Area of sphere-continuum mechanics

[merav1985]

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

 

[wrobel]

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

 

[vanhees71]

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, $\vec{a}_n=\vec{n} \otimes \vec{n} \cdot \vec{a}=\vec{n} (\vec{n} \cdot \vec{a})$.

 

[merav1985]

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?

 

[wrobel]

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is determined by the equation $\boldsymbol r=\boldsymbol r(\theta,\phi)$ then the area element can be expressed in outdated terms as follows $dA=\Big|\frac{\partial\boldsymbol r}{\partial \theta}\times\frac{\partial\boldsymbol r}{\partial \phi}\Big|d\theta d\phi$.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

 

[Chestermiller]

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.

 

[vanhees71]

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.

 

[merav1985]

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.

 

[merav1985]

thank you all very much! you have all directed me to the right answer