I'm trying to evaluate $\iint_D \langle x,y,-2\rangle \textbf{n} \cdot \textbf{dS} $, where $D$ is given by $z=1- x^2 - y^2$, $x^2+y^2 \leq 1$, oriented up.

Now, I'm trying to find a parameterization for D but I'm struggling to do so. Graphically, we have a paraboloid with maximum $z=1$ and a cylinder with radius $\leq 1$. So that means I'm trying to parameterize an area that is like a hemisphere. Can I use spherical coordinates then such that:

$$r(\phi,\theta) = \langle r\sin(\phi)\cos(\theta), r\sin(\phi) \sin(\theta), r\cos(\phi), \quad 0\leq \phi \leq \frac \pi 2, \quad 0 \leq \theta \leq 2\pi $$

Then I can find the flux using the formula $\iint\textbf{F}\cdot(r_\phi \times r_\theta) \, dA $.

P.S. I'm weak at parameterizing..

  • $\begingroup$ Proper notation is $\langle x,y\rangle,$ not $<x,y>,$ and $r\sin\alpha,$ not $r sin \alpha.$ I edited accordingly and also changed $\int\int$ to $\iint. \qquad$ $\endgroup$ – Michael Hardy Mar 24 '17 at 1:56
  • $\begingroup$ Thanks Michael! Will be more pedantic about it next time. $\endgroup$ – misheekoh Mar 24 '17 at 2:01

You have proposed spherical (and not polar) coordinates, which I do think are appropriate here. I would suggest the simple following parameterization: $$ \begin{cases} x=x\\ y=y\quad \qquad\qquad\text{with } (x,y)\in D = \left\{(r,\theta)\mid r\in [0,1], \theta \in [0,2\pi] \right\}\\ z=1-x^2-y^2 \end{cases} $$

The advantage of this parametrization is the fact that it is straightforward to compute $$ r_x \times r_y = \pmatrix{1\\0\\-2x} \times \pmatrix{0\\1\\-2y} = \pmatrix{2x\\2y \\1} $$ The integral thus simplifies to $$ \iint_D \pmatrix{x\\y\\-2} \cdot \pmatrix{2x\\2y \\1}\; dA = 2 \iint_D x^2+y^2 -1\; dA = 2 \iint_D(r^2 -1)r\; dr d\theta $$

  • $\begingroup$ Sorry! Typo. Yes, spherical coordinates. $\endgroup$ – misheekoh Mar 24 '17 at 1:52
  • $\begingroup$ @Michael Hardy I am curious: is there a difference between \mbox{} and \text{} ? I was not aware of \mid, thanks! $\endgroup$ – Kuifje Mar 24 '17 at 1:57
  • $\begingroup$ @Kuifje : A first little experiment: $$\mbox{this is only one \\ abcd $\iiint$}$$ $$\text{this is only one \\ abcd $\iiint$}$$ The code in the two lines above is identical except for \mbox{} versus \text{} $\endgroup$ – Michael Hardy Mar 24 '17 at 2:05
  • $\begingroup$ $$\int_\mbox{region} \qquad \int_\text{region}$$ Notice the way \mbox{} and \text{} are used in the foregoing line. @Kuifje $\endgroup$ – Michael Hardy Mar 24 '17 at 2:07

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