Mixing
Flux of curl of a vector field through cross section of a sphere
$begingroup$
Suppose $R>0$, $M$ be the sphere centered at the origin with radius $R$, $M=S^2(0,R)$, and $Gamma={(x,y,z)in M:x+y+z=1}$. I have to prove that, with the appropriate orientation of $Gamma$:
$$int_Gamma (ydx+zdy+xdz)=sqrt{3}pi R^2$$
My attempt:
Let $omega(x,y,z)=ydx+zdy+xdz$. $Gamma$ is a circle whose center is $Q=(frac{1}{3},frac{1}{3},frac{1}{3})$, with a radius $r=sqrt{R^2-frac{1}{3}}$ and a normal unitary vector $nu=(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}})$. Now let $X$ be a vector field whose differential work form is $omega$, then by Stokes-Kelvin theorem:
$$int_Gamma omega=int_etalangle rot(X),nu rangle dV_eta$$
where $eta={(x,y,z)in M:x+y+zgeq 1 }$, however if $A$ is the entire circle enclosed by $Gamma$, then:
$$int_Gamma omega=int_Alangle rot(X),nu rangle dV_A$$
because both have the same border. We also know that $rot(X)=(-1,-1,-1)$, thus:
$$int_Alangle rot(X),nu rangle dV_A=int_A-frac{3}{sqrt{3}} dV_A=-frac{3}{sqrt{3}}Area(A)=-frac{3}{sqrt{3}}(R^2-frac{1}{3})$$
This is clearly different but somehow close to the answer, but I don't know what I'm missing. I think maybe it has something to do with the orientation (at least maybe it explains the minus signal).
multivariable-calculus stokes-theorem
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
asked Jan 9 at 17:41
BidonBidon
967
$endgroup$
add a comment |
$begingroup$
Suppose $R>0$, $M$ be the sphere centered at the origin with radius $R$, $M=S^2(0,R)$, and $Gamma={(x,y,z)in M:x+y+z=1}$. I have to prove that, with the appropriate orientation of $Gamma$:
$$int_Gamma (ydx+zdy+xdz)=sqrt{3}pi R^2$$
My attempt:
Let $omega(x,y,z)=ydx+zdy+xdz$. $Gamma$ is a circle whose center is $Q=(frac{1}{3},frac{1}{3},frac{1}{3})$, with a radius $r=sqrt{R^2-frac{1}{3}}$ and a normal unitary vector $nu=(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}})$. Now let $X$ be a vector field whose differential work form is $omega$, then by Stokes-Kelvin theorem:
$$int_Gamma omega=int_etalangle rot(X),nu rangle dV_eta$$
where $eta={(x,y,z)in M:x+y+zgeq 1 }$, however if $A$ is the entire circle enclosed by $Gamma$, then:
$$int_Gamma omega=int_Alangle rot(X),nu rangle dV_A$$
because both have the same border. We also know that $rot(X)=(-1,-1,-1)$, thus:
$$int_Alangle rot(X),nu rangle dV_A=int_A-frac{3}{sqrt{3}} dV_A=-frac{3}{sqrt{3}}Area(A)=-frac{3}{sqrt{3}}(R^2-frac{1}{3})$$
This is clearly different but somehow close to the answer, but I don't know what I'm missing. I think maybe it has something to do with the orientation (at least maybe it explains the minus signal).
multivariable-calculus stokes-theorem
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
asked Jan 9 at 17:41
BidonBidon
967
$endgroup$
$begingroup$
I'm guessing that the curve should be given by $x+y+z=0$.
$endgroup$
– Ted Shifrin
Jan 9 at 18:44
add a comment |
$begingroup$
Suppose $R>0$, $M$ be the sphere centered at the origin with radius $R$, $M=S^2(0,R)$, and $Gamma={(x,y,z)in M:x+y+z=1}$. I have to prove that, with the appropriate orientation of $Gamma$:
$$int_Gamma (ydx+zdy+xdz)=sqrt{3}pi R^2$$
My attempt:
Let $omega(x,y,z)=ydx+zdy+xdz$. $Gamma$ is a circle whose center is $Q=(frac{1}{3},frac{1}{3},frac{1}{3})$, with a radius $r=sqrt{R^2-frac{1}{3}}$ and a normal unitary vector $nu=(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}})$. Now let $X$ be a vector field whose differential work form is $omega$, then by Stokes-Kelvin theorem:
$$int_Gamma omega=int_etalangle rot(X),nu rangle dV_eta$$
where $eta={(x,y,z)in M:x+y+zgeq 1 }$, however if $A$ is the entire circle enclosed by $Gamma$, then:
$$int_Gamma omega=int_Alangle rot(X),nu rangle dV_A$$
because both have the same border. We also know that $rot(X)=(-1,-1,-1)$, thus:
$$int_Alangle rot(X),nu rangle dV_A=int_A-frac{3}{sqrt{3}} dV_A=-frac{3}{sqrt{3}}Area(A)=-frac{3}{sqrt{3}}(R^2-frac{1}{3})$$
This is clearly different but somehow close to the answer, but I don't know what I'm missing. I think maybe it has something to do with the orientation (at least maybe it explains the minus signal).
multivariable-calculus stokes-theorem
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
asked Jan 9 at 17:41
BidonBidon
967
$endgroup$
Suppose $R>0$, $M$ be the sphere centered at the origin with radius $R$, $M=S^2(0,R)$, and $Gamma={(x,y,z)in M:x+y+z=1}$. I have to prove that, with the appropriate orientation of $Gamma$:
$$int_Gamma (ydx+zdy+xdz)=sqrt{3}pi R^2$$
My attempt:
Let $omega(x,y,z)=ydx+zdy+xdz$. $Gamma$ is a circle whose center is $Q=(frac{1}{3},frac{1}{3},frac{1}{3})$, with a radius $r=sqrt{R^2-frac{1}{3}}$ and a normal unitary vector $nu=(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}})$. Now let $X$ be a vector field whose differential work form is $omega$, then by Stokes-Kelvin theorem:
$$int_Gamma omega=int_etalangle rot(X),nu rangle dV_eta$$
where $eta={(x,y,z)in M:x+y+zgeq 1 }$, however if $A$ is the entire circle enclosed by $Gamma$, then:
$$int_Gamma omega=int_Alangle rot(X),nu rangle dV_A$$
because both have the same border. We also know that $rot(X)=(-1,-1,-1)$, thus:
$$int_Alangle rot(X),nu rangle dV_A=int_A-frac{3}{sqrt{3}} dV_A=-frac{3}{sqrt{3}}Area(A)=-frac{3}{sqrt{3}}(R^2-frac{1}{3})$$
This is clearly different but somehow close to the answer, but I don't know what I'm missing. I think maybe it has something to do with the orientation (at least maybe it explains the minus signal).
multivariable-calculus stokes-theorem
multivariable-calculus stokes-theorem
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
asked Jan 9 at 17:41
BidonBidon
967
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
asked Jan 9 at 17:41
BidonBidon
967
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
edited Jan 9 at 18:42
Ted Shifrin
63.6k44591
63.6k44591
asked Jan 9 at 17:41
BidonBidon
967
asked Jan 9 at 17:41
BidonBidon
967
asked Jan 9 at 17:41
BidonBidon
967
967
$begingroup$
I'm guessing that the curve should be given by $x+y+z=0$.
$endgroup$
– Ted Shifrin
Jan 9 at 18:44
add a comment |
$begingroup$
I'm guessing that the curve should be given by $x+y+z=0$.
$endgroup$
– Ted Shifrin
Jan 9 at 18:44
$begingroup$
I'm guessing that the curve should be given by $x+y+z=0$.
$endgroup$
– Ted Shifrin
Jan 9 at 18:44
$begingroup$
I'm guessing that the curve should be given by $x+y+z=0$.
$endgroup$
– Ted Shifrin
Jan 9 at 18:44
add a comment |
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$begingroup$
I'm guessing that the curve should be given by $x+y+z=0$.
$endgroup$
– Ted Shifrin
Jan 9 at 18:44