Mixing
Let $F ( x ) = frac { x } { | x | ^ { 3 } }$, if $0 in Omega$ show $iint _ { partial Omega } ( F cdot nu ) d...
$begingroup$
Problem: Let $Omega subset mathbb { R } ^ { 3 }$ a regular domain and $nu$ the unit external vector of $partial Omega$. Let $F ( x ) = frac { x } { | x | ^ { 3 } }$.
Prove that if $0 in Omega$, then $iint _ { partial Omega } ( F cdot nu ) d s = 4 pi$.
I tried to apply the divergence theorem, $$iiint _ { Omega } operatorname { div } F mathrm { d } x _ { 1 } mathrm { d } x _ { 2 } mathrm { d } x _ { 3 } = iint _ { partial Omega } ( F cdot nu ) mathrm { d } s$$
But without getting interesting results since I have so few informations about $Omega$.
calculus multivariable-calculus vector-analysis
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
$endgroup$
add a comment |
$begingroup$
Problem: Let $Omega subset mathbb { R } ^ { 3 }$ a regular domain and $nu$ the unit external vector of $partial Omega$. Let $F ( x ) = frac { x } { | x | ^ { 3 } }$.
Prove that if $0 in Omega$, then $iint _ { partial Omega } ( F cdot nu ) d s = 4 pi$.
I tried to apply the divergence theorem, $$iiint _ { Omega } operatorname { div } F mathrm { d } x _ { 1 } mathrm { d } x _ { 2 } mathrm { d } x _ { 3 } = iint _ { partial Omega } ( F cdot nu ) mathrm { d } s$$
But without getting interesting results since I have so few informations about $Omega$.
calculus multivariable-calculus vector-analysis
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
$endgroup$
add a comment |
$begingroup$
Problem: Let $Omega subset mathbb { R } ^ { 3 }$ a regular domain and $nu$ the unit external vector of $partial Omega$. Let $F ( x ) = frac { x } { | x | ^ { 3 } }$.
Prove that if $0 in Omega$, then $iint _ { partial Omega } ( F cdot nu ) d s = 4 pi$.
I tried to apply the divergence theorem, $$iiint _ { Omega } operatorname { div } F mathrm { d } x _ { 1 } mathrm { d } x _ { 2 } mathrm { d } x _ { 3 } = iint _ { partial Omega } ( F cdot nu ) mathrm { d } s$$
But without getting interesting results since I have so few informations about $Omega$.
calculus multivariable-calculus vector-analysis
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
$endgroup$
Problem: Let $Omega subset mathbb { R } ^ { 3 }$ a regular domain and $nu$ the unit external vector of $partial Omega$. Let $F ( x ) = frac { x } { | x | ^ { 3 } }$.
Prove that if $0 in Omega$, then $iint _ { partial Omega } ( F cdot nu ) d s = 4 pi$.
I tried to apply the divergence theorem, $$iiint _ { Omega } operatorname { div } F mathrm { d } x _ { 1 } mathrm { d } x _ { 2 } mathrm { d } x _ { 3 } = iint _ { partial Omega } ( F cdot nu ) mathrm { d } s$$
But without getting interesting results since I have so few informations about $Omega$.
calculus multivariable-calculus vector-analysis
calculus multivariable-calculus vector-analysis
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
asked Jan 12 at 14:10
NotAbelianGroupNotAbelianGroup
15511
15511
add a comment |
add a comment |
1 Answer
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$begingroup$
You cannot apply divergence theorem directly to the domain $Omega$ since $F$ is not $C^1$ on the whole domain $Omega$ due to $0inOmega$. Instead, let $B_epsilon ={xinmathbb{R}^n;|;|x|<epsilon}$. Pick sufficiently small $epsilon>0$ such that $B_{2epsilon}subset Omega$ and let $Omega_epsilon = Omega setminus B_epsilon$. Then, $partialOmega_epsilon= partial Omega -partial B_epsilon$ and it holds
$$
int_{Omega_epsilon}text{div}F;dxdy=int_{partialOmega_epsilon}Fcdotnu ;dS =int_{partialOmega}Fcdotnu ;dS-int_{partial B_epsilon}Fcdotnu ;dS.
$$ Note that $text{div}F=0$ on $Omega_epsilon$ and hence
$$
int_{partialOmega}Fcdotnu ;dS=int_{partial B_epsilon}Fcdotnu ;dS.
$$ Now, since outward unit normal vector $nu(x)$ on $partial B_epsilon$ is $frac{x}{epsilon}$, we get
$$
int_{partial B_epsilon}Fcdotnu ;dS=int_{partial B_epsilon}frac{x}{r^3}cdotfrac{x}{epsilon} ;dS=frac{1}{epsilon^2}text{vol}(partial B_epsilon)=4pi.
$$
answered Jan 12 at 14:25
SongSong
12.4k630
$endgroup$
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
1
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You cannot apply divergence theorem directly to the domain $Omega$ since $F$ is not $C^1$ on the whole domain $Omega$ due to $0inOmega$. Instead, let $B_epsilon ={xinmathbb{R}^n;|;|x|<epsilon}$. Pick sufficiently small $epsilon>0$ such that $B_{2epsilon}subset Omega$ and let $Omega_epsilon = Omega setminus B_epsilon$. Then, $partialOmega_epsilon= partial Omega -partial B_epsilon$ and it holds
$$
int_{Omega_epsilon}text{div}F;dxdy=int_{partialOmega_epsilon}Fcdotnu ;dS =int_{partialOmega}Fcdotnu ;dS-int_{partial B_epsilon}Fcdotnu ;dS.
$$ Note that $text{div}F=0$ on $Omega_epsilon$ and hence
$$
int_{partialOmega}Fcdotnu ;dS=int_{partial B_epsilon}Fcdotnu ;dS.
$$ Now, since outward unit normal vector $nu(x)$ on $partial B_epsilon$ is $frac{x}{epsilon}$, we get
$$
int_{partial B_epsilon}Fcdotnu ;dS=int_{partial B_epsilon}frac{x}{r^3}cdotfrac{x}{epsilon} ;dS=frac{1}{epsilon^2}text{vol}(partial B_epsilon)=4pi.
$$
answered Jan 12 at 14:25
SongSong
12.4k630
$endgroup$
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
1
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
add a comment |
$begingroup$
You cannot apply divergence theorem directly to the domain $Omega$ since $F$ is not $C^1$ on the whole domain $Omega$ due to $0inOmega$. Instead, let $B_epsilon ={xinmathbb{R}^n;|;|x|<epsilon}$. Pick sufficiently small $epsilon>0$ such that $B_{2epsilon}subset Omega$ and let $Omega_epsilon = Omega setminus B_epsilon$. Then, $partialOmega_epsilon= partial Omega -partial B_epsilon$ and it holds
$$
int_{Omega_epsilon}text{div}F;dxdy=int_{partialOmega_epsilon}Fcdotnu ;dS =int_{partialOmega}Fcdotnu ;dS-int_{partial B_epsilon}Fcdotnu ;dS.
$$ Note that $text{div}F=0$ on $Omega_epsilon$ and hence
$$
int_{partialOmega}Fcdotnu ;dS=int_{partial B_epsilon}Fcdotnu ;dS.
$$ Now, since outward unit normal vector $nu(x)$ on $partial B_epsilon$ is $frac{x}{epsilon}$, we get
$$
int_{partial B_epsilon}Fcdotnu ;dS=int_{partial B_epsilon}frac{x}{r^3}cdotfrac{x}{epsilon} ;dS=frac{1}{epsilon^2}text{vol}(partial B_epsilon)=4pi.
$$
answered Jan 12 at 14:25
SongSong
12.4k630
$endgroup$
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
1
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
add a comment |
$begingroup$
You cannot apply divergence theorem directly to the domain $Omega$ since $F$ is not $C^1$ on the whole domain $Omega$ due to $0inOmega$. Instead, let $B_epsilon ={xinmathbb{R}^n;|;|x|<epsilon}$. Pick sufficiently small $epsilon>0$ such that $B_{2epsilon}subset Omega$ and let $Omega_epsilon = Omega setminus B_epsilon$. Then, $partialOmega_epsilon= partial Omega -partial B_epsilon$ and it holds
$$
int_{Omega_epsilon}text{div}F;dxdy=int_{partialOmega_epsilon}Fcdotnu ;dS =int_{partialOmega}Fcdotnu ;dS-int_{partial B_epsilon}Fcdotnu ;dS.
$$ Note that $text{div}F=0$ on $Omega_epsilon$ and hence
$$
int_{partialOmega}Fcdotnu ;dS=int_{partial B_epsilon}Fcdotnu ;dS.
$$ Now, since outward unit normal vector $nu(x)$ on $partial B_epsilon$ is $frac{x}{epsilon}$, we get
$$
int_{partial B_epsilon}Fcdotnu ;dS=int_{partial B_epsilon}frac{x}{r^3}cdotfrac{x}{epsilon} ;dS=frac{1}{epsilon^2}text{vol}(partial B_epsilon)=4pi.
$$
answered Jan 12 at 14:25
SongSong
12.4k630
$endgroup$
You cannot apply divergence theorem directly to the domain $Omega$ since $F$ is not $C^1$ on the whole domain $Omega$ due to $0inOmega$. Instead, let $B_epsilon ={xinmathbb{R}^n;|;|x|<epsilon}$. Pick sufficiently small $epsilon>0$ such that $B_{2epsilon}subset Omega$ and let $Omega_epsilon = Omega setminus B_epsilon$. Then, $partialOmega_epsilon= partial Omega -partial B_epsilon$ and it holds
$$
int_{Omega_epsilon}text{div}F;dxdy=int_{partialOmega_epsilon}Fcdotnu ;dS =int_{partialOmega}Fcdotnu ;dS-int_{partial B_epsilon}Fcdotnu ;dS.
$$ Note that $text{div}F=0$ on $Omega_epsilon$ and hence
$$
int_{partialOmega}Fcdotnu ;dS=int_{partial B_epsilon}Fcdotnu ;dS.
$$ Now, since outward unit normal vector $nu(x)$ on $partial B_epsilon$ is $frac{x}{epsilon}$, we get
$$
int_{partial B_epsilon}Fcdotnu ;dS=int_{partial B_epsilon}frac{x}{r^3}cdotfrac{x}{epsilon} ;dS=frac{1}{epsilon^2}text{vol}(partial B_epsilon)=4pi.
$$
answered Jan 12 at 14:25
SongSong
12.4k630
answered Jan 12 at 14:25
SongSong
12.4k630
answered Jan 12 at 14:25
SongSong
12.4k630
answered Jan 12 at 14:25
SongSong
12.4k630
12.4k630
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
1
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
add a comment |
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
1
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
$begingroup$
Thank you for your great and clear answer.
$endgroup$
– NotAbelianGroup
Jan 12 at 14:58
1
1
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:16
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
$begingroup$
I had my final exam this morning and one of the problems was pretty much the same. Thank you again!
$endgroup$
– NotAbelianGroup
Jan 14 at 17:05
add a comment |
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