another proof of divergence theorem
$begingroup$
The following is a simple proof of divergence theorem :
$$
begin{align}
& phantom{={}} iiint (nablacdot F) , dV \
& = iiint frac{(∂F_x)}{∂x} , text{dx dy dz} + iiint frac{(∂F_y)}{∂y} , text{dy dx dz} + iiint frac{(∂F_z)}{∂z} text{dz dx dy} \
& =iint F_x, text{dy dz} + iint F_y , text{dx dz} + iint F_z text{dx dy} \
& =iint F_x dS_x +iint F_y , dS_y +iint F_z , dS_z
end{align}
$$
(since the projected area onto (for eg) $x$-$y$ plane is the $z$-component of area vector)
$$ =iint (F_x+F_y+F_z)cdot( dS_x+ dS_y+ dS_z) $$
$$ =iint Fcdot dS$$
Is this proof enough? Then why we use the long proof given in most textbooks?
multivariable-calculus
edited Oct 12 '13 at 17:04
Michael Hardy
1
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
$endgroup$
add a comment |
$begingroup$
The following is a simple proof of divergence theorem :
$$
begin{align}
& phantom{={}} iiint (nablacdot F) , dV \
& = iiint frac{(∂F_x)}{∂x} , text{dx dy dz} + iiint frac{(∂F_y)}{∂y} , text{dy dx dz} + iiint frac{(∂F_z)}{∂z} text{dz dx dy} \
& =iint F_x, text{dy dz} + iint F_y , text{dx dz} + iint F_z text{dx dy} \
& =iint F_x dS_x +iint F_y , dS_y +iint F_z , dS_z
end{align}
$$
(since the projected area onto (for eg) $x$-$y$ plane is the $z$-component of area vector)
$$ =iint (F_x+F_y+F_z)cdot( dS_x+ dS_y+ dS_z) $$
$$ =iint Fcdot dS$$
Is this proof enough? Then why we use the long proof given in most textbooks?
multivariable-calculus
edited Oct 12 '13 at 17:04
Michael Hardy
1
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
$endgroup$
1
$begingroup$
Because what you've written doesn't mean anything, sadly.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 14:58
1
$begingroup$
There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) leq x leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible.
$endgroup$
– Paul Siegel
Oct 12 '13 at 14:59
1
$begingroup$
@PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 15:00
$begingroup$
That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs.
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:01
1
$begingroup$
It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.)
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:02
add a comment |
$begingroup$
The following is a simple proof of divergence theorem :
$$
begin{align}
& phantom{={}} iiint (nablacdot F) , dV \
& = iiint frac{(∂F_x)}{∂x} , text{dx dy dz} + iiint frac{(∂F_y)}{∂y} , text{dy dx dz} + iiint frac{(∂F_z)}{∂z} text{dz dx dy} \
& =iint F_x, text{dy dz} + iint F_y , text{dx dz} + iint F_z text{dx dy} \
& =iint F_x dS_x +iint F_y , dS_y +iint F_z , dS_z
end{align}
$$
(since the projected area onto (for eg) $x$-$y$ plane is the $z$-component of area vector)
$$ =iint (F_x+F_y+F_z)cdot( dS_x+ dS_y+ dS_z) $$
$$ =iint Fcdot dS$$
Is this proof enough? Then why we use the long proof given in most textbooks?
multivariable-calculus
edited Oct 12 '13 at 17:04
Michael Hardy
1
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
$endgroup$
The following is a simple proof of divergence theorem :
$$
begin{align}
& phantom{={}} iiint (nablacdot F) , dV \
& = iiint frac{(∂F_x)}{∂x} , text{dx dy dz} + iiint frac{(∂F_y)}{∂y} , text{dy dx dz} + iiint frac{(∂F_z)}{∂z} text{dz dx dy} \
& =iint F_x, text{dy dz} + iint F_y , text{dx dz} + iint F_z text{dx dy} \
& =iint F_x dS_x +iint F_y , dS_y +iint F_z , dS_z
end{align}
$$
(since the projected area onto (for eg) $x$-$y$ plane is the $z$-component of area vector)
$$ =iint (F_x+F_y+F_z)cdot( dS_x+ dS_y+ dS_z) $$
$$ =iint Fcdot dS$$
Is this proof enough? Then why we use the long proof given in most textbooks?
multivariable-calculus
multivariable-calculus
edited Oct 12 '13 at 17:04
Michael Hardy
1
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
edited Oct 12 '13 at 17:04
Michael Hardy
1
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
edited Oct 12 '13 at 17:04
Michael Hardy
1
edited Oct 12 '13 at 17:04
Michael Hardy
1
edited Oct 12 '13 at 17:04
Michael Hardy
1
1
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
asked Oct 12 '13 at 14:47
faheemahmed400faheemahmed400
112
112
1
$begingroup$
Because what you've written doesn't mean anything, sadly.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 14:58
1
$begingroup$
There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) leq x leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible.
$endgroup$
– Paul Siegel
Oct 12 '13 at 14:59
1
$begingroup$
@PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 15:00
$begingroup$
That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs.
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:01
1
$begingroup$
It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.)
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:02
add a comment |
1
$begingroup$
Because what you've written doesn't mean anything, sadly.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 14:58
1
$begingroup$
There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) leq x leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible.
$endgroup$
– Paul Siegel
Oct 12 '13 at 14:59
1
$begingroup$
@PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 15:00
$begingroup$
That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs.
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:01
1
$begingroup$
It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.)
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:02
1
1
$begingroup$
Because what you've written doesn't mean anything, sadly.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 14:58
$begingroup$
Because what you've written doesn't mean anything, sadly.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 14:58
1
1
$begingroup$
There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) leq x leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible.
$endgroup$
– Paul Siegel
Oct 12 '13 at 14:59
$begingroup$
There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) leq x leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible.
$endgroup$
– Paul Siegel
Oct 12 '13 at 14:59
1
1
$begingroup$
@PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 15:00
$begingroup$
@PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 15:00
$begingroup$
That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs.
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:01
$begingroup$
That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs.
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:01
1
1
$begingroup$
It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.)
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:02
$begingroup$
It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.)
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:02
add a comment |
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1
$begingroup$
Because what you've written doesn't mean anything, sadly.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 14:58
1
$begingroup$
There is a problem in the very first step. To express a triple integral as a single iterated integral with, say, the order of integration $dx dy dz$, you must assume that the region of integration has the form $f(y,z) leq x leq g(y,z)$ for $y,z$ in some domain in the plane. This is rarely possible.
$endgroup$
– Paul Siegel
Oct 12 '13 at 14:59
1
$begingroup$
@PaulSiegel: Even if you assume the region is $x$-, $y$-, and $z$-"simple," this isn't a proof without a whole lot more.
$endgroup$
– Ted Shifrin
Oct 12 '13 at 15:00
$begingroup$
That said, any suitably nice region can be approximated arbitrarily well by the union of cubes, and some version of your calculation works on a cube. This is the idea behind many of those long proofs.
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:01
1
$begingroup$
It looks more or less correct to me on a simple region. (Though the last step should be written out correctly using a parametrization of the boundary.)
$endgroup$
– Paul Siegel
Oct 12 '13 at 15:02