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A Area of sphere-continuum mechanics

  1. Jul 10, 2016 #1
    suppose I have a sphere that has radius with the vlaue of 1.
    the integral is:∫∫n⊗ndA
    where dA = sinθdθdφ

    what is n⊗n????
    I'm supposed to the area of the sphere.

    This question was in an exam in continuum mechanics in the Technion
     
    Last edited: Jul 10, 2016
  2. jcsd
  3. Jul 10, 2016 #2
    It looks like the tensor product of the unit normal vector by itself. For which purpose such an expression is considered depends on the environment
     
  4. Jul 10, 2016 #3

    vanhees71

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    Welcome to Physics Forums!

    What's the context? Usually ##\vec{a} \otimes \vec{b}## denotes the tensor product of two vectors. In Cartesian components it denotes the tensor ##T_{jk}=a_j b_k##. For a unit vector ##\vec{n}## it's also the projection operator to this direction, ##\vec{a}_n=\vec{n} \otimes \vec{n} \cdot \vec{a}=\vec{n} (\vec{n} \cdot \vec{a})##.
     
  5. Jul 10, 2016 #4
    I'm supposed to find the area of the sphere.

    This question was in an exam in continuum mechanics in the Technion

    what is n in sphere coordinates?
     
  6. Jul 10, 2016 #5
    If the sphere is determined by the equation ##\boldsymbol r=\boldsymbol r(\theta,\phi)## then the area element can be expressed in outdated terms as follows ##dA=\Big|\frac{\partial\boldsymbol r}{\partial \theta}\times\frac{\partial\boldsymbol r}{\partial \phi}\Big|d\theta d\phi##. I can not find another reason to mention the normal vector in this context
     
  7. Jul 10, 2016 #6
    On the surface of a sphere with its center at the origin, n is the same as the unit vector in the spherical coordinate radial direction.
     
  8. Jul 10, 2016 #7

    vanhees71

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    A sphere can be parametrized by spherical coordinates (with the exception of the two points at the polar axis):
    $$\vec{x}(\vartheta,\varphi)=R \begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix}.$$
    The coordinate lines build a mesh, defining tangent vectors which span the surface-vector elements
    $$\mathrm{d} \vec{F} = \mathrm{d} \vartheta \mathrm{d} \varphi \frac{\partial \vec{x}}{\partial \vartheta} \times \frac{\partial \vec{x}}{\partial \varphi} = R^2 \sin \vartheta \begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix}.$$
    The surface normal vectors are the unit vectors in direction of ##\vec{x}##, i.e.,
    $$\vec{n} = \frac{\vec{x}}{|\vec{x}|}=\frac{\vec{x}}{R} = \begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix}.$$
    To evaluate the surface of the sphere you just take the modulus of the surface vectors and integrate over ##\vartheta \in [0,\pi]##, ##\varphi \in [0,2 \pi]##:
    $$S=\int_{0}^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{\varphi} R^2 \sin \vartheta=4 \pi R^2,$$
    as it should be.

    I'm still not sure, what all this has to do with your original question, where you look at a dyad (Kornecker product ##\vec{n} \otimes \vec{n}##), which is a 2nd-rank tensor by definition.
     
  9. Jul 10, 2016 #8
     
  10. Jul 10, 2016 #9
    thank you all very much! you have all directed me to the right answer
     
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