engineering mathematics iiocw.snu.ac.kr/sites/default/files/note/4335.pdf · 2018-01-30 ·...
TRANSCRIPT
![Page 1: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/1.jpg)
EEngineering Mathematics IIngineering Mathematics II
Prof. Dr. Yong-Su Na g(32-206, Tel. 880-7204)
Text book: Erwin Kreyszig, Advanced Engineering Mathematics,
9th Edition, Wiley (2006)
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Ch. 10 Vector Integral Calculus. Ch. 10 Vector Integral Calculus. Ch. 10 Vector Integral Calculus. Ch. 10 Vector Integral Calculus. Integral TheoremsIntegral Theorems
10.1 Line Integrals
10.2 Path Independence of Line Integralsp g
10.3 Calculus Review: Double Integrals
10 4 Green’s Theorem in the Plane10.4 Green s Theorem in the Plane
10.5 Surfaces for Surface Integrals
10.6 Surface Integrals
10.7 Triple Integrals. Divergence Theorem of Gauss
10.8 Further Applications of the Divergence Theorem
10.9 Stokes’s Theorem10.9 Stokes s Theorem
2
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Ch. 10 Vector Integral Calculus. Ch. 10 Vector Integral Calculus. Ch. 10 Vector Integral Calculus. Ch. 10 Vector Integral Calculus. Integral TheoremsIntegral Theorems
10.1 Line Integrals
10.2 Path Independence of Line Integralsp g
10.3 Calculus Review: Double Integrals
10 4 Green’s Theorem in the Plane10.4 Green s Theorem in the Plane
10.5 Surfaces for Surface Integrals
10.6 Surface Integrals
10.7 Triple Integrals. Divergence Theorem of Gauss
10.8 Further Applications of the Divergence Theorem
10.9 Stokes’s Theorem10.9 Stokes s Theorem
3
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
면적분
1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
면적분
( ) ( ) ( ) ( )[ ] ( ) ( ) ( ) . ,, : 매끄럽다구분적으로는곡면 kjir u,vzu,vyu,vxu,vzu,vyu,vxu,vS ++==
vu rrN ×=∗ : 법선벡터
( )( ) ( )∫∫∫∫
vuvu
rrrr
NN
n ××
==∗11 : 단위법선벡터
( )( ) ( )면적분에서대해에벡터함수
,, :
NNnn
NrFnFF
==∗
•=• ∫∫∫∫
dudvdudvdA
dudvvuvudASRS
법선성분의 : FnF •∗
[ ] [ ] [ ] ( )=== NNNFFF 321321 , , cos,cos,cos ,,, ,,, γβαγβα nnNF 각도사이의좌표축과는
( )
( ) ( )∫∫∫∫
∫∫∫∫
++=++=
++=•⇒SS
dxdyFdzdxFdydzFdudvNFNFNF
dAFFFdA 321 coscoscos γβαnF
( ) ( )∫∫∫∫ ++=++=SR
dxdyFdzdxFdydzFdudvNFNFNF 321332211
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))Ex. 1 곡면을 통과하는 유출량
.30202 계산하라유출량을물의통과하는을기둥면포물 z,x,xy:S ≤≤≤≤=
1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
[ ] . sm ,663
. 30 20 2 측정된다로속도는이고속도벡터
계산하라유출량을물의통과하는을기둥면포물
xz,,z
z,x,xy:S
==
≤≤≤≤
Fv
22 ==⇒== uxyvzux
[ ] ( ) ( ) [ ][ ] [ ] [ ]0 ,1 ,21 ,0 ,00 ,2 ,1
6 ,6 ,3 ,30 ,20
,
22
−=−×=×=
=≤≤≤≤=⇒
==⇒==
uu
uvvSvu,vu,uS :
uxyvzux
vu rrN
Fr
[ ] [ ] [ ]
( ) 66
,,,,,,
2 −=•∴ uvS
vu
NF
( ) ( ) ( ) ( ) ⎤⎡∫∫∫ ∫∫∫333
32
3 2223 2
2( ) ( ) ( ) ( ) ⎥⎦⎤
⎢⎣⎡=−=−=−=−=•
== ∫∫∫ ∫∫∫ secm7212412126366
33
03
0
2
0
2
022
0 0
2vu
S
vvdvvdvuvududvuvdAnF
해법다른
[ ] [ ] [ ]
( ) ( ) 7223634363463
0cos ,0cos ,0cos 0 ,1 ,20 ,1 ,2cos,cos,cos
323
22 33 4
2
=<>⇒−=−===
∫∫∫ ∫∫ ∫∫∫ ddddd ddA
xu
F
NnNN γβαγβα
해법다른
( ) ( ) 7223634363463 3
00
2
0 00 0
2 =⋅⋅−⋅=⋅−=−=• ∫∫∫ ∫∫ ∫∫∫ dxdzzdxdzdydzzdAS
nF
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))Ex. 1 곡면을 통과하는 유출량
1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
.30202 계산하라유출량을물의통과하는을기둥면포물 z,x,xy:S ≤≤≤≤=
22⇒
[ ] . sm ,663
. 30 20 2 측정된다로속도는이고속도벡터
계산하라유출량을물의통과하는을기둥면포물
xz,,z
z,x,xy:S
==
≤≤≤≤
Fv
[ ] ( ) ( ) [ ][ ] [ ] [ ]0 ,1 ,21 ,0 ,00 ,2 ,1
6 ,6 ,3 ,30 ,20 ,
22
22
−=×=×==≤≤≤≤=⇒
==⇒==
uuuvvSvu,vu,uS :
uxyvzux
vu rrNFr
( ) 66 2 −=•∴ uvS NF
( ) ( ) ( ) ( ) ⎤⎡∫∫∫ ∫∫∫333
32
3 2223 2
2( ) ( ) ( ) ( ) ⎥⎦⎤
⎢⎣⎡=−=−=−=−=•
== ∫∫∫ ∫∫∫ secm7212412126366
33
03
0
2
0
2
022
0 0
2vu
S
vvdvvdvuvududvuvdAnF
해법다른
[ ] [ ] [ ]
( ) ( ) 7223634363463
0cos ,0cos ,0cos 0 ,1 ,20 ,1 ,2cos,cos,cos
323
22 33 4
2
=<>⇒−=−===
∫∫∫ ∫∫ ∫∫∫ ddddd ddA
xu
F
NnNN γβαγβα
해법다른
( ) ( ) 7223634363463 3
00
2
0 00 0
2 =⋅⋅−⋅=⋅−=−=• ∫∫∫ ∫∫ ∫∫∫ dxdzzdxdzdydzzdAS
nF
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))Ex. 1 곡면을 통과하는 유출량
1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
.30202 계산하라유출량을물의통과하는을기둥면포물 z,x,xy:S ≤≤≤≤=
[ ] . sm ,663
. 30 20 2 측정된다로속도는이고속도벡터
계산하라유출량을물의통과하는을기둥면포물
xz,,z
z,x,xy:S
==
≤≤≤≤
Fv
22⇒
[ ] ( ) ( ) [ ][ ] [ ] [ ]0 ,1 ,21 ,0 ,00 ,2 ,1
6 ,6 ,3 ,30 ,20 ,
22
22
−=×=×==≤≤≤≤=⇒
==⇒==
uuuvvSvu,vu,uS :
uxyvzux
vu rrNFr
( ) ( ) ( ) ( ) ⎤⎡∫∫∫ ∫∫∫333
32
3 2223 2
2
( ) 66 2 −=•∴ uvS NF
( ) ( ) ( ) ( ) ⎥⎦⎤
⎢⎣⎡=−=−=−=−=•
== ∫∫∫ ∫∫∫ secm7212412126366
33
03
0
2
0
2
022
0 0
2vu
S
vvdvvdvuvududvuvdAnF
해법다른
[ ] [ ] [ ]
( ) ( ) 7223634363463
0cos ,0cos ,0cos 0 ,1 ,20 ,1 ,2cos,cos,cos
323
22 33 4
2
=<>⇒−=−===
∫∫∫ ∫∫ ∫∫∫ ddddd ddA
xu
F
NnNN γβαγβα
해법다른
( ) ( ) 7223634363463 3
00
2
0 00 0
2 =⋅⋅−⋅=⋅−=−=• ∫∫∫ ∫∫ ∫∫∫ dxdzzdxdzdydzzdAS
nF
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))곡면의 방향
• 방향의 변경 : 1 일치한다것과곱하는을면적분에것은대체하는으로을 nn방향의 변경
• 매끄러운 곡면은 방향을 가질 수 있다 (Orientable).
• 구분적으로 매끄러운 곡면도 방향을 가질 수 있다.
. 1 일치한다것과곱하는을면적분에것은대체하는으로을 −− nn
구분적 매 러운 곡면 방향을 가질 수 있다.
• 매끄러운 곡면의 충분히 작은 조각도 항상 방향을 가진다. 그러나 전체 곡면에 대해서는
성립하지 않을 수도 있다. 예. 뫼비우스의 띠
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
방향을 고려하지 않는 면적분
• ( ) ( )( ) ( )dudvvuvuGdAGRS
,, : = ∫∫∫∫ Nrr형식다른면적분의
( ) [ ] [ ] [ ] ffffffyxfzS vuvuvuvu
22
2211,,,1,0,0,1 , :
⎞⎛⎞⎛
++=−−=×=×=⇒= rrN
( ) ( )( ) dxdyyf
xfyxfyxGdAG
RS
22
*
1,,, ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
+=⇒ ∫∫∫∫ r
• ( ) dudvdASASR
vuS
∫∫∫∫ ×== : rr면적의
( ) ( ) dxdyyf
xfSAyxfzS
RS
∫∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
+=⇒=22
1 , : yxR
∫∫ ⎟⎠
⎜⎝ ∂⎠⎝ ∂*
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
Ex. 4 구의 면적
( ) [ ] . 22 ,20 ,sin,sincos,coscos, 계산하여라직접대해에구 πππ ≤≤−≤≤= vuvauvauvavur
[ ]22222 ii[ ]=+=+
=×
2222
22222
1sincos ,1sincos
sincos ,sincos ,coscos
vvuu
vvauvauvavu rr
( ) =++=×⇒ 2212224242 cossincossincoscoscos vavvuvuvavu rr
( ) ∫∫ ∫ ===⇒2
222 2
2 4cos2cos ππ
π
ππ avdvadudvvaSA−− 22
0 ππ
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
Ex. 4 구의 면적
( ) [ ] . 22 ,20 ,sin,sincos,coscos, 계산하여라직접대해에구 πππ ≤≤−≤≤= vuvauvauvavur
[ ]22222 ii[ ]=+=+
=×
2222
22222
1sincos ,1sincos
sincos ,sincos ,coscos
vvuu
vvauvauvavu rr
( ) =++=×⇒ 2212224242 cossincossincoscoscos vavvuvuvavu rr
( ) ∫∫ ∫ ===⇒2
222 2
2 4cos2cos ππ
π
ππ avdvadudvvaSA−− 22
0 ππ
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1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))1010.6 .6 Surface Surface Integrals Integrals ((면면적분적분))
PROBLEM SET 10.6
HW 30 ( ) (b) ( ) (d)HW: 30 (a), (b), (c), (d)
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1010.7 .7 Triple Triple Integrals. Divergence Integrals. Divergence pp g gg gTheorem of Gauss Theorem of Gauss ((삼중적분삼중적분. Gauss. Gauss의의 발산정리발산정리))
삼중적분 : 공간의 닫힌 유한한 영역에서 함수의 적분
. )( 분할한다를평면으로평행한에삼차원좌표평면 T
( )
( ) ) :.( ,
,
부피상자의번째만든다합을형태의같은와
택하여을점한상자에서각
kΔVΔVz,yxfJ
z,yx
k
n
kkkkn
kkk
∑= ( ) )(,1
yf kk
kkkkn ∑=
( ) ,, 연속이고영역에서포함하는를가 Tzyxf
( ) . ,, ,
한다삼중적분이라의에서의영역극한을수렴이수열
가정제한된다고의해곡선에매끄러운유한개의는
zyxfTJ
T
n⇒
![Page 14: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/14.jpg)
1010.7 .7 Triple Triple Integrals. Divergence Integrals. Divergence pp g gg gTheorem of Gauss Theorem of Gauss ((삼중적분삼중적분. Gauss. Gauss의의 발산정리발산정리))
Gauss의 발산정리(삼중적분과 면적분 간의 변환)
T : 입체유한한닫혀있고
T
TS
F 1 :
:
벡터함수가지는편도함수를차연속인영역에서포함하는를연속이며
표면의곡면으로가지는방향을매끄러우며구분적으로경계가
[ ] [ ]== FFF coscoscosS γβα 이면법선벡터외향의이고 nF
dAdVST∫∫∫∫∫ •=⇒ nFFdiv
[ ] [ ]
( ) ( )∫∫∫∫∫∫∫ ++=++=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
==
SST
dxdyFdzdxFdydzFdAFFFdxdydzz
FyF
xF
FFF
321321321
321
coscoscos
cos,cos ,cos S , ,
γβα
γβα 이면법선벡터외향의이고 nF
⎠⎝ SST y
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1010.7 .7 Triple Triple Integrals. Divergence Integrals. Divergence pp g gg gTheorem of Gauss Theorem of Gauss ((삼중적분삼중적분. Gauss. Gauss의의 발산정리발산정리))
Ex. 1 발산정리에 의한 면적분의 계산Ex. 1 발산정리에 의한 면적분의 계산
다음 적분을 계산하라.
( )223 zdxdyxydzdxxdydzxI ++= ∫∫
( ) ( ) 표면닫힌이루어진으로과원판와원기둥 0 0 : 222222 ayxbzzbzayxS
S
≤+==≤≤=+
( ) dd dd d d
xxxxzxFyxFxF 222223
22
31
i
53div , ,
θθθ
=++=⇒===
적용극좌표
F
( )
( ) dzrdrdrdxdydzxI
dzrdrddxdydzryrx
b a2222 cos55
sin ,cos
θθ
θθθ
π
==
=⇒==
∫ ∫ ∫∫∫∫
적용극좌표
( )
badzadzda
y
bb
z rT
442
24
0 0 0
55cos5 ππθθπ
θ
=== ∫∫ ∫
∫ ∫ ∫∫∫∫= = =
badzdzdzz 00 0 44
5cos4
5 θθθ
∫∫ ∫== =
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1010.7 .7 Triple Triple Integrals. Divergence Integrals. Divergence pp g gg gTheorem of Gauss Theorem of Gauss ((삼중적분삼중적분. Gauss. Gauss의의 발산정리발산정리))
Ex. 1 발산정리에 의한 면적분의 계산Ex. 1 발산정리에 의한 면적분의 계산
다음 적분을 계산하라.
( )223 zdxdyxydzdxxdydzxI ++= ∫∫
( ) ( ) 표면닫힌이루어진으로과원판와원기둥 0 0 : 222222 ayxbzzbzayxS
S
≤+==≤≤=+
( ) dd dd d d
xxxxzxFyxFxF 222223
22
31
i
53div , ,
θθθ
=++=⇒===
적용극좌표
F
( )
( ) dzrdrdrdxdydzxI
dzrdrddxdydzryrx
b a2222 cos55
sin ,cos
θθ
θθθ
π
==
=⇒==
∫ ∫ ∫∫∫∫
적용극좌표
( )
badzadzda
y
bb
z rT
442
24
0 0 0
55cos5 ππθθπ
θ
=== ∫∫ ∫
∫ ∫ ∫∫∫∫= = =
badzdzdzz 00 0 44
5cos4
5 θθθ
∫∫ ∫== =
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1010.7 .7 Triple Triple Integrals. Divergence Integrals. Divergence pp g gg gTheorem of Gauss Theorem of Gauss ((삼중적분삼중적분. Gauss. Gauss의의 발산정리발산정리))
Example 2
발산의 좌표계 불변
• 삼중적분의 평균값 정리
( )
( ) ( ) ( ) ( )( ):
,
체적의
에서대해에연속함수의영역단순연결된유한하고
TTVTVzyxfdVx y zf
Tx, y, zfT
=∫∫∫ ( ) ( ) ( ) ( )( )
( ) .
:
000
000
있다가점만족하는
체적의
, z, yx
TTVTV, z, yxfdVx, y, zfT
=∫∫∫
• 발산의 불변
영역에서 1차 편도함수가 연속인 벡터함수의 발산은 직각 좌표계의 특별한 선택에 독립이다.
![Page 18: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/18.jpg)
1010.7 .7 Triple Triple Integrals. Divergence Integrals. Divergence pp g gg gTheorem of Gauss Theorem of Gauss ((삼중적분삼중적분. Gauss. Gauss의의 발산정리발산정리))
PROBLEM SET 10.7
HW 8 24HW: 8, 24
![Page 19: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/19.jpg)
1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
Ex. 1 유체흐름. 발산의 물리적 해석유체 름 발산의 물리적 해석
.
1 .
보자고려해흐름을일정한유체의비압축성가지는
을밀도상수의위해이를있다수얻을해석을직관적대한벡터의발산에발산정리로부터 =ρ
( )경계면의영역공간상의
결정의해서에속도벡터장에서의점임의의흐름은
: S
T
PP v
∗
∗
법선벡터단위외향의 : Sn∗
dATS ∫∫ •nv : .1 전체질량유체의흐르는외부로로부터통하여를시간당단위
dAV
TS
S
∫∫
∫∫•nv1 : .2 평균유출량흐르는외부로의
발생강도흐름의그대응점에서의발산은의속도벡터장상류의비압축성
0v
v
=∗
∗
div :
필요충분조건없을발생점이내의
발생강도흐름의그대응점에서의발산은의속도벡터장상류의비압축성
T
( ) 1∫∫∫∫( )
( ) ( ) ( )0 1limdiv
0=•⇒•= ∫∫∫∫→
dAdATV
PSTS
Tdnvnvv
![Page 20: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/20.jpg)
1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
Ex. 2 열전도 모델화. 열전도 방정식 혹은 확산 방정식
( ) )(grad
. .
,
물체의는시간는온도는
의미한다같음을식과다음의가속도열전도물체에서이것은증명한다것을비례한다는
기울기에온도의전도율은전도되며방향으로감소하는온도가열은물체에서실험에서는물리적
열전도도KttzyxUUK=v
v
( )
.
.) , , ,,,( grad
세워라모델을수학적
대한열전도에불리는이라고혹은이용하여정보를이러한
물체의는시간는온도는
방정식 확산방정식 열전도
열전도도KttzyxUU-K=v
( ) ( ) dxdydzUKdxdydzUKdAUUUUU
dAT
zzyyxx
S
∫∫∫∫∫∫∫∫
∫∫
∇−=−=•⇒++=∇=
•
22 graddivgraddiv
: .1
nv
nv열량나가는로부터시간당단위
( ) ( ) dxdydzUKdxdydzUKdAUUUUUTTS
zzyyxx ∫∫∫∫∫∫∫∫ ∇⇒++∇ graddiv graddiv nv
( )dxdydzUHTT∫∫∫= ρσσρ : , : : .2 밀도재료의비열재료의물체전체량열의내의
dxdydzt
Ut
HHT∫∫∫ ∂
∂−=
∂∂
− σρ : 비율시간감소하는가
TH . 한다같아야양과열의나가는로부터비율은시간감소하는가
UKt
UdxdydzUKt
UdxdydzUKdxdydzt
U
TTT
222 0 ∇=∂∂
∴=⎟⎠⎞
⎜⎝⎛ ∇−
∂∂
⇒∇−=∂∂
−⇒ ∫∫∫∫∫∫∫∫∫ σρσρσρ
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1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
Ex. 2 열전도 모델화. 열전도 방정식 혹은 확산 방정식
( ) )(grad
. .
,
물체의는시간는온도는
의미한다같음을식과다음의가속도열전도물체에서이것은증명한다것을비례한다는
기울기에온도의전도율은전도되며방향으로감소하는온도가열은물체에서실험에서는물리적
열전도도KttzyxUUK=v
v
( )
.
.) , , ,,,( grad
세워라모델을수학적
대한열전도에불리는이라고혹은이용하여정보를이러한
물체의는시간는온도는
방정식 확산방정식 열전도
열전도도KttzyxUU-K=v
( ) ( ) dxdydzUKdxdydzUKdAUUUUU
dAT
zzyyxx
S
∫∫∫∫∫∫∫∫
∫∫
∇−=−=•⇒++=∇=
•
22 graddivgraddiv
: .1
nv
nv열량나가는로부터시간당단위
( ) ( ) dxdydzUKdxdydzUKdAUUUUUTTS
zzyyxx ∫∫∫∫∫∫∫∫ ∇⇒++∇ graddiv graddiv nv
( )dxdydzUHTT∫∫∫= ρσσρ : , : : .2 밀도재료의비열재료의물체전체량열의내의
dxdydzt
Ut
HHT∫∫∫ ∂
∂−=
∂∂
− σρ : 비율시간감소하는가
TH . 한다같아야양과열의나가는로부터비율은시간감소하는가
UKt
UdxdydzUKt
UdxdydzUKdxdydzt
U
TTT
222 0 ∇=∂∂
∴=⎟⎠⎞
⎜⎝⎛ ∇−
∂∂
⇒∇−=∂∂
−⇒ ∫∫∫∫∫∫∫∫∫ σρσρσρ
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1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
퍼텐셜 이론. 조화함수
• 라플라스 방정식 : 02
2
2
2
2
22 =
∂∂
+∂∂
+∂∂
=∇z
fy
fx
ff
• 퍼텐셜 이론 : 라플라스 방정식 해에 대한 이론
• 조화함수 : 연속적인 2차 편도함수를 갖는 라플라스 방정식의 해
y
조화함수의 기본성질
구분적으로 매끄럽게 닫히고 방향을 줄 수 있는 곡면에서 조화함수의 법선도함수 적분
값은 0이 된다.
![Page 23: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/23.jpg)
1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
E 4 G 정리Ex. 4 Green 정리
가정만족한다고가정들을발산정리의에서영역가함수이고스칼라가와 grad Tgfgf =F
( )zgf
ygf
xgfgf , ,divgraddivdiv ⎟⎟
⎠
⎞⎜⎜⎝
⎛⎥⎦
⎤⎢⎣
⎡∂∂
∂∂
∂∂
==F
gfgfzgf
zg
zf
ygf
yg
yf
xgf
xg
xf gradgrad 2
2
2
2
2
2
2•+∇=⎟⎟
⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
∂∂
=
( ) ( ) fggff gradgrad •=•=•=•⇒ nnFnnF 함수스칼라가
( ) ( )공식번째첫의 Green gradgrad grad 2 dAgfdVgfgfgg ∫∫∫∫∫ ∂∂
=•+∇⇒∂∂
=•n ( ) ( )gggn
fgfgfn
gST∫∫∫∫∫ ∂∂
( ) ( )공식번째두의 Green 22 dAfggfdVfggf ∫∫∫∫∫ ⎟⎠⎞
⎜⎝⎛ ∂
−∂
=∇−∇( ) ( )공식째두의n
gn
ffggfST∫∫∫∫∫ ⎟
⎠⎜⎝ ∂∂
![Page 24: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/24.jpg)
1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
조화함수
:
:
곡면있는수줄방향을닫힌매끄럽고구분적으로내의
조화함수에서영역
DS
Df
.0 0
:
이다동일하게에서는이다값이모든점에서의가
전체영역감싸는를속하는에
TfSf
SDT
⇒
라플라스 방정식에 대한 유일성 정리
: 영역만족하는가정을발산정리T
Di i hl t 문제의 유일성
.
:
결정된다유일하게내에서값으로상에서는
조화함수에서영역포함하는를경계면의와
TSf
DSTTf
⇒
Dirichlet 문제의 유일성
유일하다해는이
가진다면해를에서문제가대한방정식에라플라스만족되고가정이위의 , Dirichlet T
유일하다해는이
![Page 25: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/25.jpg)
1010.8 Further Applications.8 Further Applications of the of the ppppDivergence Theorem Divergence Theorem ((발산정리의발산정리의 응용응용))
PROBLEM SET 10.8
HW 7 8HW: 7, 8
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1010.9 Stokes’s.9 Stokes’s TheoremTheorem1010.9 Stokes s.9 Stokes s TheoremTheorem
Stokes의 정리(면적분과 선적분 간의 변환)
SC
S
:
:
곡선닫힌단순히매끄럽고구분적으로경계로의
곡면갖는방향을매끄럽고구분적으로공간에서
S
SC
:
:
F 벡터함수연속인가지는편도함수를연속인영역에서포함하는를
곡선닫힌단순히매끄럽고구분적으로경계로의
( ) ( )dssdACS
'curl ∫∫∫ •=•⇒ rFnF
표시하면성분으로
( )∫∫∫ ++=⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
−∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−∂∂
⇒CR
dzFdyFdxFdudvNyF
xFN
xF
zFN
zF
yF
321312
231
123
⎦⎣ ⎠⎝⎠⎝⎠⎝ CR yy
![Page 27: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/27.jpg)
1010.9 Stokes’s.9 Stokes’s TheoremTheorem1010.9 Stokes s.9 Stokes s TheoremTheoremEx. 1 Stokes의 정리 검증
[ ] ( ) ( )22 검 하라대해에면와
Case 1 선적분
[ ] ( ) ( ) . 0 ,1, : ,, 22 검증하라대해에포물면와 ≥+−=== zyxyxfzSxzyF
( ) [ ] ( ) [ ]−=⇒= ssssssC 0,cos,sin':0,sin,cos: rr 접선벡터단위( ) [ ] ( ) [ ]
( )( ) [ ]
( )( ) ( ) ( )( )[ ]ππ
=
⇒
∫∫∫ ddd
sss
ssssssC
22
00ii'
cos ,0 ,sin
0 ,cos ,sin : 0 ,sin ,cos :
FF
rF
rr 접선벡터단위
Case 2 면적분
( )( ) ( ) ( )( )[ ] π−=++−=•=•∴ ∫∫∫ dsssdsssdC 00
00sinsin' rrFrF
[ ] [ ] [ ]−−−===⇒=== 111curlcurlcurl 321321 xzyFFFxFzFyF F [ ] [ ] [ ]
( )( ) [ ]
( ) −−−=•⇒
=−=
⇒
122curl
1 ,2 ,2,grad :
1 ,1 ,1 , ,curl , ,curlcurl , , 321321
yx
yxyxfzS
xzyFFFxFzFyF
NF
N
F
법선벡터의
( )
( ) ( ) ( )−−−=•=•∴ ∫∫∫∫∫∫ 122curlcurl
22 1
dxdyyxdxdydA
y
RRS
NFnF
( )( ) ( ) ( ) ππθθθθθθπ
θ
π
θ
−=−+=⎟⎠⎞
⎜⎝⎛ −+−=−+−= ∫∫ ∫
== =
22100
21sincos
321sincos2
2
0
2
0
1
0
drdrdrr
![Page 28: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/28.jpg)
1010.9 Stokes’s.9 Stokes’s TheoremTheorem1010.9 Stokes s.9 Stokes s TheoremTheoremEx. 1 Stokes의 정리 검증
[ ] ( ) ( )22 검 하라대해에면와
Case 1 선적분
[ ] ( ) ( ) . 0 ,1, : ,, 22 검증하라대해에포물면와 ≥+−=== zyxyxfzSxzyF
( ) [ ] ( ) [ ]−=⇒= ssssssC 0,cos,sin':0,sin,cos: rr 접선벡터단위( ) [ ] ( ) [ ]
( )( ) [ ]
( )( ) ( ) ( )( )[ ]ππ
=
⇒
∫∫∫ ddd
sss
ssssssC
22
00ii'
cos ,0 ,sin
0 ,cos ,sin : 0 ,sin ,cos :
FF
rF
rr 접선벡터단위
Case 2 면적분
( )( ) ( ) ( )( )[ ] π−=++−=•=•∴ ∫∫∫ dsssdsssdC 00
00sinsin' rrFrF
[ ] [ ] [ ]−−−===⇒=== 111curlcurlcurl 321321 xzyFFFxFzFyF F [ ] [ ] [ ]
( )( ) [ ]
( ) −−−=•⇒
=−=
⇒
122curl
1 ,2 ,2,grad :
1 ,1 ,1 , ,curl , ,curlcurl , , 321321
yx
yxyxfzS
xzyFFFxFzFyF
NF
N
F
법선벡터의
( )
( ) ( ) ( )−−−=•=•∴ ∫∫∫∫∫∫ 122curlcurl
22 1
dxdyyxdxdydA
y
RRS
NFnF
( )( ) ( ) ( ) ππθθθθθθπ
θ
π
θ
−=−+=⎟⎠⎞
⎜⎝⎛ −+−=−+−= ∫∫ ∫
== =
22100
21sincos
321sincos2
2
0
2
0
1
0
drdrdrr
![Page 29: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/29.jpg)
1010.9 Stokes’s.9 Stokes’s TheoremTheorem1010.9 Stokes s.9 Stokes s TheoremTheoremEx. 1 Stokes의 정리 검증
[ ] ( ) ( )22 검 하라대해에면와
Case 1 선적분
[ ] ( ) ( ) . 0 ,1, : ,, 22 검증하라대해에포물면와 ≥+−=== zyxyxfzSxzyF
( ) [ ] ( ) [ ]−=⇒= ssssssC 0,cos,sin':0,sin,cos: rr 접선벡터단위( ) [ ] ( ) [ ]
( )( ) [ ]
( )( ) ( ) ( )( )[ ]ππ
=
⇒
∫∫∫ ddd
sss
ssssssC
22
00ii'
cos ,0 ,sin
0 ,cos ,sin : 0 ,sin ,cos :
FF
rF
rr 접선벡터단위
Case 2 면적분
( )( ) ( ) ( )( )[ ] π−=++−=•=•∴ ∫∫∫ dsssdsssdC 00
00sinsin' rrFrF
[ ] [ ] [ ]−−−===⇒=== 111curlcurlcurl 321321 xzyFFFxFzFyF F [ ] [ ] [ ]
( )( ) [ ]
( ) −−−=•⇒
=−=
⇒
122curl
1 ,2 ,2,grad :
1 ,1 ,1 , ,curl , ,curlcurl , , 321321
yx
yxyxfzS
xzyFFFxFzFyF
NF
N
F
법선벡터의
( )
( ) ( ) ( )−−−=•=•∴ ∫∫∫∫∫∫ 122curlcurl
22 1
dxdyyxdxdydA
y
RRS
NFnF
( )( ) ( ) ( ) ππθθθθθθπ
θ
π
θ
−=−+=⎟⎠⎞
⎜⎝⎛ −+−=−+−= ∫∫ ∫
== =
22100
21sincos
321sincos2
2
0
2
0
1
0
drdrdrr
![Page 30: Engineering Mathematics IIocw.snu.ac.kr/sites/default/files/NOTE/4335.pdf · 2018-01-30 · Engineering Mathematics II Prof. Dr. Yong-Su Na (32-206, Tel. 880-7204) Text book: Erwin](https://reader034.vdocument.in/reader034/viewer/2022042309/5ed6b47bcaf45b18673a57f2/html5/thumbnails/30.jpg)
1010.9 Stokes’s.9 Stokes’s TheoremTheorem1010.9 Stokes s.9 Stokes s TheoremTheorem
PROBLEM SET 10.9
HW 19HW: 19