Question

Find Flux

Answer 1

whr did u last see it?

Answer 2

it's written beside FIND.

Answer 3

Aeon?

Answer 4

field is
\[{<x^2 y,-x y^2,z} > \]

Answer 5

over function x^2+y^2+z^2 =1

Answer 6

i don't remember anything :/

Answer 7

Ok

Answer 8

find flux of field
{-y,x}

in function

x^2+y^2=1

you should get this

Answer 9

Compute the divergence and integrate it over the region enclosed by that surface (which is the unit sphere). It should be much easier.

Answer 10

\[\left\{x^2 y,-x y^2,z\right\}.\left\{\frac{2 x}{\sqrt{4 x^2+4 y^2+4 z^2}},\frac{2 y}{\sqrt{4 x^2+4 y^2+4 z^2}},\frac{2 z}{\sqrt{4 x^2+4 y^2+4 z^2}}\right\}\]

Answer 11

=\[\frac{x^3 y-x y^3+z^2}{\sqrt{x^2+y^2+z^2}}\]

Answer 12

\[\frac{x^3 y-x y^3+z^2}{\sqrt{x^2+y^2+z^2}} * 4 \pi \left(x^2+y^2+z^2\right)\]

\[4 \pi \sqrt{x^2+y^2+z^2} \left(x^3 y-x y^3+z^2\right)\]

Answer 13

Did you use spherical coordinates? It's not necessary. The divergence is simply\[\frac{\partial x^2 y}{\partial x}-\frac{\partial xy^2}{\partial y}+\frac{\partial z}{\partial z} = 2xy - 2xy + 1 = 1.\]Then, by the divergence theorem the flux is\[\iiint_{S} 1dV = V(S) = \frac{4}{3}\pi.\]

Answer 14

so the mix partial always cancel leaving only 1?

Answer 15

Not always. It depends on the field. To find the flux of a field over a surface you can use the definition, the divergence theorem or Stokes' theorem. In this case, the divergence of the field is incredibly simple so that seems to be the easiest way.

Answer 16

\[\left\{\frac{x}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}},\frac{y}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}},\frac{z}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\right\}\]

so flux for this would be 0?

Answer 17

Not necessarily. If the origin isn't enclosed by the surface, yes, it would be zero. If it is, then that field doesn't satisfy the hypothesis for the divergence theorem (it must be continuously differentiable over the whole volume enclosed by the surface), so the theorem is inconclusive. You must use the definition in that case.

Answer 18

for your question IM its 0
(-y,x) flux around the unit circle would be 0+0=0
0 flux which means theres circulation
i j k
Px Py Pz
-y x 0
since this is del cross F its going to be pointing in the z coord so lets just compute that
0i +0j +2k
so circulation is 2||R||=2pi

Answer 19

(-y,x)(dx,dy)=-ydx+xdy=according to brother Green thats nothing more than the double integral around the unit circle of
2||R||=2pi