2

I need help calculating $$\iint_S F\cdot ds$$ where $F=\langle z,y,x \rangle$ and $$S=\left\{(x,y,z)\mid \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\right\}$$ and is oriented outwards.

Would the divergence theorem be used here, I'm not sure how to solve this.

Michael Hardy
  • 1
user19289
  • 157
  • I took the liberty of changing $\displaystyle\int\int$ to $\displaystyle\iint$ and $F\bullet: ds$ to $F\cdot ds$. Normally I'd write f(x),dx to get $f(x),dx$ with that spacing between $f(x)$ and $dx$ added manually, but I don't think that's the right way to do it in this instance since binary operators like \cdot provide spacing on both sides. ${}\qquad{}$ – Michael Hardy Aug 10 '15 at 17:53
  • Oh. Thanks. I'll use that notation from now on. – user19289 Aug 10 '15 at 17:55
  • what $\left<z,y,x\right>$ mean ? – Surb Aug 10 '15 at 18:12

1 Answers1

1

yes we would use the divergence theorem (easier evaluation) and so we have:

$$ \operatorname{div} \vec F = 1 $$ $$\iint_S \vec{F}\cdot d\vec{S} = \iiint_E \operatorname{div} \vec F dV = \iiint_E dV =\text{volume of ellipse} = \frac{4}{3}\pi abc$$

Hope this helped

Michael Hardy
  • 1
CivilSigma
  • 1,047
  • Can you add why it is the volume of an ellipse? – user19289 Aug 10 '15 at 17:51
  • Well , we are evaluating a triple integral over the region E which is the ellipse in 3 space, and since div F = 1 , the triple integral simply evaluates to what the volume of the ellipse is. So, we can cheat a bit , google the formula for the volume of an ellipse, plug in numbers and get the answer. – CivilSigma Aug 10 '15 at 17:54
  • Because the general form of an ellipsoid is $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=1$$. Since we are then integrating inside those bounds, we will get the volume of the ellipse. – FundThmCalculus Aug 10 '15 at 17:54