How to do the Jacobian of $\vec{r}(\theta, z) = \langle 2\cos(\theta), 2\sin(\theta), z \rangle$ I'm trying to find the following surface integral over the surface named $S_1$, the lateral side of the cylinder bounded by $x^2 + y^2 = 4$ and the planes $z = 0$ and $z = 1$:
$$\iint_{S_1} (x^2 + y^2 + z^2) \text{d}S$$
I asked a similar question here, and this question is part of a similar problem I was given in a similar context.
My goal was to put the integral in the following form: $$\int_a^b \int_c^d f(\theta,z)\,\mathrm{d}z
\, \mathrm{d}\theta$$
I parameterized this surface using the multivariable vector equation $\vec{r}(\theta, z) = \langle 2\cos\theta , 2\sin\theta, z \rangle$, with the restrictions $0 \leq \theta \leq 2\pi$ and $0 \leq z \leq 1$.
Using this parameterization, it's easy to visualize how $a=0$, $b=2\pi$, $c = 0$, and $d = 1$. Furthermore, $x^2 + y^2 + z^2 = 4 + z^2$. However, the real answer turns out to be
$$\int_0^{2\pi} \int_0^1 (4 + z^2)\cdot2 \,\mathrm{d}z \,\mathrm{d}\theta$$
I can only assume that this extra $2$ attached to the $\mathrm{d}z\,\mathrm{d}\theta$ term is derived somehow using the Jacobian of $\vec{r}$, but I have no idea where to start. How could I do the Jacobian so I can account for that $2$? In other words, how can I prove that, in this case,
$$\text{d}S = 2 \, \mathrm{d}z\,\mathrm{d}\theta$$
Using the Jacobian?
 A: A parametrization of a surface in $\mathbb{R}^3$ is a map
$U \to \mathbb{R}^3$ for some domain $U \subseteq \mathbb{R}^2$, say $(s, t) \mapsto (x, y, z)$. It's natural to integrate a vector field $\vec{F}$ over this surface, and if we must integrate a scalar field $f$, then the trick is to interpret it as a vector field that is everywhere normal to the surface, i.e. integrate $\vec{F} = f \, \hat{n}$.
Your intuition is correct: in order to pull the integral back to the parameter space, we have to insert a Jacobian factor. The technique to find the outward normal as the vector product
$$
\vec{n} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}s} 
\times \frac{\mathrm{d}\vec{r}}{\mathrm{d}t},  
$$
which has components
$$
\biggl\langle 
\frac{\mathrm{d}(y, z)}{\mathrm{d}(s, t)}, 
\frac{\mathrm{d}(z, x)}{\mathrm{d}(s, t)}, 
\frac{\mathrm{d}(x, y)}{\mathrm{d}(s, t)}
\biggr\rangle, 
$$
where
$$
\frac{\mathrm{d}(y, z)}{\mathrm{d}(s, t)} 
= \frac{\mathrm{d}y}{\mathrm{d}s} \frac{\mathrm{d}z}{\mathrm{d}t}
- \frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}z}{\mathrm{d}s}, 
$$
etc.
Note: the middle component is sometimes written in the equivalent form
$$
-\frac{\mathrm{d}(x, z)}{\mathrm{d}(s, t)}, 
$$
insisting on a lexicographic order on coordinates. But, I think the cyclic ordering makes the formulas cleaner and more intuitive, without needing an extra negative. (None of this will matter in the case where we integrate a scalar, as we only need the magnitude of the Jacobian, but it's worth understanding for integrating vector fields.)

For your example, we want $(s, t) = (\theta, z)$ and
$$ 
\langle x, y, z \rangle 
= \langle 2\cos\theta,\, 2\sin\theta,\, z \rangle. 
$$
You have the polar coordinates in the opposite order, but this results in an inward-pointing normal and would yield a global sign error. Now,
$$
\frac{\mathrm{d}(y, z)}{\mathrm{d}(\theta, z)} 
= \frac{\mathrm{d}y}{\mathrm{d}\theta} \frac{\mathrm{d}z}{\mathrm{d}z}
- \frac{\mathrm{d}y}{\mathrm{d}z} \frac{\mathrm{d}z}{\mathrm{d}\theta} 
= 2\cos\theta, 
$$
and you can calculate the other two coordinates similarly, yielding
$$
\vec{n} = \langle 2\cos\theta,\, 2\sin\theta,\, 0 \rangle. 
$$
Thus, when we pull back the integral to the parameter space,
$$
\mathrm{d}S = \bigl\lVert \vec{n} \bigr\rVert\, 
\mathrm{d}\theta\, \mathrm{d}z 
= 2\, \mathrm{d}\theta\, \mathrm{d}z. 
$$
Note: In this case, the magnitude of the Jacobian is a constant, but in general it is a function of the parameters.

Putting this all together,
$$
\iint_{S_1} (x^2 + y^2 + z^2) \, \mathrm{d}S 
= \int_0^1 \int_0^{2\pi} 
(4 + z^2)\, 2\, \mathrm{d}\theta\, \mathrm{d}z, 
$$
in agreement with your answer.
