0
$\begingroup$

Since my previous, introductory question Calculate the flux through a surface S from a field described by vectors about this example raised even more questions that I had initially - I was advised to post a new question and below I present my solution of the example, just to make sure if I did that correctly.

The given data: The $F$ and $S$ are as follows ($S$ is oriented outwards): $$\vec{F}=r^2 \cdot \vec{r}$$ $$S: x^2+y^2+z^2=R^2$$ $$\iint\limits_{S} \vec{F} \vec{ds} =\text{ ?}$$

I began with rejecting the use of vector normal to the surface: $$\vec{n}= \frac{\vec{r}}{R}$$ since I have not seen it applied in any other example exploiting Divergence theorem.

Am I right with this?

My solution: (applying advice from @ $$\vec{F}=r^2 \cdot \vec{r}= (r^2x, r^2y, r^2z)$$ $$div\vec{F}=5r^2$$ Then I determined my new set of coordinates and their range: $$V: \left\{ (r, \varphi, \theta) \quad 0 \le r \le R; 0 \le \varphi \le 2\pi; \frac{-\pi}{2} \le \theta \le \frac{\pi}{2}\right\}$$ Then, I calculated the divergence of $\vec{F}$ and substituted the result into the triple integral over the volume described by $S$:

$$\iint\limits_{S} \vec{F} \vec{ds} = \iiint\limits_{V} div\vec{F}\vec{ds}=\iiint\limits_{V} 5r^2 dxdydz = \int_{0}^{R} \left[ \int_{0}^{2\pi} \left[ \int_{ \frac{-\pi}{2} }^{ \frac{\pi}{2} } 5r^2 \cdot R^{2}cos \theta d \theta \right] d \varphi \right] dr =$$

$$=\int_{0}^{R} \left[ \int_{0}^{2\pi} 10r^2 \ R^2 d \varphi \right] dr =20 \pi R^2\int_{0}^{R} r^2 dr=\frac{20}{3} \pi R^{5}$$

  1. Is it the right answer?

  2. Is the normal vector not supposed to be used here?

$\endgroup$
  • $\begingroup$ Your definition of F doesn't make any sense. It is a vector quantity yet you define it in terms of a scaler product of two things. $\endgroup$ – user204299 Sep 4 '15 at 12:25
  • $\begingroup$ @JakeLebovic How should it look like then? $\endgroup$ – Peter Cerba Sep 4 '15 at 13:40
  • $\begingroup$ As per the last thread, I believe it is just the scalar multiplied by the vector, i.e. $r^2\vec{r}$. $\endgroup$ – michaelrccurtis Sep 4 '15 at 14:57
  • $\begingroup$ One final point, the volume element given your limits is $r^2\mathrm{cos}(\theta)$ rather than $R^2$. $\endgroup$ – michaelrccurtis Sep 4 '15 at 23:21
1
$\begingroup$

Your intepretation of $\vec{F}$ is wrong. You need to multiply the scalar by each component of the vector.

$$ \vec{F} = (r^2x, r^2y, r^2z)$$

Remember that the divergence is defined for a vector field - you can't apply it to a scalar. You can think of it as taking:

$$ \nabla\vec{F} = (\partial/\partial x, \partial/\partial y, \partial/\partial z) \cdot (r^2x, r^2y, r^2z) $$

In this case, each term is similar:

$$\frac{\partial}{\partial x}(r^2x) = \frac{\partial (r^2)}{\partial x}x + r^2 $$

where

$$\frac{\partial (r^2)}{\partial x} = 2x $$

Putting this together, we get:

$$ \nabla\vec{F} = 3r^2 + 2(x^2 + y^2 + z^2) = 5r^2 $$

$\endgroup$
  • $\begingroup$ So, should the $\vec{F}$ be equal to $(x^3+y^2x+z^2x)+(x^2y+y^3+z^2y)+(x^2z+y^2z+z^3)$? $\endgroup$ – Peter Cerba Sep 4 '15 at 15:49
  • $\begingroup$ $\vec{F}$ is a vector. $\endgroup$ – michaelrccurtis Sep 4 '15 at 16:06
  • $\begingroup$ Can you please show me what should I put under $div$ operator. I am trying and still do not know how to proceed. I can't figure out how did you get that $div \vec{F}$ is $5r^2$ $\endgroup$ – Peter Cerba Sep 4 '15 at 16:41
  • $\begingroup$ Do you understand what the divergence is? $\endgroup$ – michaelrccurtis Sep 4 '15 at 17:37
  • $\begingroup$ Apparently no. Can you describe in detail how did you get from $\vec{F} = (r^2x, r^2y, r^2z)$ to $5r^2$. For me it looks like te result should be $6r^2$, since the derivative of $r^2x, r^2y, r^2z$ versus $dx, dy, dz$ respectively is $3 \cdot r^2=6r^2$. Am I right? $\endgroup$ – Peter Cerba Sep 4 '15 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.