Understanding the concept of the formula for surface area

[Xyius]

Hey all,

I was reviewing the concepts of how to derive the formula for the surface area, namely..

$$SA=\int \int \left\|\frac{\partial \widetilde{r}}{\partial u} \times \frac{\partial \widetilde{r}}{\partial v}\right\| dA$$

(I didn't know how to make vector arrows so I used the tilde sign.)

I understand the concept that in order to approximate a small section of the surface area, you compute the magnitude of the cross product to achieve the area of the parallelogram. However, the thing that is confusing me is, since the two vectors are the partial derivatives and give the rate of change at the point in question, aren't the lengths of the vectors (or the rates of change) changing at every point? So how can you calculate surface area when the parallelograms will be different sizes at each point?

 

[HallsofIvy]

Well, that is, after all, the whole point of Calculus! Using the derivatives allows us to work with things that are "different sizes at each point".

 

[Xyius]

No but I mean, at each point, it seems to me that the approximation could either be an over approximation or an under approximation depending on the lengths of the partial derivative vectors.

For example, say at point (x,y) the two partial derivatives have a HUGE rate of change, resulting in two derivative vectors that are very long. Would that not yield a HUGE area of the area of the parallelogram?

Obviously this doesn't happen but I want to understand what is wrong with my logic! :x

 

[arildno]

1. The surface area of the infinitesemal paralllellogram is
$$\left\|\frac{\partial \widetilde{r}}{\partial u} \times \frac{\partial \widetilde{r}}{\partial v}\right\| dA$$, where dA=du*dv, if a parallellogram is a mapping from a rectangle in the (u,v)-plane.

2. For a fixed angle between the two tangent vectors, this formula DOES mean that if the rates of change are large, then the same rectangle in the (u,v)-plane is mapped onto a larger parallellogram on the surface lying say, in the (x,y,z)-space.

 

[homology]

\vec{v} gives $$\vec{v}$$

 

[SteamKing]

Xylus: Remember, the integral used to calculate the surface area is the limit of a sum. The lengths of the differential vectors in the limit are not intended to be "very long", because if they are, the limit will probably not approach a fixed value.

 

[mathwonk]

I think its the dA that chops it down. I.e. dA = dX.dY and when you take the limit in the Riemann sum, the lengths dX and dY get smaller.

If you like, put the dX or I guess its the du, inside the absolute value, on the ∂r/∂u factor, and put the dv inside on the ∂r/∂v factor. then you see you are taking the area of a parallelogram whose sides are not ∂r/∂u and ∂r/dv,

but rather (∂r/∂u).du, and (∂r/∂v).dv. those are short vectors.

Basically the area of a rectangle in the u.v plane can be broken up into the sum of a lot of small rectangles, each of sides du and dv. The surface parametrized by u and v, breaks up also approximately this time, into the sum of a lot of little parallelograms, but all of different sizes. The partials ∂r/∂u, and ∂r/∂v (or their lengths) are multipliers that show how the lengths du and dv get stretched when they are mapped onto the surface.

then as halls said, the method of limits, let's you sum up all those different sized parallelograms to get the surface area in the limit.

good question by the way. to me that clunky formula has almost no visible insight in it whatsoever.

a slightly less mysterious way to write it would be as f'(du.e1)xf'(dv.e2), where f is the parametrization map, and f' is its linear approximation. I.e. the little rectangle in the u,v plane is spanned by du.e1, and dv.e2, so the image parallelogram is spanned by their images under f'. this still stinks though.

 

[Xyius]

Thanks everyone! I understand now :)