## Surface Area, Another Formula

### Another Formula

In the introduction, we developed a formula for surface area, as a double integral. The surface was defined as a function of x and y, like a roof over your head. But what if the surface is not that simple? What if z is not always a function of x and y?

Let a surface in 3 space be defined by the functions x(u,v) y(u,v) and z(u,v). Assume this map, from 2 space into 3 space, is piecewise differentiable. What is the area of this surface, floating in space?

In the original formulation, small tiles on the floor were projected onto the roof, and these small roof areas were added up. The limit of this two-dimensional sum, i.e. the double integral, gives the surface area. The process is essentially the same here, but the projection is not straight up from floor to ceiling. Instead, a small square in the uv plane is mapped onto a parallelogram in xyz space. The two vectors that define this parallelogram are determined by the partials of x y and z with respect to u and v.

x∂u, y∂u, z∂u
x∂v, y∂v, z∂v

Take the length of the cross product of these two vectors to find the area of the parallelogram. This produces the following double integral over the appropriate region in the uv plane.

Area = ∫∫sqrt( (y∂u×z∂v-z∂u×y∂v)2 +
(z∂u×x∂v-x∂u×z∂v)2 +
(x∂u×y∂v-y∂u×x∂v)2 )

If the surface is magnified by a scaling factor k, the partial derivatives in the above equation are scaled by k, and the surface area is scaled by k2 as expected.

Let's examine a special case of the general formula for surface area. Let x = u, y = v, and z = f(u,v). Substitute in the above and get the "roof" formula that was presented in the introduction.

Area = ∫∫sqrt(1 + (f∂x)2 + (f∂y)2)

As a final example, let's compute the area of the sphere, and I promise, this is the last time we're going to do this. The region in the uv plane is a rectangle, 2π by π. Here is the map.

x ← cos(u)sin(v)
y ← sin(u)sin(v)
z ← -cos(v)

Here are the vectors that build the parallelogram, along with their cross product. Following the standard convention, the cross product is the one on top.

 cross product cos(u)sin(v)2 sin(u)sin(v)2 -cos(v)sin(v) ∂u -sin(u)sin(v) cos(u)sin(v) 0 ∂v cos(u)cos(v) sin(u)cos(v) sin(v)

Remember, the top vector is the cross product. Its length is sqrt(sin(v)4+cos(v)2sin(v)2), or simply sin(v). The integral becomes -cos(v), which is the z coordinate on the sphere. Thus the surface area is -2πcos(v). We saw this before; the surface area of a band on the sphere is proportional to the difference between the z coordinates of the latitude lines.

### Reparameterization

When we defined arc length, line integral, path integral, curvature, and other properties of curves in space, we showed these properties were independent of the parameterization. The arc length depends only on the curve, not on the map that caries the unit interval onto the curve. We would like to do the same for surfaces. The area of a surface shouldn't depend on the map that carries a region in the plane onto the surface. Area is area and that's the end of it.

Let x(u,v) y(u,v) z(u,v) be a map from a region in the uv plane onto a surface in xyz space. A reparameterization replaces u with u(s,t) and v with v(s,t). Thus the same surface is represented by three functions of s and t. Will the surface area come out the same?

After reparameterization, the surface is now a function of s and t. Apply the earlier formula. When you compute the partial of x with respect to s, for instance, use the chain rule.

x∂s = x∂u×u∂s + x∂v×v∂s

Write an expression like the above for the partials of x y and z with respect to s and t. Take the cross product, square the components, add them up, and take the square root. That's a lot of algebra, and it's probably best left to a computer. So I'll jump straight to the anser. The result is the original formula for surface area, times the jacobian produced by the partials of u and v with respect to s and t. But aha, that's just integration by substitution. In other words, the surface area is the same whether we integrate in the uv plane or in the st plane. Surface area, like arc length, is well defined.

A geometric argument is somewhat more intuitive. Recall that x y and z have become functions of s and t. Start with a tiny square of size δ in the st plane. This becomes an approximate parallelogram in the uv plane, and its area is multiplied by the jacobian of the map from st to uv. As δ approaches 0, the ever-shrinking parallelograms in the uv region form a generalized Riemann net, that acts as a base for the integral of any function of u and v. Take the next step, from uv to xyz. If a tile in the uv plane is small enough, be it square or parallelogram or whatever, the integral of f over that tile is approximately the area of the tile times f. In this case f is the original surface area function. It is multiplied by the area of the parallelogram, which is the jacobian from st to uv. Therefore the surface area, evaluated as a function of s and t, is the jacobian from st to uv times the surface area evaluated as a function of u and v. Again, this is merely integration by substitution; both formulas produce the same result.

### Higher Dimensions

Surface area can be generalized to higher dimensions.

The earlier formulas examined surfaces in 3 space. Thus a region in the 2 dimensional plane was mapped onto a surface floating in space. And if that surface was a roof over the region in the plane, the formula for surface area could be simplified. Similar formulas exist for n-1 dimensional surfaces in n dimensional space.

Let f be a differentiable function that maps a reagion in n-1 space onto a surface embedded in n space. Thus there are n-1 input variables and n output variables. Take all the partials, giving an n-1 by n matrix. Complete this matrix by taking the n dimensional cross product. The resulting vector measures the size of the parallelatope, i.e. the area of the surface at that point. Take the length of the cross product and integrate over the region in n-1 space. That's all there is to it, though the algebra can get out of hand very quickly.

The earlier geometric proof, employing jacobians and volume, can be generalized to n-1 dimensions. In other words, we can reparameterize the surface and it will still have the same area. The surface area is determined by the shape, rather than the precise formula that is used to describe the shape.

As we saw before, the formula simplifies when the surface is a roof suspended over the region of integration. In other words, f maps the first n-1 coordinates onto themselves, and computes the nth coordinate as a function g of the first n-1 variables. Here g defines the roof, and the other coordinates are unchanged as the tiles on the floor are projected up onto the ceiling. Almost all partials drop to zero, and the cross product is easily calculated.

Let's illustrate with a 4 dimensional example. The surface is embedded in wxyz space, and in this case it is defined by a roof function w = g(x,y,z), over a region in xyz space. Here is a full description of the surface, followed by a matrix of its partials.

x(x,y,z) ← x
y(x,y,z) ← y
z(x,y,z) ← z
w(x,y,z) ← g(x,y,z)

x, y, z, w
1, 0, 0, g∂x
0, 1, 0, g∂y
0, 0, 1, g∂z

The last component in the cross product is 1, produced by multiplying down the main diagonal. The other components are the partials of g. Square these partials, add 1, and take the square root. This is the length of the cross product, which gives the size of the parallelatope, which measures the amount of "roof" over a small tile of size δ. Integrate this expression to find surface area.

Let's find the surface area of the 3 sphere embedded in four dimensions. Consider the northern hemisphere, a roof over the region x2+y2+z2 ≤ 1. The function is g(x,y,z) = sqrt(1-x2-y2-z2). The partial with respect to x is -x/g, and similarly for y and z. Square these, add them up, and add 1, giving 1/g2. Take the square root to get 1/g. In other words, the surface area of the top half of a sphere, defined by g(x,y,z), is equal to the volume under the inverse function 1/g. And this works in all dimensions.

Return to our four dimensional example and switch to spherical coordinates. Now the integrand becomes sin(φ)r2/g. Replace r2 with 1-g2 and integrate g and 1/g separately, as r runs from 0 to 1. The result is π/4. Bring in a factor of 2 as we integrate with respect to φ, and a factor of 2π for θ. Thus the surface area of the northern hemisphere is π2. Double this to get the surface area of the entire sphere, which agrees with the value given in the table of hyperspheres.