flux of $langle x,y,z^2rangle$ across unit sphere
$begingroup$
I'm trying to compute $iint_S Fcdot$ dS where $F=langle x,y,z^2rangle$ and $S$ is the unit sphere centered at the origin.
Here's my attempt:
On the sphere we can describe any point by $r(phi,theta)=langle sinphi costheta,sinphi costheta, costhetarangle$ and the outward normal vector to $S$ is given by $n=r.$
Therefore,
$$iint_S Fcdot dS = iint_D F(r(phi,theta))cdot n dA$$
$$=iint_D langle sinphicostheta,sinphisintheta,cos^2thetaranglecdot langle
sinphicostheta,sinphisintheta,costhetarangle dA$$
$$=iint_D sin^2phi(cos^2theta + sin^2theta)+cos^3dtheta dA = int_0^{pi}int_0^{2pi} sin^2phi +cos^3theta dtheta dphi = pi^2$$
however the answer is $frac{8}{3}pi$ and the last equals sign is correct (used Wolfram to confirm), so I guess I did something wrong in the setup.
multivariable-calculus proof-verification vector-analysis
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
$endgroup$
add a comment |
$begingroup$
I'm trying to compute $iint_S Fcdot$ dS where $F=langle x,y,z^2rangle$ and $S$ is the unit sphere centered at the origin.
Here's my attempt:
On the sphere we can describe any point by $r(phi,theta)=langle sinphi costheta,sinphi costheta, costhetarangle$ and the outward normal vector to $S$ is given by $n=r.$
Therefore,
$$iint_S Fcdot dS = iint_D F(r(phi,theta))cdot n dA$$
$$=iint_D langle sinphicostheta,sinphisintheta,cos^2thetaranglecdot langle
sinphicostheta,sinphisintheta,costhetarangle dA$$
$$=iint_D sin^2phi(cos^2theta + sin^2theta)+cos^3dtheta dA = int_0^{pi}int_0^{2pi} sin^2phi +cos^3theta dtheta dphi = pi^2$$
however the answer is $frac{8}{3}pi$ and the last equals sign is correct (used Wolfram to confirm), so I guess I did something wrong in the setup.
multivariable-calculus proof-verification vector-analysis
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
$endgroup$
add a comment |
$begingroup$
I'm trying to compute $iint_S Fcdot$ dS where $F=langle x,y,z^2rangle$ and $S$ is the unit sphere centered at the origin.
Here's my attempt:
On the sphere we can describe any point by $r(phi,theta)=langle sinphi costheta,sinphi costheta, costhetarangle$ and the outward normal vector to $S$ is given by $n=r.$
Therefore,
$$iint_S Fcdot dS = iint_D F(r(phi,theta))cdot n dA$$
$$=iint_D langle sinphicostheta,sinphisintheta,cos^2thetaranglecdot langle
sinphicostheta,sinphisintheta,costhetarangle dA$$
$$=iint_D sin^2phi(cos^2theta + sin^2theta)+cos^3dtheta dA = int_0^{pi}int_0^{2pi} sin^2phi +cos^3theta dtheta dphi = pi^2$$
however the answer is $frac{8}{3}pi$ and the last equals sign is correct (used Wolfram to confirm), so I guess I did something wrong in the setup.
multivariable-calculus proof-verification vector-analysis
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
$endgroup$
I'm trying to compute $iint_S Fcdot$ dS where $F=langle x,y,z^2rangle$ and $S$ is the unit sphere centered at the origin.
Here's my attempt:
On the sphere we can describe any point by $r(phi,theta)=langle sinphi costheta,sinphi costheta, costhetarangle$ and the outward normal vector to $S$ is given by $n=r.$
Therefore,
$$iint_S Fcdot dS = iint_D F(r(phi,theta))cdot n dA$$
$$=iint_D langle sinphicostheta,sinphisintheta,cos^2thetaranglecdot langle
sinphicostheta,sinphisintheta,costhetarangle dA$$
$$=iint_D sin^2phi(cos^2theta + sin^2theta)+cos^3dtheta dA = int_0^{pi}int_0^{2pi} sin^2phi +cos^3theta dtheta dphi = pi^2$$
however the answer is $frac{8}{3}pi$ and the last equals sign is correct (used Wolfram to confirm), so I guess I did something wrong in the setup.
multivariable-calculus proof-verification vector-analysis
multivariable-calculus proof-verification vector-analysis
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
asked Dec 28 '18 at 23:43
ClclstdntClclstdnt
473414
473414
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The position vector should be
$$r(phi,theta)=langle sintheta cosphi,sintheta sinphi, costhetarangle.$$
And the volume element should be
$$ dA = langle sintheta cosphi,sintheta sinphi, costhetarangle sin theta dtheta dphi.$$
So the integral should be
$$iint_S Fcdot dA = int_{theta = 0}^{theta = pi} int_{phi = 0}^{phi = 2pi} left( sin^2 theta + cos^3 thetaright) sin theta dtheta dphi.$$
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
$endgroup$
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
1
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
2
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
|
show 2 more comments
$begingroup$
You can use the divergence theorem to simplify things. We have
$$operatorname{div} vec{F}(x,y,z) = frac{partial F_x}{partial x} + frac{partial F_y}{partial y} + frac{partial F_z}{partial z} = 2+2z$$
so
$$I = iintlimits_{text{unit sphere}} vec{F}cdot dmathbf{S} = iiintlimits_{text{unit ball}} operatorname{div} vec{F} ,dV = int_{phi = 0}^{2pi} int_{theta = 0}^pi int_{r=0}^1 2(1+rcostheta)underbrace{r^2sintheta,dr,dtheta,dphi}_{dV}$$
The integrand is independent of $phi$ so it factors out as $2pi$, and then the second integral vanishes:
$$I = 4pileft(int_{theta = 0}^pi int_{r=0}^1 r^2sintheta,dr,dtheta + int_{theta = 0}^pi int_{r=0}^1 r^3costhetasintheta,dr,dthetaright) = frac83pi$$
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
$endgroup$
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%2f3055396%2fflux-of-langle-x-y-z2-rangle-across-unit-sphere%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The position vector should be
$$r(phi,theta)=langle sintheta cosphi,sintheta sinphi, costhetarangle.$$
And the volume element should be
$$ dA = langle sintheta cosphi,sintheta sinphi, costhetarangle sin theta dtheta dphi.$$
So the integral should be
$$iint_S Fcdot dA = int_{theta = 0}^{theta = pi} int_{phi = 0}^{phi = 2pi} left( sin^2 theta + cos^3 thetaright) sin theta dtheta dphi.$$
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
$endgroup$
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
1
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
2
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
|
show 2 more comments
$begingroup$
The position vector should be
$$r(phi,theta)=langle sintheta cosphi,sintheta sinphi, costhetarangle.$$
And the volume element should be
$$ dA = langle sintheta cosphi,sintheta sinphi, costhetarangle sin theta dtheta dphi.$$
So the integral should be
$$iint_S Fcdot dA = int_{theta = 0}^{theta = pi} int_{phi = 0}^{phi = 2pi} left( sin^2 theta + cos^3 thetaright) sin theta dtheta dphi.$$
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
$endgroup$
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
1
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
2
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
|
show 2 more comments
$begingroup$
The position vector should be
$$r(phi,theta)=langle sintheta cosphi,sintheta sinphi, costhetarangle.$$
And the volume element should be
$$ dA = langle sintheta cosphi,sintheta sinphi, costhetarangle sin theta dtheta dphi.$$
So the integral should be
$$iint_S Fcdot dA = int_{theta = 0}^{theta = pi} int_{phi = 0}^{phi = 2pi} left( sin^2 theta + cos^3 thetaright) sin theta dtheta dphi.$$
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
$endgroup$
The position vector should be
$$r(phi,theta)=langle sintheta cosphi,sintheta sinphi, costhetarangle.$$
And the volume element should be
$$ dA = langle sintheta cosphi,sintheta sinphi, costhetarangle sin theta dtheta dphi.$$
So the integral should be
$$iint_S Fcdot dA = int_{theta = 0}^{theta = pi} int_{phi = 0}^{phi = 2pi} left( sin^2 theta + cos^3 thetaright) sin theta dtheta dphi.$$
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
edited Dec 29 '18 at 1:09
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
answered Dec 28 '18 at 23:50
Kenny WongKenny Wong
19.1k21441
19.1k21441
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
1
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
2
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
|
show 2 more comments
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
1
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
2
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
$begingroup$
dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication?
$endgroup$
– Clclstdnt
Dec 29 '18 at 0:54
1
1
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
@Clclstdnt I meant ordinary multiplication. Sorry, will change the notation.
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:09
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA?
$endgroup$
– Clclstdnt
Dec 29 '18 at 1:13
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
$begingroup$
$sin theta dtheta dphi$ is the area element on the surface of a unit sphere. [… while $r^2 sin theta dr theta dphi$ is the volume element in $mathbb R^3$]
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:14
2
2
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
$begingroup$
@Clclstdnt It's a good thing to learn off by heart. You can also derive it using $dvec A = frac{dvec r}{dtheta} times frac{dvec r}{dphi}dtheta dphi$. (And this time, I really do mean the cross product.)
$endgroup$
– Kenny Wong
Dec 29 '18 at 1:18
|
show 2 more comments
$begingroup$
You can use the divergence theorem to simplify things. We have
$$operatorname{div} vec{F}(x,y,z) = frac{partial F_x}{partial x} + frac{partial F_y}{partial y} + frac{partial F_z}{partial z} = 2+2z$$
so
$$I = iintlimits_{text{unit sphere}} vec{F}cdot dmathbf{S} = iiintlimits_{text{unit ball}} operatorname{div} vec{F} ,dV = int_{phi = 0}^{2pi} int_{theta = 0}^pi int_{r=0}^1 2(1+rcostheta)underbrace{r^2sintheta,dr,dtheta,dphi}_{dV}$$
The integrand is independent of $phi$ so it factors out as $2pi$, and then the second integral vanishes:
$$I = 4pileft(int_{theta = 0}^pi int_{r=0}^1 r^2sintheta,dr,dtheta + int_{theta = 0}^pi int_{r=0}^1 r^3costhetasintheta,dr,dthetaright) = frac83pi$$
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
$endgroup$
add a comment |
$begingroup$
You can use the divergence theorem to simplify things. We have
$$operatorname{div} vec{F}(x,y,z) = frac{partial F_x}{partial x} + frac{partial F_y}{partial y} + frac{partial F_z}{partial z} = 2+2z$$
so
$$I = iintlimits_{text{unit sphere}} vec{F}cdot dmathbf{S} = iiintlimits_{text{unit ball}} operatorname{div} vec{F} ,dV = int_{phi = 0}^{2pi} int_{theta = 0}^pi int_{r=0}^1 2(1+rcostheta)underbrace{r^2sintheta,dr,dtheta,dphi}_{dV}$$
The integrand is independent of $phi$ so it factors out as $2pi$, and then the second integral vanishes:
$$I = 4pileft(int_{theta = 0}^pi int_{r=0}^1 r^2sintheta,dr,dtheta + int_{theta = 0}^pi int_{r=0}^1 r^3costhetasintheta,dr,dthetaright) = frac83pi$$
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
$endgroup$
add a comment |
$begingroup$
You can use the divergence theorem to simplify things. We have
$$operatorname{div} vec{F}(x,y,z) = frac{partial F_x}{partial x} + frac{partial F_y}{partial y} + frac{partial F_z}{partial z} = 2+2z$$
so
$$I = iintlimits_{text{unit sphere}} vec{F}cdot dmathbf{S} = iiintlimits_{text{unit ball}} operatorname{div} vec{F} ,dV = int_{phi = 0}^{2pi} int_{theta = 0}^pi int_{r=0}^1 2(1+rcostheta)underbrace{r^2sintheta,dr,dtheta,dphi}_{dV}$$
The integrand is independent of $phi$ so it factors out as $2pi$, and then the second integral vanishes:
$$I = 4pileft(int_{theta = 0}^pi int_{r=0}^1 r^2sintheta,dr,dtheta + int_{theta = 0}^pi int_{r=0}^1 r^3costhetasintheta,dr,dthetaright) = frac83pi$$
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
$endgroup$
You can use the divergence theorem to simplify things. We have
$$operatorname{div} vec{F}(x,y,z) = frac{partial F_x}{partial x} + frac{partial F_y}{partial y} + frac{partial F_z}{partial z} = 2+2z$$
so
$$I = iintlimits_{text{unit sphere}} vec{F}cdot dmathbf{S} = iiintlimits_{text{unit ball}} operatorname{div} vec{F} ,dV = int_{phi = 0}^{2pi} int_{theta = 0}^pi int_{r=0}^1 2(1+rcostheta)underbrace{r^2sintheta,dr,dtheta,dphi}_{dV}$$
The integrand is independent of $phi$ so it factors out as $2pi$, and then the second integral vanishes:
$$I = 4pileft(int_{theta = 0}^pi int_{r=0}^1 r^2sintheta,dr,dtheta + int_{theta = 0}^pi int_{r=0}^1 r^3costhetasintheta,dr,dthetaright) = frac83pi$$
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
answered Dec 29 '18 at 17:33
mechanodroidmechanodroid
28.8k62548
28.8k62548
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.
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%2f3055396%2fflux-of-langle-x-y-z2-rangle-across-unit-sphere%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
