Finding the normal of manifold in $mathbb{R}^3$ (Divergence theorem)
Calculate $oint_gamma f(X) dX$ where $f(x,y,z) = (yz,xz,xy)$,
$gamma = {X(t)| 0 le t le 2pi}$, $X(t) = (cos(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(frac{pi}{8} + frac{t(2pi-t)}{4pi}))$
I know that I need to use the Divergence theorem(that's the subject of the exercise). I want to calculate the normal $N(x)$ of the manifold $gamma$, so I could try and get anything close to $oint_gamma f(X)dX$ in the form of $int<F,N> dx$ where $F$ is some vector field (and then use the Divergence theorem).
I know that usually the normal of manifold which is given by parameterization is $N(x)=frac{r_{u_1}timesdotstimes r_{u_{n-1}}}{||r_{u_1}timesdotstimes r_{u_{n-1}}||}$ (Where $r(U)$ is the parameterization) but here I'm given a parameterization that operate over single variable, $t$.
How does that formula for the normal $N(x)$ works in that case? Or perhaps is there any other way to calculate the normal?
Or is there any other trick that I'm missing to use the Divergence theorem with the current integral?
calculus integration multivariable-calculus manifolds divergence
asked Dec 26 '18 at 22:16
ChikChak
771418
add a comment |
Calculate $oint_gamma f(X) dX$ where $f(x,y,z) = (yz,xz,xy)$,
$gamma = {X(t)| 0 le t le 2pi}$, $X(t) = (cos(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(frac{pi}{8} + frac{t(2pi-t)}{4pi}))$
I know that I need to use the Divergence theorem(that's the subject of the exercise). I want to calculate the normal $N(x)$ of the manifold $gamma$, so I could try and get anything close to $oint_gamma f(X)dX$ in the form of $int<F,N> dx$ where $F$ is some vector field (and then use the Divergence theorem).
I know that usually the normal of manifold which is given by parameterization is $N(x)=frac{r_{u_1}timesdotstimes r_{u_{n-1}}}{||r_{u_1}timesdotstimes r_{u_{n-1}}||}$ (Where $r(U)$ is the parameterization) but here I'm given a parameterization that operate over single variable, $t$.
How does that formula for the normal $N(x)$ works in that case? Or perhaps is there any other way to calculate the normal?
Or is there any other trick that I'm missing to use the Divergence theorem with the current integral?
calculus integration multivariable-calculus manifolds divergence
asked Dec 26 '18 at 22:16
ChikChak
771418
add a comment |
Calculate $oint_gamma f(X) dX$ where $f(x,y,z) = (yz,xz,xy)$,
$gamma = {X(t)| 0 le t le 2pi}$, $X(t) = (cos(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(frac{pi}{8} + frac{t(2pi-t)}{4pi}))$
I know that I need to use the Divergence theorem(that's the subject of the exercise). I want to calculate the normal $N(x)$ of the manifold $gamma$, so I could try and get anything close to $oint_gamma f(X)dX$ in the form of $int<F,N> dx$ where $F$ is some vector field (and then use the Divergence theorem).
I know that usually the normal of manifold which is given by parameterization is $N(x)=frac{r_{u_1}timesdotstimes r_{u_{n-1}}}{||r_{u_1}timesdotstimes r_{u_{n-1}}||}$ (Where $r(U)$ is the parameterization) but here I'm given a parameterization that operate over single variable, $t$.
How does that formula for the normal $N(x)$ works in that case? Or perhaps is there any other way to calculate the normal?
Or is there any other trick that I'm missing to use the Divergence theorem with the current integral?
calculus integration multivariable-calculus manifolds divergence
asked Dec 26 '18 at 22:16
ChikChak
771418
Calculate $oint_gamma f(X) dX$ where $f(x,y,z) = (yz,xz,xy)$,
$gamma = {X(t)| 0 le t le 2pi}$, $X(t) = (cos(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(t)cos(frac{pi}{8} + frac{t(2pi-t)}{4pi}),sin(frac{pi}{8} + frac{t(2pi-t)}{4pi}))$
I know that I need to use the Divergence theorem(that's the subject of the exercise). I want to calculate the normal $N(x)$ of the manifold $gamma$, so I could try and get anything close to $oint_gamma f(X)dX$ in the form of $int<F,N> dx$ where $F$ is some vector field (and then use the Divergence theorem).
I know that usually the normal of manifold which is given by parameterization is $N(x)=frac{r_{u_1}timesdotstimes r_{u_{n-1}}}{||r_{u_1}timesdotstimes r_{u_{n-1}}||}$ (Where $r(U)$ is the parameterization) but here I'm given a parameterization that operate over single variable, $t$.
How does that formula for the normal $N(x)$ works in that case? Or perhaps is there any other way to calculate the normal?
Or is there any other trick that I'm missing to use the Divergence theorem with the current integral?
calculus integration multivariable-calculus manifolds divergence
calculus integration multivariable-calculus manifolds divergence
asked Dec 26 '18 at 22:16
ChikChak
771418
asked Dec 26 '18 at 22:16
ChikChak
771418
edited Dec 28 '18 at 20:09
asked Dec 26 '18 at 22:16
ChikChak
771418
asked Dec 26 '18 at 22:16
ChikChak
771418
asked Dec 26 '18 at 22:16
ChikChak
771418
771418
add a comment |
add a comment |
2 Answers
2
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oldest
votes
The given vector field $f$ is nothing else but $f=nabla F$ for the scalar function $F(x,y,z):=xyz$. We are therefore told to integrate
$$int_gamma nabla F(X)cdot dX$$
along the curve $$gamma: quad tmapsto X(t):=bigl(cospsi(t)cos t,cospsi(t)sin t,sinpsi(t)bigr)qquad(0leq tleq2pi) ,$$
with $$psi(t):={piover8}+{t(2pi -t)over4pi} ,$$
hence $psi(0)=psi(2pi)={piover8}$. It follows that $gamma$ begins and ends at the point ${bf p}:=gamma(0)=bigl(cos{piover8},0,sin{piover8}bigr)$. Since $F$ is $C^1$ on all of ${mathbb R}^3$ this allows to conclude that
$$int_gamma nabla F(X)cdot dX=F({bf p})-F({bf p})=0 .$$
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
add a comment |
The whole point of the exercise is that there is no reason to find the normal to the surface at all! The "divergence theorem" says that
$intint vec{F}cdotvec{n}dS= intintint nablacdotvec{F} dV$.
You want to integrate $nablacdotvec{F}$ over the solid having $gamma$ as surface.
answered Dec 26 '18 at 22:32
user247327
10.4k1515
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
add a comment |
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2 Answers
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The given vector field $f$ is nothing else but $f=nabla F$ for the scalar function $F(x,y,z):=xyz$. We are therefore told to integrate
$$int_gamma nabla F(X)cdot dX$$
along the curve $$gamma: quad tmapsto X(t):=bigl(cospsi(t)cos t,cospsi(t)sin t,sinpsi(t)bigr)qquad(0leq tleq2pi) ,$$
with $$psi(t):={piover8}+{t(2pi -t)over4pi} ,$$
hence $psi(0)=psi(2pi)={piover8}$. It follows that $gamma$ begins and ends at the point ${bf p}:=gamma(0)=bigl(cos{piover8},0,sin{piover8}bigr)$. Since $F$ is $C^1$ on all of ${mathbb R}^3$ this allows to conclude that
$$int_gamma nabla F(X)cdot dX=F({bf p})-F({bf p})=0 .$$
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
add a comment |
The given vector field $f$ is nothing else but $f=nabla F$ for the scalar function $F(x,y,z):=xyz$. We are therefore told to integrate
$$int_gamma nabla F(X)cdot dX$$
along the curve $$gamma: quad tmapsto X(t):=bigl(cospsi(t)cos t,cospsi(t)sin t,sinpsi(t)bigr)qquad(0leq tleq2pi) ,$$
with $$psi(t):={piover8}+{t(2pi -t)over4pi} ,$$
hence $psi(0)=psi(2pi)={piover8}$. It follows that $gamma$ begins and ends at the point ${bf p}:=gamma(0)=bigl(cos{piover8},0,sin{piover8}bigr)$. Since $F$ is $C^1$ on all of ${mathbb R}^3$ this allows to conclude that
$$int_gamma nabla F(X)cdot dX=F({bf p})-F({bf p})=0 .$$
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
add a comment |
The given vector field $f$ is nothing else but $f=nabla F$ for the scalar function $F(x,y,z):=xyz$. We are therefore told to integrate
$$int_gamma nabla F(X)cdot dX$$
along the curve $$gamma: quad tmapsto X(t):=bigl(cospsi(t)cos t,cospsi(t)sin t,sinpsi(t)bigr)qquad(0leq tleq2pi) ,$$
with $$psi(t):={piover8}+{t(2pi -t)over4pi} ,$$
hence $psi(0)=psi(2pi)={piover8}$. It follows that $gamma$ begins and ends at the point ${bf p}:=gamma(0)=bigl(cos{piover8},0,sin{piover8}bigr)$. Since $F$ is $C^1$ on all of ${mathbb R}^3$ this allows to conclude that
$$int_gamma nabla F(X)cdot dX=F({bf p})-F({bf p})=0 .$$
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
The given vector field $f$ is nothing else but $f=nabla F$ for the scalar function $F(x,y,z):=xyz$. We are therefore told to integrate
$$int_gamma nabla F(X)cdot dX$$
along the curve $$gamma: quad tmapsto X(t):=bigl(cospsi(t)cos t,cospsi(t)sin t,sinpsi(t)bigr)qquad(0leq tleq2pi) ,$$
with $$psi(t):={piover8}+{t(2pi -t)over4pi} ,$$
hence $psi(0)=psi(2pi)={piover8}$. It follows that $gamma$ begins and ends at the point ${bf p}:=gamma(0)=bigl(cos{piover8},0,sin{piover8}bigr)$. Since $F$ is $C^1$ on all of ${mathbb R}^3$ this allows to conclude that
$$int_gamma nabla F(X)cdot dX=F({bf p})-F({bf p})=0 .$$
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
answered Dec 27 '18 at 13:31
Christian Blatter
172k7112326
172k7112326
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
add a comment |
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
I really like your answer! But unfortunately I must use the Divergence theorem. I actually havn't noticed until now that $f=nabla F$. So $f$ is the normal of the space given by $F=c$, but is it related in any way to our $gamma$?
– ChikChak
Dec 27 '18 at 13:55
add a comment |
The whole point of the exercise is that there is no reason to find the normal to the surface at all! The "divergence theorem" says that
$intint vec{F}cdotvec{n}dS= intintint nablacdotvec{F} dV$.
You want to integrate $nablacdotvec{F}$ over the solid having $gamma$ as surface.
answered Dec 26 '18 at 22:32
user247327
10.4k1515
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
add a comment |
The whole point of the exercise is that there is no reason to find the normal to the surface at all! The "divergence theorem" says that
$intint vec{F}cdotvec{n}dS= intintint nablacdotvec{F} dV$.
You want to integrate $nablacdotvec{F}$ over the solid having $gamma$ as surface.
answered Dec 26 '18 at 22:32
user247327
10.4k1515
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
add a comment |
The whole point of the exercise is that there is no reason to find the normal to the surface at all! The "divergence theorem" says that
$intint vec{F}cdotvec{n}dS= intintint nablacdotvec{F} dV$.
You want to integrate $nablacdotvec{F}$ over the solid having $gamma$ as surface.
answered Dec 26 '18 at 22:32
user247327
10.4k1515
The whole point of the exercise is that there is no reason to find the normal to the surface at all! The "divergence theorem" says that
$intint vec{F}cdotvec{n}dS= intintint nablacdotvec{F} dV$.
You want to integrate $nablacdotvec{F}$ over the solid having $gamma$ as surface.
answered Dec 26 '18 at 22:32
user247327
10.4k1515
answered Dec 26 '18 at 22:32
user247327
10.4k1515
answered Dec 26 '18 at 22:32
user247327
10.4k1515
answered Dec 26 '18 at 22:32
user247327
10.4k1515
10.4k1515
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
add a comment |
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
Yes but in order to use the theorem, I need to get first to an integral which is of the form $intint vec{F}cdotvec{n}dS$, no? and then, use the theorem and calculate $intintint nablacdotvec{F} dV$ instead. But first I how can I find what $intint vec{F}cdotvec{n}dS$ is? I mean, all I know is that I need to calculate $oint_gamma f(X)dX$, but I don't know if $intint vec{F}cdotvec{n}dS= oint_gamma f(X)dX$...
– ChikChak
Dec 26 '18 at 22:35
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
You already have that form!! That's exactly what $intint vec{F}cdotvec{n}dS$ is. You do NOT need to write it out explicitly. You just need to write $intintint nablacdot vec{F}dV$ explicitely.
– user247327
Dec 26 '18 at 22:45
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
Why is that exactly what $intint vec{F}cdotvec{n} dS$ is?
– ChikChak
Dec 29 '18 at 16:43
add a comment |
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