# Surface Area

Now that we have some understanding of integrals and of parametric surfaces, we can apply that knowledge to find the surface area of a parametric surface. To do this, we need to think of a way to split the surface up into lots of very small pieces that we can easily calculate the surface area of, then add them all up to find the total surface area over the whole surface.

To start out, consider the surface area of a very small, non-curved horizontal rectangle. That rectangle is just going to have an area of its width times its height.

Unfortunately, we cannot jump directly from knowing the surface area of a rectangle to finding the surface area of something that is not rectangular in nature. Rectangles do not really tesselate all that well into non-rectangular shapes and fixed bounds of integration only make sense if the candidate shape is rectangular. For most surfaces, the bounds of the shape change depending on where you are in it.

We can get around this problem by shifting around the orientation of the vectors making up the lines of each small patch by changing our co-ordinate system, in the same way that we define a parametric surface. However, if we do that then, it becomes harder to conceptualize what the surface area of the shape will be. The area does not remain constant, it actually changes as the shape contracts and expands as the orientation of the lines change around.

Another way to think about area, then, is with the normal vector. On face value these two things do not seem to be related. After all the normal vector is just a vector that sticks out from the shape, perpendicular to its surface.

But remember that the normal vector has a magnitude like any other vector and its magnitude is actually proportionate to the lengths of the vectors that made up its cross product.

${x}_{1}=$
${y}_{1}=$
${z}_{1}=$
${x}_{2}=$
${y}_{2}=$
${z}_{2}=$
$|N|=$1.41

As it turns out, the normal vector becomes a proxy for working out the area of a quadrialteral defined by two vectors. If the normal vector is short, then the area will be quite small. If the normal vector is long, then the area will be larger.

${x}_{1}=$
${y}_{1}=$
${z}_{1}=$
${x}_{2}=$
${y}_{2}=$
${z}_{2}=$
$|N|=$1.41

Astute readers may recognize the link between the normal vector defined as the cross product between two vectors and the determinant of a square 3x3 matrix. This is because the determinant of a 3x3 matrix will compute the volume of a parallelepiped having the basis vectors $a$, $b$ and $c$ as edges.

$det\left(\begin{array}{ccc}{a}_{1}& {a}_{2}& {a}_{3}\\ {b}_{1}& {b}_{2}& {b}_{3}\\ {c}_{1}& {c}_{2}& {c}_{3}\end{array}\right)={c}_{1}\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right)-{c}_{2}\left({a}_{1}{b}_{3}-{a}_{3}{b}_{1}\right)+{c}_{3}\left({a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right)$$‖\left(a×b\right)\cdot c‖={c}_{1}\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right)-{c}_{2}\left({a}_{1}{b}_{3}-{a}_{3}{b}_{1}\right)+{c}_{3}\left({a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right)$

So if we divide by $c$ and drop the depth dimension, we are just left with area, which is the cross product between$a$ and $b$

Back to surface areas, we reformulate the problem in another way - divide up the surface into small patches and then add up the magnitude of their normal vectors. That will give you the surface area!

Unfortunately, we now have another problem, which is how exactly those small patches should be defined. We cannot just pick one set of orientations for our vectors and expect them to tesselate perfectly throughout the shape as it is possible that the surface is curved and that tesselation will not work.

Instead, we come up with a mechanism to find the vector orientations at each point. Calculus comes to the rescue! Since our parametric surface is a function of two parameters, $u$ and $v$ for each of those two co-ordinates, outputs a three co-ordinate vector $\left(x,y,z\right)$, we can find the rate of change of both $u$ and $v$, with respect to each of $\left(x,y,z\right)$ and that will give us the two gradient vectors that we can use to calculate the normal, and thus the surface area at that point:

${S}_{u}=\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }u},\frac{\mathrm{\partial }y}{\mathrm{\partial }u},\frac{\mathrm{\partial }z}{\mathrm{\partial }u}\right)$${S}_{v}=\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }v},\frac{\mathrm{\partial }y}{\mathrm{\partial }v},\frac{\mathrm{\partial }z}{\mathrm{\partial }v}\right)$

Once we have those two, we can calculate the normal:

$N={S}_{u}×{S}_{v}$

And then from there, we can calculate the magnitude of the normal:

$\sqrt{{N}_{x}^{2}+{N}_{y}^{2}+{N}_{z}^{2}}$

All of this working then tells us how we can integrate

${\int }_{0}^{1}{\int }_{0}^{1}‖{S}_{u}×{S}_{v}‖\phantom{\rule{thinmathspace}{0ex}}du\phantom{\rule{thinmathspace}{0ex}}dv$

We will give it a try on the function defined above. Our parametric function was:

$S\left(u,v\right)=\left(3\mathrm{sin}\left(u+v{\right)}^{2},3\mathrm{sin}\left(v+\frac{u}{4}{\right)}^{2},0\right)$

First, we find the gradient vector in terms of $u$ and $v$.

${S}_{u}=\left(6\mathrm{sin}\left(u+v\right),\frac{3}{2}\mathrm{sin}\left(v+\frac{u}{4}\right),0\right)$${S}_{v}=\left(6\mathrm{sin}\left(u+v\right),6\mathrm{sin}\left(v+\frac{u}{4}\right),0\right)$

Now that we know that, we can take the cross product

${S}_{u}×{S}_{v}=\left(0,0,27\mathrm{sin}\left(u+v\right)\mathrm{sin}\left(u+\frac{v}{4}\right)\right)$

And we can then integrate over the magnitude of it:

${\int }_{0}^{1}{\int }_{0}^{1}\sqrt{\left(27\mathrm{sin}\left(u+v\right)\mathrm{sin}\left(u+\frac{v}{4}{\right)}^{2}}du\phantom{\rule{thinmathspace}{0ex}}dv=12.8456$