Why is it that the surface integral of the flux of a vector field is the same as the surface integral of the...
$begingroup$
In other words, this:
http://www.math.ucla.edu/~archristian/teaching/32b-w17/week-7.pdf
Is this just a definition because what we really care about is how much the vectors are "pushing" through the surface? Or is it an actual equality?
multivariable-calculus stokes-theorem
edited Jan 9 at 21:32
Arthur
112k7109191
asked Jan 9 at 21:13
khajiitkhajiit
111
$endgroup$
add a comment |
$begingroup$
In other words, this:
http://www.math.ucla.edu/~archristian/teaching/32b-w17/week-7.pdf
Is this just a definition because what we really care about is how much the vectors are "pushing" through the surface? Or is it an actual equality?
multivariable-calculus stokes-theorem
edited Jan 9 at 21:32
Arthur
112k7109191
asked Jan 9 at 21:13
khajiitkhajiit
111
$endgroup$
add a comment |
$begingroup$
In other words, this:
http://www.math.ucla.edu/~archristian/teaching/32b-w17/week-7.pdf
Is this just a definition because what we really care about is how much the vectors are "pushing" through the surface? Or is it an actual equality?
multivariable-calculus stokes-theorem
edited Jan 9 at 21:32
Arthur
112k7109191
asked Jan 9 at 21:13
khajiitkhajiit
111
$endgroup$
In other words, this:
http://www.math.ucla.edu/~archristian/teaching/32b-w17/week-7.pdf
Is this just a definition because what we really care about is how much the vectors are "pushing" through the surface? Or is it an actual equality?
multivariable-calculus stokes-theorem
multivariable-calculus stokes-theorem
edited Jan 9 at 21:32
Arthur
112k7109191
asked Jan 9 at 21:13
khajiitkhajiit
111
edited Jan 9 at 21:32
Arthur
112k7109191
asked Jan 9 at 21:13
khajiitkhajiit
111
edited Jan 9 at 21:32
Arthur
112k7109191
edited Jan 9 at 21:32
Arthur
112k7109191
edited Jan 9 at 21:32
Arthur
112k7109191
112k7109191
asked Jan 9 at 21:13
khajiitkhajiit
111
asked Jan 9 at 21:13
khajiitkhajiit
111
asked Jan 9 at 21:13
khajiitkhajiit
111
111
add a comment |
add a comment |
1 Answer
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$begingroup$
This is just notation: $dmathbf S = mathbf n,dS$. Both sides are computing the flux of $mathbf F$, i.e., the net amount of $mathbf F$ pushing outwards across the surface.
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
add a comment |
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$begingroup$
This is just notation: $dmathbf S = mathbf n,dS$. Both sides are computing the flux of $mathbf F$, i.e., the net amount of $mathbf F$ pushing outwards across the surface.
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
add a comment |
$begingroup$
This is just notation: $dmathbf S = mathbf n,dS$. Both sides are computing the flux of $mathbf F$, i.e., the net amount of $mathbf F$ pushing outwards across the surface.
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
add a comment |
$begingroup$
This is just notation: $dmathbf S = mathbf n,dS$. Both sides are computing the flux of $mathbf F$, i.e., the net amount of $mathbf F$ pushing outwards across the surface.
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
This is just notation: $dmathbf S = mathbf n,dS$. Both sides are computing the flux of $mathbf F$, i.e., the net amount of $mathbf F$ pushing outwards across the surface.
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
answered Jan 9 at 22:15
Ted ShifrinTed Shifrin
63.2k44489
63.2k44489
add a comment |
add a comment |
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