magnitude is directly proportional to the area of the submerged shape in the 2D case, whereas the force is proportional to the submerged volume in the 3D case. It appears to be a very viable analogy.