Find vector flux of $v=(yz,y^2z, yz^2)$ through the surface of the cylinder $x^2+y^2=1, 0 leq z leq 1$
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
547416
add a comment |
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
547416
1
I agree with everything above.
– Doug M
Dec 28 '18 at 19:42
@DougM Ah, okay. This was quick haha.
– Poujh
Dec 28 '18 at 19:43
Why do you have $4y$ on the last line?
– Andrei
Dec 28 '18 at 21:30
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
– Poujh
Dec 28 '18 at 22:28
2
@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.
– Andrei
Dec 28 '18 at 22:36
add a comment |
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
547416
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
547416
asked Dec 28 '18 at 19:39
PoujhPoujh
547416
edited yesterday
Poujh
asked Dec 28 '18 at 19:39
PoujhPoujh
547416
asked Dec 28 '18 at 19:39
PoujhPoujh
547416
asked Dec 28 '18 at 19:39
PoujhPoujh
547416
547416
1
I agree with everything above.
– Doug M
Dec 28 '18 at 19:42
@DougM Ah, okay. This was quick haha.
– Poujh
Dec 28 '18 at 19:43
Why do you have $4y$ on the last line?
– Andrei
Dec 28 '18 at 21:30
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
– Poujh
Dec 28 '18 at 22:28
2
@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.
– Andrei
Dec 28 '18 at 22:36
add a comment |
1
I agree with everything above.
– Doug M
Dec 28 '18 at 19:42
@DougM Ah, okay. This was quick haha.
– Poujh
Dec 28 '18 at 19:43
Why do you have $4y$ on the last line?
– Andrei
Dec 28 '18 at 21:30
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
– Poujh
Dec 28 '18 at 22:28
2
@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.
– Andrei
Dec 28 '18 at 22:36
1
1
I agree with everything above.
– Doug M
Dec 28 '18 at 19:42
I agree with everything above.
– Doug M
Dec 28 '18 at 19:42
@DougM Ah, okay. This was quick haha.
– Poujh
Dec 28 '18 at 19:43
@DougM Ah, okay. This was quick haha.
– Poujh
Dec 28 '18 at 19:43
Why do you have $4y$ on the last line?
– Andrei
Dec 28 '18 at 21:30
Why do you have $4y$ on the last line?
– Andrei
Dec 28 '18 at 21:30
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
– Poujh
Dec 28 '18 at 22:28
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
– Poujh
Dec 28 '18 at 22:28
2
2
@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.
– Andrei
Dec 28 '18 at 22:36
@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.
– Andrei
Dec 28 '18 at 22:36
add a comment |
1 Answer
1
active
oldest
votes
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
11.5k21026
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055227%2ffind-vector-flux-of-v-yz-y2z-yz2-through-the-surface-of-the-cylinder-x2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
11.5k21026
add a comment |
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
11.5k21026
add a comment |
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
11.5k21026
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
11.5k21026
answered Dec 28 '18 at 22:04
AndreiAndrei
11.5k21026
answered Dec 28 '18 at 22:04
AndreiAndrei
11.5k21026
answered Dec 28 '18 at 22:04
AndreiAndrei
11.5k21026
11.5k21026
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055227%2ffind-vector-flux-of-v-yz-y2z-yz2-through-the-surface-of-the-cylinder-x2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
I agree with everything above.
– Doug M
Dec 28 '18 at 19:42
@DougM Ah, okay. This was quick haha.
– Poujh
Dec 28 '18 at 19:43
Why do you have $4y$ on the last line?
– Andrei
Dec 28 '18 at 21:30
@Andrei It’s 4yz with z= 1 (the circle on top of the cylinder)
– Poujh
Dec 28 '18 at 22:28
2
@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.
– Andrei
Dec 28 '18 at 22:36