My vector calculus professor told me not to include jacobian variables when my double integrals were in polar coordinates, but I stumbled upon a problem that is solved using a jacobian anyway and I can't understand why that is.

The problem:

Find the surface area of the part of the sphere x^2 + y^2 +z^2 = 4 that lies aove the cone z = sqrt(x^2 + y^2)

I took the projection of the part of the sphere and found the region D x^2 + y^2 = 4 and used polar coordinates to find the limits of integration, but the solution still requires the "r" to be placed in the integral.

  • $\begingroup$ Did the below response help? $\endgroup$ – Faraad Armwood Apr 3 '17 at 2:56

$\textbf{Comment}$: A very simple answer to your question is, once you switched to polar coordinates to describe the domain of $(x,y)$, you switched coordinate systems. I give a more elaborate answer below, however here is an example in which no coordinate change occurs.

$\textbf{Problem}$: Calculate the area of the closed unit disk $E$. We can parametrize the disk using the domain $D = \{(r, \theta): r \in [0,1], \theta \in [0, 2 \pi]\}$. Thus, the area is given by the integral below.

$$\int_{r=0}^1 \int_{\theta = 0}^{2\pi} 1 \ d\theta \ dr = 2\pi$$

Pay close attention to this calculation. Observe that we can use $\sigma(r, \theta) = (r, \theta)$ (the identity) as a "parametrization" of $E$ since,

$$\left(\int_{\partial E} \|\sigma_r \times \sigma_{\theta}\| = 1 \right) = 0$$

Your teacher is correct. If you already parametrize a surface with some smooth map $\sigma(u,v)$ where $(u,v) \in D$ then we have,

$$\textbf{Surface Area}(S) = \int_D \|\sigma_u \times \sigma_v\| \ du \ dv$$

$\textbf{Integrals using parametrizations}$: In most calculus books, they allow parametrizations to be non-injective on $\partial D$ i.e the boundary of $D$. In essence this would mean $\sigma(D) = S$ where $S$ is your surface, but no compact surface arises in this fashion. However, most compact surfaces in these calculus texts will have parameterizations that vanish on the boundary i.e ,

$$\int_{\partial D} \|\sigma_u \times \sigma_v\| \ du \ dv = 0$$

and so $\sigma(\textbf{int}(D)) \subset S$ and $\sigma(D) = S$. The above now says that for special surfaces (almost all the ones in a calculus book) you can use $\sigma$ to parametrize all of $S$ since the boundary won't contribute to the integral.

\begin{align*} \int_D \|\sigma_u \times \sigma_v\| \ du \ dv &=\int_{\partial{D}} \|\sigma_u \times \sigma_v\| \ du \ dv + \int_{\textbf{int}(D)} \|\sigma_u \times \sigma_v\| \ du \ dv \\ \\ &=\int_{\textbf{int}(D)} \|\sigma_u \times \sigma_v\| \ du \ dv\end{align*}

$\textbf{Using Jacobian}$: Let us define $\textbf{n}(u,v) = \sigma_u \times \sigma_v$ i.e $\textbf{n}: D \to \mathbb{R}^3$. Suppose we start with $\sigma(u,v)$ as a parametrization of $S$ and want to switch to another $\psi: V \to S$ which also parametrizes $S$. The change of variables map is given by $ \phi:=\sigma \circ \psi^{-1} : (u,v) \mapsto (x,y)$ i.e $\phi: D \to V$ and hence by the change of variables formula,

$$\textbf{Surface Area}(S) = \int_{V} (\textbf{n} \circ \phi) \ \|\textbf{Jac}(D\phi)\| \ dx \ dy$$


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