- #1

jonjacson

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I will use an example:

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

y+z=2

-And the infinite cilinder:

x

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:

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

I will show what I mean.

1.- Using the first method:

S

S

After doing the cross product I get:

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

2.- WHat if I simply differenciate

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

We had

dS= (dr cos θ - sin θ dθ r)

I don't need to continue since the

I thought the differential of a parametrized surface is just the differential of its components but I don't know what to think now.

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

y+z=2

-And the infinite cilinder:

x

^{2}+y^{2}<=1We 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:

**= r**

SS

_{r}x S_{θ}**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.

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