Understanding how to calculate surface area of parametrized surfaces

Question

I am trying to follow a derivation for surface area of a parameterized surface and my book does not explain the reasoning behind different steps.

I understand the derivation for surface area for a surface where $z=f(x,y)$ in which one can create a small parallelogram in the tangent plane to approximate the surface area beneath it. In this derivation, two vectors are crossed and those vectors are $\mathbf a=\Delta x\mathbf i + \frac{\partial f(x,y)}{\partial x}\Delta x\mathbf k$ and $\mathbf b =\Delta y\mathbf j + \frac{\partial f(x,y)}{\partial y}\Delta y\mathbf k$. I can follow this because I have done the cross product to get $\sqrt{1+\left(\frac{\partial f(x,y)}{\partial x}\right)^2 +\left(\frac{\partial f(x,y)}{\partial y}\right)^2 +1}$.

However I get confused with the derivation for the parameterized surface area. I know it has the same idea to use a cross product giving area of a parallelogram. My book says the 2 vectors that I cross are defined as $\frac{\partial}{\partial u}\Delta u$ and $\frac{\partial}{\partial v}\Delta v$ when $\mathbf r(u,v)=x(u,v)\mathbf i + y(u,v)\mathbf j + z(u,v)\mathbf k$.

I don't realy see where this is coming from because my book has two pictures... one of the graph of the surface which is in $x,y,z$ 3d space and then a view of the region below the surface which is depicted in terms of $u$ and $v$. I don't see how they connect.

In this first i can clearly see that when one variable is held constant we get a plane view of the other variables and I can use logic of differentials where $dy=f'(x)dx$ to find the $2$ vectors. I however can't seem to get this in the parameterized version. I know it should be simple because the first case is just a special case of the second but something isn't clicking. Any help is appreciated.

Answer 1

First we parameterize our surface by two families of curves $\{u=c\}$ and $\{v=d\}$:

The we can see that $$\mathbf r_u = \frac{\partial \mathbf r(u_0,v_0)}{\partial u} = \lim_{h\to 0} \frac{\mathbf r(u_0+h,v_0)-\mathbf r(u_0,v_0)}{h}$$

will be tangent to the curve $u=u_0$ at the point $(u_0,v_0)$ and thus also tangent to the surface. Likewise for $\mathbf r_v = \dfrac{\partial \mathbf r(u_0,v_0)}{\partial v}$.

Also notice that because of the way we choose our charts $\mathbf r_u$ and $\mathbf r_v$ will point in different directions. Thus they will be a basis for the tangent space at that point.

Then we realize that, in analogy to how every differentiable curve looks locally like a straight line, every surface looks locally like a plane:

And we know how to find the area of a planar region: we chop it into looks rectangles (or parallelograms) and find the area of each of those.

So that's what we do in the tangent space.

In the above image, we see a $\mathbf r(u_0 + \delta u, v_0) - \mathbf r(u_0,v_0)$. Using what we know of single variable calculus, partial derivatives, and vectors we can see that this is just $\mathbf r_u(u_0,v_0)du$. Likewise the other side of that parallelogram is $\mathbf r_v(u_0,v_0)dv$. Then to find the area of that we just take the norm of the cross product:

$$dA = \|\mathbf r_u(u_0,v_0)du\times \mathbf r_v(u_0,v_0)dv\| = \|\mathbf r_u(u_0,v_0)\times \mathbf r_v(u_0,v_0)(dudv)\| \\ = \|\mathbf r_u(u_0,v_0)\times \mathbf r_v(u_0,v_0)\||dudv| = \|\mathbf r_u(u_0,v_0)\times \mathbf r_v(u_0,v_0)\|dudv$$

where I was able to move the $du$ and $dv$ around and outside the norm because they are (positive) scalar differentials.

Then finally, we know that the surface area will just be the integral of all of the differential area elements:

$$A = \int_D dA = \int_D \|\mathbf r_u\times \mathbf r_v\|dudv$$

and we have our expresssion for surface area of a parameterized surface.

Answer 2

No no no! Everyone is talking about the perpendicular vector to the surface. What does this have to do with surface area.

If you have two perpendicular vectors $a\frac{d}{dx}$ and $b\frac{d}{dy}$ on the plane, then they make a rectangle of area $ab$.

What if you have two vectors $ a_1\frac{d}{dx}+b_1\frac{d}{dy}$ and $a_2\frac{d}{dx}+b_2\frac{d}{dy}$, then the parallelogram they make (on the plane) has area $\|(a_1\frac{d}{dx}+b_1\frac{d}{dy})\times(a_2\frac{d}{dx}+b_2\frac{d}{dy})\|, $ that is the length of their cross-product. This is high-school geometry, and does not deeply depend on our definition of cross-product or anything.

Case of a plane-to-plane mapping Now assume you have a small area on the plane, being distorted (smoothly). Say a circle being elongated, a material being sucked in by heat, or a fluid spreading out. This is saying that you have a map $(u,v) \longrightarrow (X(u,v),Y(u,v))$, $u,v$ being thought as the initial position of a point, ending up at position $(X(u,v),Y(u,v))$.

What happens for a small, tiny square with its lower-left corner at $(u,v)$? Look at the horizontal line passing through $(u,v)$, ie. fix $v$ and move to the right by varying $u$. The very small segment of the lower side of the square gets expanded/shrunk, and bent into a possibly curved segment on the $(X,Y)$ plane. But if we still approximate that little curve by a straight line, which is no more horizontal and ask: what is the ratio of the length of the new segment to the side of the square that gave birth to it? Well, by definition of partial derivative, the new segment agrees infinitely closely, both in size and direction, with the vector $X_u\vec{i}+Y_u\vec{j}$.

Similarly, the vertical side of the square gets mapped to a curve well approximated by the vector $X_v\vec{i}+Y_v\vec{j}$.

Now, how about the whole square? What happens to that? It gets mapped into a non-square region, but well approximated by the parallelogram formed by those two vectors above. And, how do we find the area of the parallelogram? As we said above, it equals the length of their crossproduct.

So, to calculate the surface area in $(X,Y)$ plane of the image of a $(u,v)$ region, we break up the $(u,v)$ region into these tiny squares, find the area that their image spans by above argument, and then add these all up. Then to better approximate, we break up into even smaller squares. This reminds us of integration:

$$Add \ up \ all \ areas \ over \ all \ (u,v) \ squares = \int_{uv-region}\ (the \ cross \ product) \ dudv$$.

Now, in case of mapping a $uv$-region into $R^3$, rather than $R^2$, notice that locally, we can approximate a part of the graph of the mapping by its tangent plane. So, essentially, we go back to plane-to-plane situation again.

I didn't intend to write this much. But I hope will help.