Find vector flux of $v=(yz,y^2z, yz^2)$ through the surface of the cylinder $x^2+y^2=1, 0 leq z leq 1$
$begingroup$
I'm given the following vector field : $v=(yz,y^2z, yz^2)$ and I need to find its flux through the cylinder $x^2+y^2=1, 0 leq z leq 1$. I don't have solutions to this exercise, so I don't know if my results are correct.
The divergence of $v$ is simply $4yz$. So if we set up our integral in cylindrical coordinates, we get $$int_{0}^{1}int_{0}^{2pi}int_{0}^{1}4zrsin(theta) r drdtheta dz$$. Now, if we evaluate this, we get $0$ because $int_{0}^{2pi}sin(theta)=0$. So the result will simply be $0$ even if the original vector field could seem a bit intimidating. We could also use a symmetry argument : integrating the variable $y$, which is an odd function, over a symmetric interval, here a circle, will always yield $0$.
Also, the integrals over the two "surface circles", where $z=0$ and $z=1$, also yield $0$ because for the first one, we integrate over $0$ and for the second, we have $4y$ which is again odd over a symmetric interval.
Do you agree with this ? Thanks for your help !
Edit : as was pointed out in the comments, if we use the divergence theorem, we don't need to evaluate over surfaces, and thus the integrals over the circles where z=0 resp. z=1 are not required at all. This step would be required if we proceed without using the divergence theorem. In this case, we would also need to evaluate over the circular surface of the cylinder.
real-analysis multivariable-calculus vector-analysis vector-fields divergence
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
$endgroup$
add a comment |
$begingroup$
I'm given the following vector field : $v=(yz,y^2z, yz^2)$ and I need to find its flux through the cylinder $x^2+y^2=1, 0 leq z leq 1$. I don't have solutions to this exercise, so I don't know if my results are correct.
The divergence of $v$ is simply $4yz$. So if we set up our integral in cylindrical coordinates, we get $$int_{0}^{1}int_{0}^{2pi}int_{0}^{1}4zrsin(theta) r drdtheta dz$$. Now, if we evaluate this, we get $0$ because $int_{0}^{2pi}sin(theta)=0$. So the result will simply be $0$ even if the original vector field could seem a bit intimidating. We could also use a symmetry argument : integrating the variable $y$, which is an odd function, over a symmetric interval, here a circle, will always yield $0$.
Also, the integrals over the two "surface circles", where $z=0$ and $z=1$, also yield $0$ because for the first one, we integrate over $0$ and for the second, we have $4y$ which is again odd over a symmetric interval.
Do you agree with this ? Thanks for your help !
Edit : as was pointed out in the comments, if we use the divergence theorem, we don't need to evaluate over surfaces, and thus the integrals over the circles where z=0 resp. z=1 are not required at all. This step would be required if we proceed without using the divergence theorem. In this case, we would also need to evaluate over the circular surface of the cylinder.
real-analysis multivariable-calculus vector-analysis vector-fields divergence
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
$endgroup$
1
$begingroup$
I agree with everything above.
$endgroup$
– Doug M
Dec 28 '18 at 19:42
$begingroup$
@DougM Ah, okay. This was quick haha.
$endgroup$
– Poujh
Dec 28 '18 at 19:43
$begingroup$
Why do you have $4y$ on the last line?
$endgroup$
– Andrei
Dec 28 '18 at 21:30
$begingroup$
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
$endgroup$
– Poujh
Dec 28 '18 at 22:28
2
$begingroup$
@Poujh You either do volume integral, so then you don't integrate the divergence over the surface, or you do a surface integral, and then the flux on top surface is $yz^2$ with $z=1$, as in my reply below. Be careful not to mix the two approaches.
$endgroup$
– Andrei
Dec 28 '18 at 22:36
add a comment |
$begingroup$
I'm given the following vector field : $v=(yz,y^2z, yz^2)$ and I need to find its flux through the cylinder $x^2+y^2=1, 0 leq z leq 1$. I don't have solutions to this exercise, so I don't know if my results are correct.
The divergence of $v$ is simply $4yz$. So if we set up our integral in cylindrical coordinates, we get $$int_{0}^{1}int_{0}^{2pi}int_{0}^{1}4zrsin(theta) r drdtheta dz$$. Now, if we evaluate this, we get $0$ because $int_{0}^{2pi}sin(theta)=0$. So the result will simply be $0$ even if the original vector field could seem a bit intimidating. We could also use a symmetry argument : integrating the variable $y$, which is an odd function, over a symmetric interval, here a circle, will always yield $0$.
Also, the integrals over the two "surface circles", where $z=0$ and $z=1$, also yield $0$ because for the first one, we integrate over $0$ and for the second, we have $4y$ which is again odd over a symmetric interval.
Do you agree with this ? Thanks for your help !
Edit : as was pointed out in the comments, if we use the divergence theorem, we don't need to evaluate over surfaces, and thus the integrals over the circles where z=0 resp. z=1 are not required at all. This step would be required if we proceed without using the divergence theorem. In this case, we would also need to evaluate over the circular surface of the cylinder.
real-analysis multivariable-calculus vector-analysis vector-fields divergence
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
$endgroup$
I'm given the following vector field : $v=(yz,y^2z, yz^2)$ and I need to find its flux through the cylinder $x^2+y^2=1, 0 leq z leq 1$. I don't have solutions to this exercise, so I don't know if my results are correct.
The divergence of $v$ is simply $4yz$. So if we set up our integral in cylindrical coordinates, we get $$int_{0}^{1}int_{0}^{2pi}int_{0}^{1}4zrsin(theta) r drdtheta dz$$. Now, if we evaluate this, we get $0$ because $int_{0}^{2pi}sin(theta)=0$. So the result will simply be $0$ even if the original vector field could seem a bit intimidating. We could also use a symmetry argument : integrating the variable $y$, which is an odd function, over a symmetric interval, here a circle, will always yield $0$.
Also, the integrals over the two "surface circles", where $z=0$ and $z=1$, also yield $0$ because for the first one, we integrate over $0$ and for the second, we have $4y$ which is again odd over a symmetric interval.
Do you agree with this ? Thanks for your help !
Edit : as was pointed out in the comments, if we use the divergence theorem, we don't need to evaluate over surfaces, and thus the integrals over the circles where z=0 resp. z=1 are not required at all. This step would be required if we proceed without using the divergence theorem. In this case, we would also need to evaluate over the circular surface of the cylinder.
real-analysis multivariable-calculus vector-analysis vector-fields divergence
real-analysis multivariable-calculus vector-analysis vector-fields divergence
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
edited Jan 7 at 19:22
Poujh
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
asked Dec 28 '18 at 19:39
PoujhPoujh
6111516
6111516
1
$begingroup$
I agree with everything above.
$endgroup$
– Doug M
Dec 28 '18 at 19:42
$begingroup$
@DougM Ah, okay. This was quick haha.
$endgroup$
– Poujh
Dec 28 '18 at 19:43
$begingroup$
Why do you have $4y$ on the last line?
$endgroup$
– Andrei
Dec 28 '18 at 21:30
$begingroup$
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
$endgroup$
– Poujh
Dec 28 '18 at 22:28
2
$begingroup$
@Poujh You either do volume integral, so then you don't integrate the divergence over the surface, or you do a surface integral, and then the flux on top surface is $yz^2$ with $z=1$, as in my reply below. Be careful not to mix the two approaches.
$endgroup$
– Andrei
Dec 28 '18 at 22:36
add a comment |
1
$begingroup$
I agree with everything above.
$endgroup$
– Doug M
Dec 28 '18 at 19:42
$begingroup$
@DougM Ah, okay. This was quick haha.
$endgroup$
– Poujh
Dec 28 '18 at 19:43
$begingroup$
Why do you have $4y$ on the last line?
$endgroup$
– Andrei
Dec 28 '18 at 21:30
$begingroup$
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
$endgroup$
– Poujh
Dec 28 '18 at 22:28
2
$begingroup$
@Poujh You either do volume integral, so then you don't integrate the divergence over the surface, or you do a surface integral, and then the flux on top surface is $yz^2$ with $z=1$, as in my reply below. Be careful not to mix the two approaches.
$endgroup$
– Andrei
Dec 28 '18 at 22:36
1
1
$begingroup$
I agree with everything above.
$endgroup$
– Doug M
Dec 28 '18 at 19:42
$begingroup$
I agree with everything above.
$endgroup$
– Doug M
Dec 28 '18 at 19:42
$begingroup$
@DougM Ah, okay. This was quick haha.
$endgroup$
– Poujh
Dec 28 '18 at 19:43
$begingroup$
@DougM Ah, okay. This was quick haha.
$endgroup$
– Poujh
Dec 28 '18 at 19:43
$begingroup$
Why do you have $4y$ on the last line?
$endgroup$
– Andrei
Dec 28 '18 at 21:30
$begingroup$
Why do you have $4y$ on the last line?
$endgroup$
– Andrei
Dec 28 '18 at 21:30
$begingroup$
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
$endgroup$
– Poujh
Dec 28 '18 at 22:28
$begingroup$
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
$endgroup$
– Poujh
Dec 28 '18 at 22:28
2
2
$begingroup$
@Poujh You either do volume integral, so then you don't integrate the divergence over the surface, or you do a surface integral, and then the flux on top surface is $yz^2$ with $z=1$, as in my reply below. Be careful not to mix the two approaches.
$endgroup$
– Andrei
Dec 28 '18 at 22:36
$begingroup$
@Poujh You either do volume integral, so then you don't integrate the divergence over the surface, or you do a surface integral, and then the flux on top surface is $yz^2$ with $z=1$, as in my reply below. Be careful not to mix the two approaches.
$endgroup$
– Andrei
Dec 28 '18 at 22:36
add a comment |
1 Answer
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You can get the same answer even without the divergence theorem. The normal to the bottom surface is $(0,0,-1)$, so the flux is $$Phi_B=int_B(0,0,-1)cdot(0,0,0)dS=0$$ The normal to the top ($z=1$) is $(0,0,1)$, so the flux is $$Phi_T=int_T(0,0,1)cdot(y,y^2,y)dS=int_TydS=0$$
In the last step I've relied on symmetry.
If you draw a picture of a circle with radius $1$, for any point $(x,y)$ on the circle, the normal to the circle is $(x,y)$, so in the case of the cylinder, the normal to the side is $(x,y,0)$. Then the flux through the side is $$Phi_S=int_S(x,y,0)cdot(yz,y^2z,yz^2)dS=int_S(xyz+y^3z)dS=int_0^1zdzint_0^{2pi}dtheta(sinthetacostheta+sin^3theta)$$
It's trivial to show that the last integral is zero as well. So the total flux is zero, as you have found from the divergence theorem.
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
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You can get the same answer even without the divergence theorem. The normal to the bottom surface is $(0,0,-1)$, so the flux is $$Phi_B=int_B(0,0,-1)cdot(0,0,0)dS=0$$ The normal to the top ($z=1$) is $(0,0,1)$, so the flux is $$Phi_T=int_T(0,0,1)cdot(y,y^2,y)dS=int_TydS=0$$
In the last step I've relied on symmetry.
If you draw a picture of a circle with radius $1$, for any point $(x,y)$ on the circle, the normal to the circle is $(x,y)$, so in the case of the cylinder, the normal to the side is $(x,y,0)$. Then the flux through the side is $$Phi_S=int_S(x,y,0)cdot(yz,y^2z,yz^2)dS=int_S(xyz+y^3z)dS=int_0^1zdzint_0^{2pi}dtheta(sinthetacostheta+sin^3theta)$$
It's trivial to show that the last integral is zero as well. So the total flux is zero, as you have found from the divergence theorem.
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
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add a comment |
$begingroup$
You can get the same answer even without the divergence theorem. The normal to the bottom surface is $(0,0,-1)$, so the flux is $$Phi_B=int_B(0,0,-1)cdot(0,0,0)dS=0$$ The normal to the top ($z=1$) is $(0,0,1)$, so the flux is $$Phi_T=int_T(0,0,1)cdot(y,y^2,y)dS=int_TydS=0$$
In the last step I've relied on symmetry.
If you draw a picture of a circle with radius $1$, for any point $(x,y)$ on the circle, the normal to the circle is $(x,y)$, so in the case of the cylinder, the normal to the side is $(x,y,0)$. Then the flux through the side is $$Phi_S=int_S(x,y,0)cdot(yz,y^2z,yz^2)dS=int_S(xyz+y^3z)dS=int_0^1zdzint_0^{2pi}dtheta(sinthetacostheta+sin^3theta)$$
It's trivial to show that the last integral is zero as well. So the total flux is zero, as you have found from the divergence theorem.
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
$endgroup$
add a comment |
$begingroup$
You can get the same answer even without the divergence theorem. The normal to the bottom surface is $(0,0,-1)$, so the flux is $$Phi_B=int_B(0,0,-1)cdot(0,0,0)dS=0$$ The normal to the top ($z=1$) is $(0,0,1)$, so the flux is $$Phi_T=int_T(0,0,1)cdot(y,y^2,y)dS=int_TydS=0$$
In the last step I've relied on symmetry.
If you draw a picture of a circle with radius $1$, for any point $(x,y)$ on the circle, the normal to the circle is $(x,y)$, so in the case of the cylinder, the normal to the side is $(x,y,0)$. Then the flux through the side is $$Phi_S=int_S(x,y,0)cdot(yz,y^2z,yz^2)dS=int_S(xyz+y^3z)dS=int_0^1zdzint_0^{2pi}dtheta(sinthetacostheta+sin^3theta)$$
It's trivial to show that the last integral is zero as well. So the total flux is zero, as you have found from the divergence theorem.
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
$endgroup$
You can get the same answer even without the divergence theorem. The normal to the bottom surface is $(0,0,-1)$, so the flux is $$Phi_B=int_B(0,0,-1)cdot(0,0,0)dS=0$$ The normal to the top ($z=1$) is $(0,0,1)$, so the flux is $$Phi_T=int_T(0,0,1)cdot(y,y^2,y)dS=int_TydS=0$$
In the last step I've relied on symmetry.
If you draw a picture of a circle with radius $1$, for any point $(x,y)$ on the circle, the normal to the circle is $(x,y)$, so in the case of the cylinder, the normal to the side is $(x,y,0)$. Then the flux through the side is $$Phi_S=int_S(x,y,0)cdot(yz,y^2z,yz^2)dS=int_S(xyz+y^3z)dS=int_0^1zdzint_0^{2pi}dtheta(sinthetacostheta+sin^3theta)$$
It's trivial to show that the last integral is zero as well. So the total flux is zero, as you have found from the divergence theorem.
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
answered Dec 28 '18 at 22:04
AndreiAndrei
13.7k21230
13.7k21230
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1
$begingroup$
I agree with everything above.
$endgroup$
– Doug M
Dec 28 '18 at 19:42
$begingroup$
@DougM Ah, okay. This was quick haha.
$endgroup$
– Poujh
Dec 28 '18 at 19:43
$begingroup$
Why do you have $4y$ on the last line?
$endgroup$
– Andrei
Dec 28 '18 at 21:30
$begingroup$
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
$endgroup$
– Poujh
Dec 28 '18 at 22:28
2
$begingroup$
@Poujh You either do volume integral, so then you don't integrate the divergence over the surface, or you do a surface integral, and then the flux on top surface is $yz^2$ with $z=1$, as in my reply below. Be careful not to mix the two approaches.
$endgroup$
– Andrei
Dec 28 '18 at 22:36