Attachment 31313

Printable View

- Jul 20th 2014, 05:37 AMkaemperDivF, Flux integral, centroid
- Jul 20th 2014, 12:36 PMHallsofIvyRe: DivF, Flux integral, centroid
In the first problem you are given $\displaystyle \vec{F}= x^2\vec{i}+ y^2\vec{j}+ z^2\vec{k}$ and are asked to find it flux out of the ball $\displaystyle (x- 2)^2+ y^2+ (z- 3)^2\le 9$. $\displaystyle div F= \nabla\cdot F= 2x+ 2y+ 2z$ (you have $\displaystyle 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}$ but that is corrected in the integral so I assume that was a typo).

You do the integral by writing it as $\displaystyle 2V\overline{x}+ 2V\overline{y}+ 2V\overline{z}$ and then try to calculate $\displaystyle \overline{x}$, $\displaystyle \overline{y}$, and $\displaystyle \overline{z}$. You realize, do you not that those are just the coordinates of the**center**of the figure? Here that figure is a sphere with center at (2, 0, 3) so you should be able to write down, with no calculation at all, that $\displaystyle \overline{x}= 2$, $\displaystyle \overline{y}= 0$, and $\displaystyle \overline{z}= 3$.

I have no idea at all how you got "$\displaystyle \overline{x}= x^2$"! You understand, don't you, that the x coordinate of the centroid of a figure is a**number**not a function of x? And it certainly is NOT the case that $\displaystyle \int\int\int xdV= x^2V$! - Jul 27th 2014, 11:37 AMkaemperRe: DivF, Flux integral, centroid
I give sincere thanks to HallsofIvy who has generously presented me with the most promising answer imaginable. I address you directly, HallsofIvy:

I do have received teaching in pre-calculus math: I was just insecure about the term "centroid".

Yours,

Student, Chemistry (Bachelor) at University of Southern Denmark (SDU). - Jul 27th 2014, 12:45 PMkaemperRe: DivF, Flux integral, centroid
So the flux out of the ball is the following:

Attachment 31355