Mixing
Derive the volume of a torus using the divergence theorem and Fubini's theorem.
$begingroup$
Given is the vector field $f=(x,y,0)$ and the set $Omega :=((sqrt{x^2+y^2}-2)^2+z^2 <1)$ (which is obviously a torus)
I'm asked to :
a) Calculate the outward unit normal field $v: delta Omega to mathbb{R}^3$
b) Calculate $int_{deltaOmega}langle f,vrangle dS$ using the Divergence Theorem. Hint : It is expected that you again derive the volume of a torus. For that, calculate first the area enclosed of the set $A_z:=((sqrt{x^2+y^2}-2)^2leq1-z^2) $for$ -1leq z leq 1$ and then use Fubini's theorem.
a) So here I think that the outward unit field vector is just the gradient of $f$ normalized. So the gradient is $(1,1,0)$ and when normalized, it is $(frac{1}{sqrt{2}},frac{1}{sqrt{2}},0)$.
b) Now, here I'm not really sure. For the area enclosed, I'm obviously supposed to use polar coordinates. So $(r-2)^2leq1-z^2$. So $pm(r-2)=pmsqrt{(1-z^2)}$.
So $r=2+sqrt{1-z^2}$ or $r=2-sqrt{1-z^2}$.
Now, the area is $int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta=int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$.
And so the volume will be $int_{-1}^{1}int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$
And our divergence is $2$, so we will just multiply the result by $2$.
Now, my questions are:
Is my procedure for a) correct ?
Is my procedure for b) correct and if yes, am I now just supposed to evaluate that integral ?
Thanks for you help !
real-analysis calculus integration multivariable-calculus vector-analysis
asked Jan 12 at 9:18
PoujhPoujh
609516
$endgroup$
add a comment |
$begingroup$
Given is the vector field $f=(x,y,0)$ and the set $Omega :=((sqrt{x^2+y^2}-2)^2+z^2 <1)$ (which is obviously a torus)
I'm asked to :
a) Calculate the outward unit normal field $v: delta Omega to mathbb{R}^3$
b) Calculate $int_{deltaOmega}langle f,vrangle dS$ using the Divergence Theorem. Hint : It is expected that you again derive the volume of a torus. For that, calculate first the area enclosed of the set $A_z:=((sqrt{x^2+y^2}-2)^2leq1-z^2) $for$ -1leq z leq 1$ and then use Fubini's theorem.
a) So here I think that the outward unit field vector is just the gradient of $f$ normalized. So the gradient is $(1,1,0)$ and when normalized, it is $(frac{1}{sqrt{2}},frac{1}{sqrt{2}},0)$.
b) Now, here I'm not really sure. For the area enclosed, I'm obviously supposed to use polar coordinates. So $(r-2)^2leq1-z^2$. So $pm(r-2)=pmsqrt{(1-z^2)}$.
So $r=2+sqrt{1-z^2}$ or $r=2-sqrt{1-z^2}$.
Now, the area is $int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta=int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$.
And so the volume will be $int_{-1}^{1}int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$
And our divergence is $2$, so we will just multiply the result by $2$.
Now, my questions are:
Is my procedure for a) correct ?
Is my procedure for b) correct and if yes, am I now just supposed to evaluate that integral ?
Thanks for you help !
real-analysis calculus integration multivariable-calculus vector-analysis
asked Jan 12 at 9:18
PoujhPoujh
609516
$endgroup$
add a comment |
$begingroup$
Given is the vector field $f=(x,y,0)$ and the set $Omega :=((sqrt{x^2+y^2}-2)^2+z^2 <1)$ (which is obviously a torus)
I'm asked to :
a) Calculate the outward unit normal field $v: delta Omega to mathbb{R}^3$
b) Calculate $int_{deltaOmega}langle f,vrangle dS$ using the Divergence Theorem. Hint : It is expected that you again derive the volume of a torus. For that, calculate first the area enclosed of the set $A_z:=((sqrt{x^2+y^2}-2)^2leq1-z^2) $for$ -1leq z leq 1$ and then use Fubini's theorem.
a) So here I think that the outward unit field vector is just the gradient of $f$ normalized. So the gradient is $(1,1,0)$ and when normalized, it is $(frac{1}{sqrt{2}},frac{1}{sqrt{2}},0)$.
b) Now, here I'm not really sure. For the area enclosed, I'm obviously supposed to use polar coordinates. So $(r-2)^2leq1-z^2$. So $pm(r-2)=pmsqrt{(1-z^2)}$.
So $r=2+sqrt{1-z^2}$ or $r=2-sqrt{1-z^2}$.
Now, the area is $int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta=int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$.
And so the volume will be $int_{-1}^{1}int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$
And our divergence is $2$, so we will just multiply the result by $2$.
Now, my questions are:
Is my procedure for a) correct ?
Is my procedure for b) correct and if yes, am I now just supposed to evaluate that integral ?
Thanks for you help !
real-analysis calculus integration multivariable-calculus vector-analysis
asked Jan 12 at 9:18
PoujhPoujh
609516
$endgroup$
Given is the vector field $f=(x,y,0)$ and the set $Omega :=((sqrt{x^2+y^2}-2)^2+z^2 <1)$ (which is obviously a torus)
I'm asked to :
a) Calculate the outward unit normal field $v: delta Omega to mathbb{R}^3$
b) Calculate $int_{deltaOmega}langle f,vrangle dS$ using the Divergence Theorem. Hint : It is expected that you again derive the volume of a torus. For that, calculate first the area enclosed of the set $A_z:=((sqrt{x^2+y^2}-2)^2leq1-z^2) $for$ -1leq z leq 1$ and then use Fubini's theorem.
a) So here I think that the outward unit field vector is just the gradient of $f$ normalized. So the gradient is $(1,1,0)$ and when normalized, it is $(frac{1}{sqrt{2}},frac{1}{sqrt{2}},0)$.
b) Now, here I'm not really sure. For the area enclosed, I'm obviously supposed to use polar coordinates. So $(r-2)^2leq1-z^2$. So $pm(r-2)=pmsqrt{(1-z^2)}$.
So $r=2+sqrt{1-z^2}$ or $r=2-sqrt{1-z^2}$.
Now, the area is $int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta=int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$.
And so the volume will be $int_{-1}^{1}int_{-sqrt{1-z^2}}^{sqrt{1-z^2}}int_{0}^{2pi}1rdrdtheta$
And our divergence is $2$, so we will just multiply the result by $2$.
Now, my questions are:
Is my procedure for a) correct ?
Is my procedure for b) correct and if yes, am I now just supposed to evaluate that integral ?
Thanks for you help !
real-analysis calculus integration multivariable-calculus vector-analysis
real-analysis calculus integration multivariable-calculus vector-analysis
asked Jan 12 at 9:18
PoujhPoujh
609516
asked Jan 12 at 9:18
PoujhPoujh
609516
asked Jan 12 at 9:18
PoujhPoujh
609516
asked Jan 12 at 9:18
PoujhPoujh
609516
asked Jan 12 at 9:18
PoujhPoujh
609516
609516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $a)$, we can calculate $v:deltaOmega to mathbb{R}^3$ using parametrization of $delta Omega$. Note that $v$ depends only on $Omega$, and is completely irrelevant to the given vector field $f$. (So your calculation is not valid.) Let $sqrt{x^2+y^2} = r$. Then we can write
$$
Omega : 2-sqrt{1-z^2}<r<2+sqrt{1-z^2}.
$$ Hence $Omega = Omega_1 -Omega_2$ where $Omega_1:r<2+sqrt{1-z^2}$ and $Omega_2: r<2-sqrt{1-z^2}$. We can calculate $v$ on $deltaOmega_1$ as follows:
$$
nabla_{x,y,z}(r-2-sqrt{1-z^2})=(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})',
$$and by normalizing $$
v(x,y,z) = sqrt{1-z^2}(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})'.
$$ We can check that $v$ is outward.
Similarly for $deltaOmega_2$,
$$
nabla_{x,y,z}(r-2+sqrt{1-z^2})=(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'
$$ and
$$
v(x,y,z) = -sqrt{1-z^2}(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'.
$$ (The sign is reversed.)
For $b)$, it is true that $int_{deltaOmega}langle f,vrangle dS=int_Omega nablacdot f;dxdydz=2text{vol}(Omega).$ We can calculate this using cylindrical coordinate
$$
(x,y,z)=(rcostheta, rsintheta, z).
$$ Then, $dxdydz = rdrdtheta dz$ and the given integral becomes
$$begin{eqnarray}
2int_Omega dxdydz &=&2int_{-1}^1 int_0^{2pi}int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}rdrdtheta dz\
&=&2piint_{-1}^1 r^2big|^{2+sqrt{1-z^2}}_{2-sqrt{1-z^2}} ;dz\
&=&16pi int_{-1}^1 sqrt{1-z^2} ;dz\
&=&16pi int_{-frac{pi}{2}}^frac{pi}{2} cos^2 u ;du=8pi^2.
end{eqnarray}$$
answered Jan 12 at 9:59
SongSong
12.3k630
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add a comment |
$begingroup$
For a) it is best to turn to cylindrical coordinates. Note that for a relation $f(x,y,z)=0$ the gradient $nabla f(x,y,z)$ is always normal to it, therefore if the equation of torus is $$f(r,phi ,z)=(r-2)^2+z^2-1=0$$then we have$$vec v={nabla f(r,phi , z)over |nabla f(r,phi , z)|_2}={(r-2,0,z)over sqrt{(r-2)^2+z^2}}$$
b) $$int_{deltaOmega}langle f,vrangle dS{=int_V nablacdot f(x,y,z)dV\=int_V 2dV\=2int_{(r-2)^2+z^2<1}rdrdphi dz}$$note that $1<r<3$ and $-1<z<1$ therefore by the definition of torus we conclude that $$2-sqrt{1-z^2}<r<2+sqrt{1-z^2}$$and $$5-z^2-4sqrt{1-z^2}<r^2<5-z^2+4sqrt{1-z^2}$$so we can write$$2int_{(r-2)^2+z^2<1}rdrdphi dz{=int_{(r-2)^2+z^2<1}dr^2dphi dz\=2pi int_{(r-2)^2+z^2<1}dr^2 dz\=2pi int_{-1}^1int_{r^2=5-z^2-4sqrt{1-z^2}}^{r^2=5-z^2+4sqrt{1-z^2}}dr^2 dz\=16piint_{-1}^1sqrt{1-z^2}dz\=8pi ^2}$$therefore$$int_{deltaOmega}langle f,vrangle dS=8pi^2$$
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
For $a)$, we can calculate $v:deltaOmega to mathbb{R}^3$ using parametrization of $delta Omega$. Note that $v$ depends only on $Omega$, and is completely irrelevant to the given vector field $f$. (So your calculation is not valid.) Let $sqrt{x^2+y^2} = r$. Then we can write
$$
Omega : 2-sqrt{1-z^2}<r<2+sqrt{1-z^2}.
$$ Hence $Omega = Omega_1 -Omega_2$ where $Omega_1:r<2+sqrt{1-z^2}$ and $Omega_2: r<2-sqrt{1-z^2}$. We can calculate $v$ on $deltaOmega_1$ as follows:
$$
nabla_{x,y,z}(r-2-sqrt{1-z^2})=(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})',
$$and by normalizing $$
v(x,y,z) = sqrt{1-z^2}(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})'.
$$ We can check that $v$ is outward.
Similarly for $deltaOmega_2$,
$$
nabla_{x,y,z}(r-2+sqrt{1-z^2})=(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'
$$ and
$$
v(x,y,z) = -sqrt{1-z^2}(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'.
$$ (The sign is reversed.)
For $b)$, it is true that $int_{deltaOmega}langle f,vrangle dS=int_Omega nablacdot f;dxdydz=2text{vol}(Omega).$ We can calculate this using cylindrical coordinate
$$
(x,y,z)=(rcostheta, rsintheta, z).
$$ Then, $dxdydz = rdrdtheta dz$ and the given integral becomes
$$begin{eqnarray}
2int_Omega dxdydz &=&2int_{-1}^1 int_0^{2pi}int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}rdrdtheta dz\
&=&2piint_{-1}^1 r^2big|^{2+sqrt{1-z^2}}_{2-sqrt{1-z^2}} ;dz\
&=&16pi int_{-1}^1 sqrt{1-z^2} ;dz\
&=&16pi int_{-frac{pi}{2}}^frac{pi}{2} cos^2 u ;du=8pi^2.
end{eqnarray}$$
answered Jan 12 at 9:59
SongSong
12.3k630
$endgroup$
add a comment |
$begingroup$
For $a)$, we can calculate $v:deltaOmega to mathbb{R}^3$ using parametrization of $delta Omega$. Note that $v$ depends only on $Omega$, and is completely irrelevant to the given vector field $f$. (So your calculation is not valid.) Let $sqrt{x^2+y^2} = r$. Then we can write
$$
Omega : 2-sqrt{1-z^2}<r<2+sqrt{1-z^2}.
$$ Hence $Omega = Omega_1 -Omega_2$ where $Omega_1:r<2+sqrt{1-z^2}$ and $Omega_2: r<2-sqrt{1-z^2}$. We can calculate $v$ on $deltaOmega_1$ as follows:
$$
nabla_{x,y,z}(r-2-sqrt{1-z^2})=(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})',
$$and by normalizing $$
v(x,y,z) = sqrt{1-z^2}(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})'.
$$ We can check that $v$ is outward.
Similarly for $deltaOmega_2$,
$$
nabla_{x,y,z}(r-2+sqrt{1-z^2})=(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'
$$ and
$$
v(x,y,z) = -sqrt{1-z^2}(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'.
$$ (The sign is reversed.)
For $b)$, it is true that $int_{deltaOmega}langle f,vrangle dS=int_Omega nablacdot f;dxdydz=2text{vol}(Omega).$ We can calculate this using cylindrical coordinate
$$
(x,y,z)=(rcostheta, rsintheta, z).
$$ Then, $dxdydz = rdrdtheta dz$ and the given integral becomes
$$begin{eqnarray}
2int_Omega dxdydz &=&2int_{-1}^1 int_0^{2pi}int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}rdrdtheta dz\
&=&2piint_{-1}^1 r^2big|^{2+sqrt{1-z^2}}_{2-sqrt{1-z^2}} ;dz\
&=&16pi int_{-1}^1 sqrt{1-z^2} ;dz\
&=&16pi int_{-frac{pi}{2}}^frac{pi}{2} cos^2 u ;du=8pi^2.
end{eqnarray}$$
answered Jan 12 at 9:59
SongSong
12.3k630
$endgroup$
add a comment |
$begingroup$
For $a)$, we can calculate $v:deltaOmega to mathbb{R}^3$ using parametrization of $delta Omega$. Note that $v$ depends only on $Omega$, and is completely irrelevant to the given vector field $f$. (So your calculation is not valid.) Let $sqrt{x^2+y^2} = r$. Then we can write
$$
Omega : 2-sqrt{1-z^2}<r<2+sqrt{1-z^2}.
$$ Hence $Omega = Omega_1 -Omega_2$ where $Omega_1:r<2+sqrt{1-z^2}$ and $Omega_2: r<2-sqrt{1-z^2}$. We can calculate $v$ on $deltaOmega_1$ as follows:
$$
nabla_{x,y,z}(r-2-sqrt{1-z^2})=(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})',
$$and by normalizing $$
v(x,y,z) = sqrt{1-z^2}(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})'.
$$ We can check that $v$ is outward.
Similarly for $deltaOmega_2$,
$$
nabla_{x,y,z}(r-2+sqrt{1-z^2})=(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'
$$ and
$$
v(x,y,z) = -sqrt{1-z^2}(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'.
$$ (The sign is reversed.)
For $b)$, it is true that $int_{deltaOmega}langle f,vrangle dS=int_Omega nablacdot f;dxdydz=2text{vol}(Omega).$ We can calculate this using cylindrical coordinate
$$
(x,y,z)=(rcostheta, rsintheta, z).
$$ Then, $dxdydz = rdrdtheta dz$ and the given integral becomes
$$begin{eqnarray}
2int_Omega dxdydz &=&2int_{-1}^1 int_0^{2pi}int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}rdrdtheta dz\
&=&2piint_{-1}^1 r^2big|^{2+sqrt{1-z^2}}_{2-sqrt{1-z^2}} ;dz\
&=&16pi int_{-1}^1 sqrt{1-z^2} ;dz\
&=&16pi int_{-frac{pi}{2}}^frac{pi}{2} cos^2 u ;du=8pi^2.
end{eqnarray}$$
answered Jan 12 at 9:59
SongSong
12.3k630
$endgroup$
For $a)$, we can calculate $v:deltaOmega to mathbb{R}^3$ using parametrization of $delta Omega$. Note that $v$ depends only on $Omega$, and is completely irrelevant to the given vector field $f$. (So your calculation is not valid.) Let $sqrt{x^2+y^2} = r$. Then we can write
$$
Omega : 2-sqrt{1-z^2}<r<2+sqrt{1-z^2}.
$$ Hence $Omega = Omega_1 -Omega_2$ where $Omega_1:r<2+sqrt{1-z^2}$ and $Omega_2: r<2-sqrt{1-z^2}$. We can calculate $v$ on $deltaOmega_1$ as follows:
$$
nabla_{x,y,z}(r-2-sqrt{1-z^2})=(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})',
$$and by normalizing $$
v(x,y,z) = sqrt{1-z^2}(frac{x}{r},frac{y}{r},frac{z}{sqrt{1-z^2}})'.
$$ We can check that $v$ is outward.
Similarly for $deltaOmega_2$,
$$
nabla_{x,y,z}(r-2+sqrt{1-z^2})=(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'
$$ and
$$
v(x,y,z) = -sqrt{1-z^2}(frac{x}{r},frac{y}{r},-frac{z}{sqrt{1-z^2}})'.
$$ (The sign is reversed.)
For $b)$, it is true that $int_{deltaOmega}langle f,vrangle dS=int_Omega nablacdot f;dxdydz=2text{vol}(Omega).$ We can calculate this using cylindrical coordinate
$$
(x,y,z)=(rcostheta, rsintheta, z).
$$ Then, $dxdydz = rdrdtheta dz$ and the given integral becomes
$$begin{eqnarray}
2int_Omega dxdydz &=&2int_{-1}^1 int_0^{2pi}int_{2-sqrt{1-z^2}}^{2+sqrt{1-z^2}}rdrdtheta dz\
&=&2piint_{-1}^1 r^2big|^{2+sqrt{1-z^2}}_{2-sqrt{1-z^2}} ;dz\
&=&16pi int_{-1}^1 sqrt{1-z^2} ;dz\
&=&16pi int_{-frac{pi}{2}}^frac{pi}{2} cos^2 u ;du=8pi^2.
end{eqnarray}$$
answered Jan 12 at 9:59
SongSong
12.3k630
edited Jan 12 at 10:06
answered Jan 12 at 9:59
SongSong
12.3k630
answered Jan 12 at 9:59
SongSong
12.3k630
answered Jan 12 at 9:59
SongSong
12.3k630
12.3k630
add a comment |
add a comment |
$begingroup$
For a) it is best to turn to cylindrical coordinates. Note that for a relation $f(x,y,z)=0$ the gradient $nabla f(x,y,z)$ is always normal to it, therefore if the equation of torus is $$f(r,phi ,z)=(r-2)^2+z^2-1=0$$then we have$$vec v={nabla f(r,phi , z)over |nabla f(r,phi , z)|_2}={(r-2,0,z)over sqrt{(r-2)^2+z^2}}$$
b) $$int_{deltaOmega}langle f,vrangle dS{=int_V nablacdot f(x,y,z)dV\=int_V 2dV\=2int_{(r-2)^2+z^2<1}rdrdphi dz}$$note that $1<r<3$ and $-1<z<1$ therefore by the definition of torus we conclude that $$2-sqrt{1-z^2}<r<2+sqrt{1-z^2}$$and $$5-z^2-4sqrt{1-z^2}<r^2<5-z^2+4sqrt{1-z^2}$$so we can write$$2int_{(r-2)^2+z^2<1}rdrdphi dz{=int_{(r-2)^2+z^2<1}dr^2dphi dz\=2pi int_{(r-2)^2+z^2<1}dr^2 dz\=2pi int_{-1}^1int_{r^2=5-z^2-4sqrt{1-z^2}}^{r^2=5-z^2+4sqrt{1-z^2}}dr^2 dz\=16piint_{-1}^1sqrt{1-z^2}dz\=8pi ^2}$$therefore$$int_{deltaOmega}langle f,vrangle dS=8pi^2$$
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
For a) it is best to turn to cylindrical coordinates. Note that for a relation $f(x,y,z)=0$ the gradient $nabla f(x,y,z)$ is always normal to it, therefore if the equation of torus is $$f(r,phi ,z)=(r-2)^2+z^2-1=0$$then we have$$vec v={nabla f(r,phi , z)over |nabla f(r,phi , z)|_2}={(r-2,0,z)over sqrt{(r-2)^2+z^2}}$$
b) $$int_{deltaOmega}langle f,vrangle dS{=int_V nablacdot f(x,y,z)dV\=int_V 2dV\=2int_{(r-2)^2+z^2<1}rdrdphi dz}$$note that $1<r<3$ and $-1<z<1$ therefore by the definition of torus we conclude that $$2-sqrt{1-z^2}<r<2+sqrt{1-z^2}$$and $$5-z^2-4sqrt{1-z^2}<r^2<5-z^2+4sqrt{1-z^2}$$so we can write$$2int_{(r-2)^2+z^2<1}rdrdphi dz{=int_{(r-2)^2+z^2<1}dr^2dphi dz\=2pi int_{(r-2)^2+z^2<1}dr^2 dz\=2pi int_{-1}^1int_{r^2=5-z^2-4sqrt{1-z^2}}^{r^2=5-z^2+4sqrt{1-z^2}}dr^2 dz\=16piint_{-1}^1sqrt{1-z^2}dz\=8pi ^2}$$therefore$$int_{deltaOmega}langle f,vrangle dS=8pi^2$$
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
For a) it is best to turn to cylindrical coordinates. Note that for a relation $f(x,y,z)=0$ the gradient $nabla f(x,y,z)$ is always normal to it, therefore if the equation of torus is $$f(r,phi ,z)=(r-2)^2+z^2-1=0$$then we have$$vec v={nabla f(r,phi , z)over |nabla f(r,phi , z)|_2}={(r-2,0,z)over sqrt{(r-2)^2+z^2}}$$
b) $$int_{deltaOmega}langle f,vrangle dS{=int_V nablacdot f(x,y,z)dV\=int_V 2dV\=2int_{(r-2)^2+z^2<1}rdrdphi dz}$$note that $1<r<3$ and $-1<z<1$ therefore by the definition of torus we conclude that $$2-sqrt{1-z^2}<r<2+sqrt{1-z^2}$$and $$5-z^2-4sqrt{1-z^2}<r^2<5-z^2+4sqrt{1-z^2}$$so we can write$$2int_{(r-2)^2+z^2<1}rdrdphi dz{=int_{(r-2)^2+z^2<1}dr^2dphi dz\=2pi int_{(r-2)^2+z^2<1}dr^2 dz\=2pi int_{-1}^1int_{r^2=5-z^2-4sqrt{1-z^2}}^{r^2=5-z^2+4sqrt{1-z^2}}dr^2 dz\=16piint_{-1}^1sqrt{1-z^2}dz\=8pi ^2}$$therefore$$int_{deltaOmega}langle f,vrangle dS=8pi^2$$
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
For a) it is best to turn to cylindrical coordinates. Note that for a relation $f(x,y,z)=0$ the gradient $nabla f(x,y,z)$ is always normal to it, therefore if the equation of torus is $$f(r,phi ,z)=(r-2)^2+z^2-1=0$$then we have$$vec v={nabla f(r,phi , z)over |nabla f(r,phi , z)|_2}={(r-2,0,z)over sqrt{(r-2)^2+z^2}}$$
b) $$int_{deltaOmega}langle f,vrangle dS{=int_V nablacdot f(x,y,z)dV\=int_V 2dV\=2int_{(r-2)^2+z^2<1}rdrdphi dz}$$note that $1<r<3$ and $-1<z<1$ therefore by the definition of torus we conclude that $$2-sqrt{1-z^2}<r<2+sqrt{1-z^2}$$and $$5-z^2-4sqrt{1-z^2}<r^2<5-z^2+4sqrt{1-z^2}$$so we can write$$2int_{(r-2)^2+z^2<1}rdrdphi dz{=int_{(r-2)^2+z^2<1}dr^2dphi dz\=2pi int_{(r-2)^2+z^2<1}dr^2 dz\=2pi int_{-1}^1int_{r^2=5-z^2-4sqrt{1-z^2}}^{r^2=5-z^2+4sqrt{1-z^2}}dr^2 dz\=16piint_{-1}^1sqrt{1-z^2}dz\=8pi ^2}$$therefore$$int_{deltaOmega}langle f,vrangle dS=8pi^2$$
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 9:54
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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