Confused on the $dS$ component of a surface integral

Question

I feel bad about posting an external link, but I was watching a video and the instructor parametrized a hemisphere using spherical coordinates.
Now of course, if we were to transform a region of integration, we would multiply the jacobian in the integrand (namely, $\sin \phi$ for a hemisphere of radius 1).
But... in a surface integral, when we parametrize, we don't require multiplying the jacboian right?
As in: Evaluating a surface integral which simplified to
$$\iint_{S} z dS$$
where we are in spherical coordinates and so $z = \cos \phi$,
we would simply "expand" the differential $dS$ to be equal to $1 d\theta d\phi$ right?
The video suggested otherwise and expanded it to $dS = \sin \phi d\theta d\phi$ which is the jacobian.
Video in reference: https://www.youtube.com/watch?v=2ieG1ka5pBw
at 13:40.

Answer 1

Let's approach it another way. The surface of the unit sphere can be parametrised as

$$ z=g(x,y)=\sqrt{1-x^2-y^2}=\sqrt{1-\sin^2\phi} $$

the differential area element is

$$ {\text d}S=\|{\bf r_x}\times{\bf r_y}\|{\text d}x{\text d}y $$ where $\bf r$ is the position vector of any point on the surface of the unit sphere:

$$ {\bf r}=x{\bf i}+y{\bf i}+z(x,y){\bf k} $$

and ${\bf r_x}$ and ${\bf r_y}$ are the tangent vectors at this point - partial derivatives of the position vector with respect to the parameters $x$ and $y$ which are in turn parametrised according to (in your notation)

$$x=\sin\phi\cos\theta$$

and

$$y=\sin\phi\sin\theta$$

If you carefully work through the logic above, you'll get the result in the MIT lecture.

Answer 2

It depends which Jacobian you mean. We parametrise the surface with a map $s: \mathbb{R}^2 \supseteq U \to S \subset \mathbb{R}^3 $, and the the surface integral has $dS = \det{J} \, du \, dv$, where $J$ is the Jacobian of $s$. It must be this: a surface integral still involves integrating over a region, it's just a region with different dimension from a volume. Equally, a line integral $\int_C f(\mathbf{x}) \, d\mathbf{x} $ is defined by using a parametrisation $\gamma: [a,b] \to C$; here the Jacobian is just the derivative $\gamma'$.