Calculate a Line Integral of differentiable form of degree $1$.integral of a 2-form over an oriented...
Having the player face themselves after the mid-game
Was this cameo in Captain Marvel computer generated?
Is it appropriate to ask a former professor to order a library book for me through ILL?
Precision notation for voltmeters
Did Amazon pay $0 in taxes last year?
Giving a career talk in my old university, how prominently should I tell students my salary?
What should I do when a paper is published similar to my PhD thesis without citation?
Short story about cities being connected by a conveyor belt
Does the US political system, in principle, allow for a no-party system?
How spaceships determine each other's mass in space?
How do you use environments that have the same name within a single latex document?
When Central Limit Theorem breaks down
Ultrafilters as a double dual
Why aren't there more Gauls like Obelix?
How can I portion out frozen cookie dough?
Should I apply for my boss's promotion?
Insult for someone who "doesn't know anything"
How to write a chaotic neutral protagonist and prevent my readers from thinking they are evil?
How to make sure I'm assertive enough in contact with subordinates?
How does learning spells work when leveling a multiclass character?
Can I challenge the interviewer to give me a proper technical feedback?
If nine coins are tossed, what is the probability that the number of heads is even?
Help! My Character is too much for her story!
Why does a car's steering wheel get lighter with increasing speed
Calculate a Line Integral of differentiable form of degree $1$.
integral of a 2-form over an oriented manifoldUsing Stokes's theorem to calculate a value of integralDifferential forms and determinantsAn Integral InequalityProving a formula for the exterior derivative of a specific $k$-form, given in base representationCalculate the integral using Green's TheoremCalculate $domega$ of an $n-1$ forms $omega$$d(Iomega) + I(domega) = omega$ for differential formsCalculus with Differential Forms.$f^{*}omega = det(df)dx_{1}wedge cdots wedge dx_{n}$ for $f(x_{1},…,x_{n}) = (y_{1},…,y_{n})$ and $omega = dy_{1}wedge cdots dy_{n}$
$begingroup$
Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.
My attempt.
$begin{eqnarray*}
domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
&=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
& = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
&=& 0.
end{eqnarray*}$
Now, by definition
$$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$
where $omega = sum_{i}a_{i}dx_{i}$. Then
$$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$
We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have
$begin{eqnarray*}
int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
&=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
& = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
& = & t^{8}Bigg|_{0}^{x}\
&=& x^{8}
end{eqnarray*}$
Is correct? This is my first problem about Line Integrals on $1$-forms.
real-analysis integration differential-forms
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
$endgroup$
add a comment |
$begingroup$
Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.
My attempt.
$begin{eqnarray*}
domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
&=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
& = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
&=& 0.
end{eqnarray*}$
Now, by definition
$$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$
where $omega = sum_{i}a_{i}dx_{i}$. Then
$$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$
We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have
$begin{eqnarray*}
int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
&=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
& = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
& = & t^{8}Bigg|_{0}^{x}\
&=& x^{8}
end{eqnarray*}$
Is correct? This is my first problem about Line Integrals on $1$-forms.
real-analysis integration differential-forms
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
$endgroup$
add a comment |
$begingroup$
Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.
My attempt.
$begin{eqnarray*}
domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
&=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
& = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
&=& 0.
end{eqnarray*}$
Now, by definition
$$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$
where $omega = sum_{i}a_{i}dx_{i}$. Then
$$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$
We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have
$begin{eqnarray*}
int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
&=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
& = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
& = & t^{8}Bigg|_{0}^{x}\
&=& x^{8}
end{eqnarray*}$
Is correct? This is my first problem about Line Integrals on $1$-forms.
real-analysis integration differential-forms
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
$endgroup$
Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.
My attempt.
$begin{eqnarray*}
domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
&=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
& = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
&=& 0.
end{eqnarray*}$
Now, by definition
$$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$
where $omega = sum_{i}a_{i}dx_{i}$. Then
$$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$
We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have
$begin{eqnarray*}
int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
&=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
& = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
& = & t^{8}Bigg|_{0}^{x}\
&=& x^{8}
end{eqnarray*}$
Is correct? This is my first problem about Line Integrals on $1$-forms.
real-analysis integration differential-forms
real-analysis integration differential-forms
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
asked yesterday
Lucas CorrêaLucas Corrêa
1,6321321
1,6321321
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The differential form is even exact:
$$
omega = 2xy^3,dx + 3x^2y^2,dy
= frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
= d(x^2y^3).
$$
From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$
Therefore we can calculate the integral as
$$
int_c omega
= int_{(0,0)}^{(x,x^2)} d(x^2y^3)
= x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
= x^2(x^2)^3 - 0^2 cdot 0^3
= x^8.
$$
Thus, your result is correct.
answered yesterday
md2perpemd2perpe
8,21111028
$endgroup$
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
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%2f3139355%2fcalculate-a-line-integral-of-differentiable-form-of-degree-1%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
$begingroup$
The differential form is even exact:
$$
omega = 2xy^3,dx + 3x^2y^2,dy
= frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
= d(x^2y^3).
$$
From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$
Therefore we can calculate the integral as
$$
int_c omega
= int_{(0,0)}^{(x,x^2)} d(x^2y^3)
= x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
= x^2(x^2)^3 - 0^2 cdot 0^3
= x^8.
$$
Thus, your result is correct.
answered yesterday
md2perpemd2perpe
8,21111028
$endgroup$
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
add a comment |
$begingroup$
The differential form is even exact:
$$
omega = 2xy^3,dx + 3x^2y^2,dy
= frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
= d(x^2y^3).
$$
From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$
Therefore we can calculate the integral as
$$
int_c omega
= int_{(0,0)}^{(x,x^2)} d(x^2y^3)
= x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
= x^2(x^2)^3 - 0^2 cdot 0^3
= x^8.
$$
Thus, your result is correct.
answered yesterday
md2perpemd2perpe
8,21111028
$endgroup$
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
add a comment |
$begingroup$
The differential form is even exact:
$$
omega = 2xy^3,dx + 3x^2y^2,dy
= frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
= d(x^2y^3).
$$
From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$
Therefore we can calculate the integral as
$$
int_c omega
= int_{(0,0)}^{(x,x^2)} d(x^2y^3)
= x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
= x^2(x^2)^3 - 0^2 cdot 0^3
= x^8.
$$
Thus, your result is correct.
answered yesterday
md2perpemd2perpe
8,21111028
$endgroup$
The differential form is even exact:
$$
omega = 2xy^3,dx + 3x^2y^2,dy
= frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
= d(x^2y^3).
$$
From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$
Therefore we can calculate the integral as
$$
int_c omega
= int_{(0,0)}^{(x,x^2)} d(x^2y^3)
= x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
= x^2(x^2)^3 - 0^2 cdot 0^3
= x^8.
$$
Thus, your result is correct.
answered yesterday
md2perpemd2perpe
8,21111028
answered yesterday
md2perpemd2perpe
8,21111028
answered yesterday
md2perpemd2perpe
8,21111028
answered yesterday
md2perpemd2perpe
8,21111028
8,21111028
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
add a comment |
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
$begingroup$
Thank you! Your approach is very clever!
$endgroup$
– Lucas Corrêa
yesterday
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%2f3139355%2fcalculate-a-line-integral-of-differentiable-form-of-degree-1%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
