Undergrad Surface parametrization and its differential

[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:

S_r =(cos θ, sin θ,-sin θ)
S_θ =(-sinθ r, r cosθ, -r cos θ)

After doing the cross product I get:

S_r x S_θ = 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.

 

[Orodruin]

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

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.

 

[jonjacson]

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.

Maybe I didn't explain it well.

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

 

[jonjacson]

I add a question.

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?

 

[jonjacson]

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

 

[Orodruin]

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

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.

 

[jonjacson]

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.

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

Thanks