# I Surface parametrization and its differential

1. Jan 29, 2017

### jonjacson

I will use an example:

-The surface is given by the intersection of the plane:

y+z=2

-And the infinite cilinder:

x2+y2<=1

We want to parametrize this surface, it could be done easily with:

x=r cosθ
y=r sin θ
z=2 - r cos θ

Then this surface could be written using vector notation:

S= r cosθ i + r sin θ j + (2 - r sin θ)k

I understand that these are a set of vectors starting from the origin and ending at the surface, so nothing has changed here, it is just a compact form of the same surface.

Then the normal vector is calculated first finding the tangent vectors on r and θ and then taking the cross product and multiplying this by dr dθ.

What I don't understand is, if I simply differenciate S, Why don't I get the same dS?

I will show what I mean.
1.- Using the first method:

Sr =(cos θ, sin θ,-sin θ)
Sθ =(-sinθ r, r cosθ, -r cos θ)

After doing the cross product I get:

Sr x Sθ
= r j + r k

This vector is normal to the surface, and it must be multiplied by dr dθ.

2.- WHat if I simply differenciate S? Shouldn't I get the same result as in part 1?

The differential of the vectors i, j, k is zero because they don't change.

We had S= r cosθ i + r sin θ j + (2 - r sin θ)k, so:

dS= (dr cos θ - sin θ dθ r) i + ...

I don't need to continue since the i part is not zero, but in the previous calculation it was zero. What is wrong with the normal process of differenciation? Shouldn't it give us a differential of surface?
I thought the differential of a parametrized surface is just the differential of its components but I don't know what to think now.

Last edited: Jan 29, 2017
2. Jan 29, 2017

### Orodruin

Staff Emeritus
This is not describing a surface. It is the intersection between two two-dimensional surfaces in three dimensions - which is typically a one-dimensional object.

3. Jan 29, 2017

### jonjacson

Maybe I didn't explain it well.

The cilinder is solid, is not empty, so the intersection is a kind of inclined circle.

4. Jan 29, 2017

### jonjacson

If we have an area in the plane, the differential of this area is simply ds * dt, if there is a parametrized surface that arises from this plane area to form a 2d manifold in 3d space, the surface corresponding to ds * dt is:

Modulus (Tangent vector to s x Tangent vector to t) * ds * dt

Where x is the cross product.

What I don't understand is, Why is it not enough with the modulus of the cross product? Why do we multiply by ds and dt?

If I calculate the tangent vectors at a point and I calculate the cross product I get an area right?

5. Jan 30, 2017

### jonjacson

In other words, Why tangent vectors are not divided by their own modulus so their modulus becomes 1?

6. Jan 30, 2017

### Orodruin

Staff Emeritus
Because the modulus of the tangent vectors tell you how far you move when the parameters change and the area you get by a certain parameter change is proportional to how far the parameter change takes you.

It is the same reason that you get
$$d\vec x = \frac{d\vec x}{dt} dt$$
in a line integral.

7. Jan 30, 2017

### jonjacson

I understand, the tangent vectors are the rate of change, multiplied by a dx you get a dy.

Thanks