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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.

Martin Sleziak
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George1811
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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.

Andrew D. Hwang
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