Surface Area, An Introduction

Introduction

You are in a rectangular room with a curvy roof overhead.  You want to paint the roof, or apply shingles.  What is the surface area of the roof?

Start by looking at the tiles on the floor.  Each tile is one centimeter square, and all the tiles together establish the area of the floor.  That was easy, but what about the roof?

Project each tile up to the roof.  If the roof happens to be flat and level above a certain tile, its area is one square centimeter.  If the roof is relatively flat, but tilted at an angle of 45, the amount of "roof" over that tile is sqrt(2)cm2.  If the roof is nearly vertical, heading straight up, the area over that small tile could be quite large.  Add up the areas above all the tiles and find an approximation to the area of the roof.

Choose smaller tiles, and the approximation improves.  Take the limit, and the surface area becomes a double integral.

But the roof, a function of x and y, has to have a few nice properties for this to work.  specifically, the function must be piecewise differentiable.  Don't worry about the piecewise adverbe; that just means you can do the roof in sections.  Find the area of the main roof, then the area of the section over the attic, and add them together.

Differentiability is a little trickier, because we are talking about a two dimensional function.  Partial derivatives are not sufficient; we need a function that is totally differentiable.  This means every point on the root looks like a plane, if you are small enough.  The plane might be tilted relative to the ground, i.e. the roof could be sloping upward at that point, but there are no obvious wiggles or bumps.  It looks like a flat plane.  Consider a sphere, like the dome of the U.S. capitol building.  From far away it is obviously curved, but if you are a fly, sitting on the dome, it appears flat, without bumps or curves.

Since the surface function looks like a smooth plane close up, our approximations, based on smaller and smaller tiles, do indeed approach the surface area.  There's a lot of δ ε algebra behind this statement, and I'm not going to present it here.  You saw some of it when we defined the length of a curve in space.  Well this is even more complicated, because it is a 2 dimensional surface, and I don't think the detailed proof gives that much more insight.

So, how much "surface" is over a given tile?  Let f(x,y) define the surface, where f is everywhere differentiable.  Consider a small tile, δ on a side, with its lower left corner at x,y.  Project this tile up to the surface.  Its four corners map to f(x,y), f(x+δ,y), f(x,y+δ), and f(x+δ,y+δ).  But at these dimensions, f is practically a plane.  The point f(x+δ,y) is basically f(x,y) plus δ times the partial of f with respect to x.  Similarly, f(x,y+δ) is f(x,y) + δ times the partial of f with respect to y.  When projected up to the surface, our small square, δ on a side, becomes a parallelogram in 3 space.  The two vectors that define this parallelogram are presented below.

δ, 0, δf∂x
0, δ, δf∂y

As you may recall, the area of a parallelogram is the length of the cross product of the two vectors that define it.  Therefore the area of the surface over a given tile of size δ is approximately δ2 times the length of the vector (f∂x,f∂y,1).  Square these components, add them up, and take the square root to get the length of the cross product.  In the limit, the surface area is given by the following integral.

Area = ∫∫sqrt(1 + (f∂x)2 + (f∂y)2)

Let's try an example.  What is the area of the unit sphere?  The surface has the equation f(x,y) = sqrt(1-x2-y2).  Now the partials are -x/f and -y/f.  Square these and add them together, add 1, and take the square root, giving 1/f.  In other words, we are integrating 1/sqrt(1-x2-y2) over the unit circle.  Switch to polar coordinates and integrate r/sqrt(1-r2), giving -sqrt(1-r2).  Evaluate as r runs from 0 to 1, and θ runs from 0 to 2π.  The result is 2π, and this is the upper half of the sphere.  Therefore the surface area of the entire sphere is 4π.  If the sphere has radius a, scale the integral, and the surface area is 4πa2.

Let's find the area of the arctic circle (on the unit sphere).  find the radius r that correspondes to 66.5 north latitude, and the answer is 2π times 1-sqrt(1-r2).  But aha, the second factor is merely the difference in z coordinates.  The surface area trapped between to lines of latitude is 2π times their difference in z coordinates.  Go from the south pole at z = -1 to the north pole at z = 1 and get 4π, the area of the entire sphere.  If the sphere has radius a, make these calculations on the unit sphere, then multiply by a2.

Iimagine slowly lowering a sphere into a lake.  Surface area disappears under the water at a constant rate.