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Here's the problem: Use Gauss's divergence theorem to calculate the total flux through the solid,$V$, enclosed by the set $M=\{(x,y,z)\in \mathbb{R^3}: z=x^2+y^2, 0\leq z\leq 1\}$ and the vector field $X:\mathbb{R^3} \to \mathbb{R^{3\times 1}}, (x,y,z)\to \begin{bmatrix} xz \\ z \\ -\frac{z^2}{2}\end{bmatrix}$.

My attempt: $$\Phi_X(M)=\int_Vdiv(X)dV$$

A parametrization for M is: $$\psi :]-\pi;\pi[\times]0;1[\to\mathbb{R^3} \\ (\theta,v)\to (\sqrt{v}cos\theta,\sqrt{v}sin(\theta),v)$$

The divergence of $X$ is (And it I think it is here where I got it wrong): $$div(X)=\frac{1}{\sqrt{det(G(\psi;(\theta,v))}}tr\bigg(J\big(\sqrt{det(G(\psi;(\theta,v)))}X\circ\psi;(\theta,v)\big)\bigg)$$

Where $G(\psi;(\theta,v))$ is the Gram matrix of $\psi$ at $(\theta,v)$ and $J$ denotes the Jacobian matrix.

But if this definition is correct then it is not clear how I can compute the trace of that Jacobian matrix because it is not a square matrix, and if the definion is incorrect how do I compute the divergent of $X$ when using the parametrization $\psi$?

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Calculate the divergence first

$$ \nabla \cdot F = \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(z) - \frac{\partial}{\partial z}\left(\frac{z^2}{2}\right) = z - z = 0 $$

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  • $\begingroup$ And then I make a sort of representation of it in the parametrization $\psi$? And is that definition wrong? and why? $\endgroup$
    – Bidon
    Dec 17, 2018 at 20:52
  • $\begingroup$ @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: \mathbb{R}^\color{red}{3} \to \mathbb{R}^\color{red}{3}$, that's what is giving you the trouble with the dimensions $\endgroup$
    – caverac
    Dec 17, 2018 at 20:57
  • $\begingroup$ How come? I think it's quite common to that kind of vector fields, gravitational and electric for example... $\endgroup$
    – Bidon
    Dec 17, 2018 at 23:08
  • $\begingroup$ @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then? $\endgroup$
    – caverac
    Dec 17, 2018 at 23:14
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    $\begingroup$ @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it) $\endgroup$
    – caverac
    Dec 18, 2018 at 18:37

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