1
$\begingroup$

Calculate the flux of the vector field $F(x, y, z) = (2x, 3y, 4z)$ through the surface of the ellipsoid $9x^2 + 4y^2 + z^2 ≤ 36$.

The divergence of $F$ is $9$. And then I changed coordinates (using spherical coordinates) and my bounds are then $0 \leq r \leq 1$ and $0 \leq \theta \leq 2 \pi$ and $0 \leq \phi \leq \pi$. Then I got $$\iiint_{\gamma} 9r^2\sin(\phi) \ d\theta d\phi dr = 12 \pi$$ I am wrong and the answer is actually $432 \pi$. How come?

$\endgroup$
1
  • 1
    $\begingroup$ You need to multiply the divergence (a constant) by the volume of the ellipsoid. $\endgroup$ May 6, 2023 at 21:05

1 Answer 1

2
$\begingroup$

The overall approach is good - using the divergence theorem is a nice idea, and switching to polar coordinates is a smart move too.

However, we've got to be careful in determining the volume element. Indeed, the volume $\gamma$ is not a sphere, but rather, an ellipsoid: $$ \gamma = \left\{ (x, y, z) \in \mathbb R^3 : \left( \frac x 2\right)^2 + \left( \frac y 3\right)^2 + \left( \frac z 6\right)^2 < 1\right\}.$$

Therefore, the coordinates $(r, \phi, \theta)$ are not the standard spherical polar coordinates, but rather, a "stretched" version of spherical polar coordinates: \begin{align} x & = 2r \sin \phi \cos \theta \\ y & = 3r \sin \phi \sin \theta \\ z & = 6r \cos \phi \end{align} With $(r, \phi, \theta)$ defined in this way, the range $$ 0 \leq r < 1, \ \ \ \ 0 \leq \phi < \pi, \ \ \ \ 0 \leq \theta < 2\pi$$ corresponds precisely to the volume $\gamma$. (If, on the other hand, $(r, \phi, \theta)$ were defined to be the standard spherical polar coordinates, which don't have the factors of $2$, $3$ and $6$, then this statement is no longer true.)

So the volume element is given by \begin{align} dxdydz & = \left| \text{det} \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial \theta} \end{bmatrix} \right| \\ & = \left| \text{det} \begin{bmatrix} 2\sin\phi\cos\theta & 2r\cos\phi\cos\theta & -2r\sin\phi\sin\theta \\ 3\sin\phi\sin\theta & 3r\cos\phi\sin\theta & 3r\sin\phi\cos\theta \\ 6\cos\phi & -6r\sin\phi & 0 \end{bmatrix} \right| \\ & = 36r^2 \sin\phi \ d\theta d\phi dr. \end{align}

Therefore, the integral you need to evaluate is $$ \int_{r=0}^{r=1} \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta = 2\pi} 9 \times 36r^2 \sin\phi \ d\theta d\phi dr . $$

My extra factor of $36$ is responsible for the discrepancy you observed.


Aside: Personally, I would have used different notation.

  • Rather than using $\gamma$ to denote the volume, I would have used $V$. The symbol $\gamma$ is suggestive of a $1$-dimensional curve rather than a $3$-dimensional volume.
  • Personally, I would have swapped $\theta$ and $\phi$, since this is more conventional as far as I'm aware.
$\endgroup$
3
  • $\begingroup$ Thank you so much for the help. But I forgot why we had to take 2,3,6 into account, could you maybe explain that? $\endgroup$ May 6, 2023 at 21:26
  • 1
    $\begingroup$ @Need_MathHelp Please see the edit - does this help? $\endgroup$
    – Kenny Wong
    May 6, 2023 at 21:40
  • $\begingroup$ Yes absolutely. Thank you so much again for clearing that up! $\endgroup$ May 6, 2023 at 21:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .