How to Calculate the flux of the Vector Field on the surface $z = 1-x^2-y^2$ ( getting normal vector...
$begingroup$
Let $S$ be the surface $z = 1-x^2-y^2 , 0leq z$.
Find $int_{S} x^2z~dydz + y^2z~dzdx + (x^2+y^2)~dxdy$.
Choose the direction of the normal upwards.
so i calculated the flux and i got that it is $boxed{frac{pi}{2}}$. i am not sure if that is right
Parametrization $R(theta,r) = (rcos(theta),rsin(theta),1-r^2)$
and the normal is $(2r^2cos(theta),2r^2sin(theta),r)$ i feel like its not right but i calculated it many times at the point $(0,0,1) ~ r = 0$ but the normal is $(0,0,0)$
$int_{S}vec{F} _dot{} vec{n}$ = $int_{theta = 0}^{theta = 2pi}int_{r=0}^{r=1}~~(2r^4-2r^6)(sin^3{theta}+cos^3{theta}) + r^3~~drdtheta$
multivariable-calculus vector-analysis surface-integrals
asked Jan 15 at 21:35
Mather Mather
4088
$endgroup$
add a comment |
$begingroup$
Let $S$ be the surface $z = 1-x^2-y^2 , 0leq z$.
Find $int_{S} x^2z~dydz + y^2z~dzdx + (x^2+y^2)~dxdy$.
Choose the direction of the normal upwards.
so i calculated the flux and i got that it is $boxed{frac{pi}{2}}$. i am not sure if that is right
Parametrization $R(theta,r) = (rcos(theta),rsin(theta),1-r^2)$
and the normal is $(2r^2cos(theta),2r^2sin(theta),r)$ i feel like its not right but i calculated it many times at the point $(0,0,1) ~ r = 0$ but the normal is $(0,0,0)$
$int_{S}vec{F} _dot{} vec{n}$ = $int_{theta = 0}^{theta = 2pi}int_{r=0}^{r=1}~~(2r^4-2r^6)(sin^3{theta}+cos^3{theta}) + r^3~~drdtheta$
multivariable-calculus vector-analysis surface-integrals
asked Jan 15 at 21:35
Mather Mather
4088
$endgroup$
$begingroup$
anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0)
$endgroup$
– Mather
Jan 15 at 22:47
1
$begingroup$
Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$.
$endgroup$
– Maxim
Jan 16 at 4:30
$begingroup$
But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ?
$endgroup$
– Mather
Jan 16 at 7:40
1
$begingroup$
The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r cos theta, 2 r sin theta, 1)/sqrt{1 + 4 r^2}$.
$endgroup$
– Maxim
Jan 16 at 8:04
add a comment |
$begingroup$
Let $S$ be the surface $z = 1-x^2-y^2 , 0leq z$.
Find $int_{S} x^2z~dydz + y^2z~dzdx + (x^2+y^2)~dxdy$.
Choose the direction of the normal upwards.
so i calculated the flux and i got that it is $boxed{frac{pi}{2}}$. i am not sure if that is right
Parametrization $R(theta,r) = (rcos(theta),rsin(theta),1-r^2)$
and the normal is $(2r^2cos(theta),2r^2sin(theta),r)$ i feel like its not right but i calculated it many times at the point $(0,0,1) ~ r = 0$ but the normal is $(0,0,0)$
$int_{S}vec{F} _dot{} vec{n}$ = $int_{theta = 0}^{theta = 2pi}int_{r=0}^{r=1}~~(2r^4-2r^6)(sin^3{theta}+cos^3{theta}) + r^3~~drdtheta$
multivariable-calculus vector-analysis surface-integrals
asked Jan 15 at 21:35
Mather Mather
4088
$endgroup$
Let $S$ be the surface $z = 1-x^2-y^2 , 0leq z$.
Find $int_{S} x^2z~dydz + y^2z~dzdx + (x^2+y^2)~dxdy$.
Choose the direction of the normal upwards.
so i calculated the flux and i got that it is $boxed{frac{pi}{2}}$. i am not sure if that is right
Parametrization $R(theta,r) = (rcos(theta),rsin(theta),1-r^2)$
and the normal is $(2r^2cos(theta),2r^2sin(theta),r)$ i feel like its not right but i calculated it many times at the point $(0,0,1) ~ r = 0$ but the normal is $(0,0,0)$
$int_{S}vec{F} _dot{} vec{n}$ = $int_{theta = 0}^{theta = 2pi}int_{r=0}^{r=1}~~(2r^4-2r^6)(sin^3{theta}+cos^3{theta}) + r^3~~drdtheta$
multivariable-calculus vector-analysis surface-integrals
multivariable-calculus vector-analysis surface-integrals
asked Jan 15 at 21:35
Mather Mather
4088
asked Jan 15 at 21:35
Mather Mather
4088
edited Jan 15 at 22:48
Mather
asked Jan 15 at 21:35
Mather Mather
4088
asked Jan 15 at 21:35
Mather Mather
4088
asked Jan 15 at 21:35
Mather Mather
4088
4088
$begingroup$
anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0)
$endgroup$
– Mather
Jan 15 at 22:47
1
$begingroup$
Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$.
$endgroup$
– Maxim
Jan 16 at 4:30
$begingroup$
But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ?
$endgroup$
– Mather
Jan 16 at 7:40
1
$begingroup$
The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r cos theta, 2 r sin theta, 1)/sqrt{1 + 4 r^2}$.
$endgroup$
– Maxim
Jan 16 at 8:04
add a comment |
$begingroup$
anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0)
$endgroup$
– Mather
Jan 15 at 22:47
1
$begingroup$
Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$.
$endgroup$
– Maxim
Jan 16 at 4:30
$begingroup$
But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ?
$endgroup$
– Mather
Jan 16 at 7:40
1
$begingroup$
The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r cos theta, 2 r sin theta, 1)/sqrt{1 + 4 r^2}$.
$endgroup$
– Maxim
Jan 16 at 8:04
$begingroup$
anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0)
$endgroup$
– Mather
Jan 15 at 22:47
$begingroup$
anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0)
$endgroup$
– Mather
Jan 15 at 22:47
1
1
$begingroup$
Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$.
$endgroup$
– Maxim
Jan 16 at 4:30
$begingroup$
Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$.
$endgroup$
– Maxim
Jan 16 at 4:30
$begingroup$
But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ?
$endgroup$
– Mather
Jan 16 at 7:40
$begingroup$
But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ?
$endgroup$
– Mather
Jan 16 at 7:40
1
1
$begingroup$
The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r cos theta, 2 r sin theta, 1)/sqrt{1 + 4 r^2}$.
$endgroup$
– Maxim
Jan 16 at 8:04
$begingroup$
The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r cos theta, 2 r sin theta, 1)/sqrt{1 + 4 r^2}$.
$endgroup$
– Maxim
Jan 16 at 8:04
add a comment |
0
active
oldest
votes
Your Answer
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%2f3075000%2fhow-to-calculate-the-flux-of-the-vector-field-on-the-surface-z-1-x2-y2-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3075000%2fhow-to-calculate-the-flux-of-the-vector-field-on-the-surface-z-1-x2-y2-g%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
$begingroup$
anyone ? i need help , i feel like the normal is Wrong at the point (0,0,1) i get the normal vector (0,0,0)
$endgroup$
– Mather
Jan 15 at 22:47
1
$begingroup$
Your calculations and the result are correct. The magnitude of the normal vector is zero when $r = 0$, but the $x$ and $y$ components are infinitesimals of higher order for small $r$, so the vector is approximately vertical and pointing upwards, and the unit normal is close to $(0, 0, 1)$.
$endgroup$
– Maxim
Jan 16 at 4:30
$begingroup$
But without parametrizing the normal is $(0,0,1)$ and with it is (0,0,0) why ?
$endgroup$
– Mather
Jan 16 at 7:40
1
$begingroup$
The unit normal is close to $(0, 0, 1)$. The unit normal is $(2 r cos theta, 2 r sin theta, 1)/sqrt{1 + 4 r^2}$.
$endgroup$
– Maxim
Jan 16 at 8:04