Integral of Gaussian curvature over S
$begingroup$
Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
asked Dec 4 '18 at 19:45
jman63jman63
33
$endgroup$
add a comment |
$begingroup$
Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
asked Dec 4 '18 at 19:45
jman63jman63
33
$endgroup$
add a comment |
$begingroup$
Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
asked Dec 4 '18 at 19:45
jman63jman63
33
$endgroup$
Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
differential-geometry
asked Dec 4 '18 at 19:45
jman63jman63
33
asked Dec 4 '18 at 19:45
jman63jman63
33
asked Dec 4 '18 at 19:45
jman63jman63
33
asked Dec 4 '18 at 19:45
jman63jman63
33
asked Dec 4 '18 at 19:45
jman63jman63
33
33
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
$endgroup$
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
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%2f3026055%2fintegral-of-gaussian-curvature-over-s%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
$begingroup$
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
$endgroup$
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
add a comment |
$begingroup$
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
$endgroup$
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
add a comment |
$begingroup$
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
$endgroup$
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
edited Dec 4 '18 at 19:57
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
answered Dec 4 '18 at 19:52
MisterRiemannMisterRiemann
5,8451624
5,8451624
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
add a comment |
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
$endgroup$
– jman63
Dec 7 '18 at 0:52
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
$begingroup$
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
$endgroup$
– MisterRiemann
Dec 7 '18 at 7:20
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%2f3026055%2fintegral-of-gaussian-curvature-over-s%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
