[Math] Surface Element ($dS$) of a surface integral over a sphere

multivariable-calculus

I am having difficulty with the following snippet from my multi-variable calculus textbook:
enter image description here

From what I understand, if $f(x,y,z)$ is a vector field on $\Bbb R^3$, then the surface integral of $f$ over a sphere of radius $R$ (centered at the origin), parametrized with spherical co-ordinates with the function $g(\theta,\phi)$, is $\int \int_{S} f \cdot dS$, where $dS = \frac {\partial g}{\partial \theta} \times \frac {\partial g} {\partial \phi} d\theta d\phi $

( ** in the textbook above it is written $T_u , T_v$ which are equivalent to my notation of $ \frac {\partial g}{\partial \theta}, \frac {\partial g} {\partial \phi} $ **). I know that $T_u \times T_v$ is equal to $rR sin \phi$, where $r (x,y,z) = xi + yj + zk$.

Is there an error with this last line since it seems to me that second equality is a scalar, whereas the first, third and fourth are vectors. Also how can I relate my definition of $dS$ with the one described in the picture above. Thanks.

Best Answer

Short answer: yes, something got muddled in the text you quoted. For the purposes of integrating using spherical coordinates, stick in a "$\mathbf{n}\ \cdot\ $" into the third and fourth expression.

Are you sure about your formula? I think the text is correct, with $T_\theta \times T_\phi = rR\sin\phi$ (quick sanity check: the result should have units of length squared).

Long answer: the real story here is that $dS$ is a two-form on the surface of the sphere. It can also be represented by a scalar (via the Hodge dual) or a vector in $\mathbb{R}^3$ (by pulling back to the ambient space and taking the Hodge dual there). Somehow the text is mixing together all of these options.