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I Surface parametrization and its differential

  1. Jan 29, 2017 #1
    I will use an example:

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


    -And the infinite cilinder:


    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. jcsd
  3. Jan 29, 2017 #2


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    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.
  4. Jan 29, 2017 #3
    Maybe I didn't explain it well.

    The cilinder is solid, is not empty, so the intersection is a kind of inclined circle.
  5. Jan 29, 2017 #4
    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?
  6. Jan 30, 2017 #5
    In other words, Why tangent vectors are not divided by their own modulus so their modulus becomes 1?
  7. Jan 30, 2017 #6


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    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.
  8. Jan 30, 2017 #7
    I understand, the tangent vectors are the rate of change, multiplied by a dx you get a dy.

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