a personal proof of the stokes' theorem-una dimostrazione personale del teorema di stokes
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7/23/2019 A PERSONAL PROOF of the STOKES' THEOREM-Una Dimostrazione Personale Del Teorema Di Stokes
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A PERSONAL PROOF OF THE STOKES' THEOREM:
Leonardo Rubino [email protected]
February 2012For www.vixra.org
Abstract: in this paper you will find a personal, practical and direct demonstration of theStokes’ Theorem.
The Stokes’ Theorem (practical proof-by Rubino!):
If we have a volume, we can hold it as made of many small volumes, as that in Fig. 1; forevery small volume, the following holds: (and so it holds also for the whole volume…)
Fig. 1: For the Stokes’ Theorem (proof by Rubino).
Let’s figure out l d B ⋅ :
On dz B is Bz; on dx B is Bx; on dy B is By;
on -dz B is dy y
Bdx
x
B B z z
z ∂
∂−
∂
∂+ , for 3-D Taylor’s development and also because to go
from the center of dz to that of –dz we go up along x, then we go down along y andnothing along z itself.
Similarly, on -dx B is dy y
B
dz z
B
B
x x
x∂
∂+
∂
∂−
and on -dy B is dz z
B
dx x
B
B
y y
y∂
∂+
∂
∂−
.
By summing up all contributions:
whereas here S d has got components [ )(ˆ dydz x , )(ˆ dxdz y , )(ˆ dxdy z ]
that is, the statement: ∫ ∫ ∫ ⋅×∇=⋅=⋅S S l
S d BS d Brot l d B , after having reminded of:
S d BS d Brot
dxdy y
B
x
Bdxdz
x
B
z
Bdydz
z
B
y
Bdydz
z
Bdx
x
B B
dy Bdxdy y
Bdz
z
B Bdx Bdz dy
y
Bdx
x
B Bdz Bl d B
x y z x y z y y
y
y x x
x x z z
z z
rrrvr
rr
⋅×∇=⋅=
=∂
∂−
∂
∂+
∂
∂−
∂
∂+
∂
∂−
∂
∂=
∂
∂+
∂
∂−+
−+∂
∂+
∂
∂−−+
∂
∂−
∂
∂+−=⋅
)()()()(
)()(
z
x
y
dx-dy
dz
Walk l d
B
0
-dz
dy-dx
∫ ∫ ∫ ⋅×∇=⋅=⋅S S l
S d BS d Brot l d B
7/23/2019 A PERSONAL PROOF of the STOKES' THEOREM-Una Dimostrazione Personale Del Teorema Di Stokes
http://slidepdf.com/reader/full/a-personal-proof-of-the-stokes-theorem-una-dimostrazione-personale-del 2/2
z y x B B B
z y x
z y x
B Brot ∂
∂
∂
∂
∂
∂=×∇=
ˆˆˆrrr
.
---------------------------------
Appendix) Divergence Theorem (a well known and practical proof):
Fig. 2: For the Divergence Theorem.
Nameφ the flux of the vector E ; we have:
dydz z y x E S d E d x ABCD ),,(−=⋅=φ ( means y “mean”)
dydz z ydx x E d x EFGH ),,( +=φ , but we obviously know that also: (as a development):
dx
z y x E
z y x E z ydx x E x
x x∂
∂
+=+
),,(
),,(),,( so:
dxdydz x
z y x E dydz z y x E d x
x EFGH ∂
∂+=
),,(),,(φ and so:
dV E
d d x EFGH ABCD
∂
∂=+ φφ . We similarly act on axes y and z:
dV y
E d d
y
BCGF AEHD∂
∂=+ φφ
dV z
E d d z
CGHD ABFE ∂
∂=+ φφ
And then we sum up the fluxes so found, having totally:
dV E dV E divdV z
E
y
E
x
E d z y x )()()(
rrr
⋅∇=⋅=∂
∂+
∂
∂+
∂
∂=φ therefore:
∫ ∫ ∫ ∫ ⋅⋅∇=⋅=⋅==S V V
S dV E dV E divS d E d E )()(φ
φφ that is the statement.
---------------------------------Bibliography:
1) ( L. Rubino) http://vixra.org/pdf/1201.0002v1.pdf
Thank you for your attention.Leonardo [email protected]
∫ ∫ ⋅=⋅S V
dV E divS d E
z
x
y
dx
dy
dz
A E
H
GC
B
D
E
0
τd
F