# Surface integrals and Jacobian

When do we need to use the Jacobian for surface integrals? I'm asking because when you parametrize a surface $S\in \mathbb{R}^3$ then surely there is no need to change coordinates (essentially parametrising again) to find a simple integral. Can anyone give me an instance where the intial parametrisation does not give a 'simple' enough integral. For example:

Calculate the flux of $\textbf{v}=x\textbf{i}+y\textbf{j}+z\textbf{k}$ out of the surface $S$ where $S$ is the portion of the elliptic paraboloid $z= 9-(x^2+y^2)$ for which $z\geq 0$

Parametrising $\textbf{t}(r,\theta)=\begin{pmatrix} r\cos{\theta} \\ r\sin{\theta} \\ 9-r^2 \end{pmatrix}$. We find the normal as $$\textbf{N}(r,\theta)=\begin{pmatrix} 2r^2\cos{\theta} \\ 2r^2\sin{\theta} \\ r \end{pmatrix}$$ Therefore the integral becomes: $$\iint_\Omega r^3+9r \;drd\theta$$ Now I'm not quite sure of the limits for the parametrised surface $\Omega$. I'd say they are $r:[0,3]$ and $\theta:[0,2\pi]$. This then gives and answer of $$\frac{243\pi}{2}$$ Is this right? And when would I need the Jacobian? That example (provided I've done it right) didn't need it.

Your calculation is correct. The Jacobian is "hidden" in your choice of normal vector. Using subscripts to denote partial derivatives, you've used $$\mathbf{N}(r, \theta) = \mathbf{t}_{r} \times \mathbf{t}_{\theta} = \frac{\mathbf{t}_{r} \times \mathbf{t}_{\theta}} {\|\mathbf{t}_{r} \times \mathbf{t}_{\theta}\|} \cdot \|\mathbf{t}_{r} \times \mathbf{t}_{\theta}\|.$$
If you had used a unit normal field $\mathbf{n}(r, \theta)$ to express the integrand $\mathbf{v} \cdot \mathbf{n}$, the Jacobian factor $\|\mathbf{t}_{r} \times \mathbf{t}_{\theta}\|$ would have had to be included manually.