Calculating integral over manifold given by equations
$begingroup$
Describe the area of the following manifold: $M={(x,y,z)|x^2+y^2=f(z)^2, 0lt z lt 1}$, as a one dimensional integral.
My attempt -
I define $F(x,y,z) = x^2+y^2-f(z)^2$ and therefore $M={(x,y,z)| F(x,y,z) = 0}$.
The formula(as much as I know) for such defintion of manifolds over $mathbb{R}^3$ is: $int_M dA = int F(x,y,z)frac{sqrt{|nabla F|}}{|F_z|}dxdydz$.
In our case: $F_z = -2f(z)f'(z)$, and $|nabla F|= |4x^2+4y^2+4f(z)^2f'(z)^2|$.
Putting it all together: $int_M dA = int_0^1int_0^{f(z)^2}int_0^{f(z)^2} (x^2+y^2-f(z)^2)frac{sqrt{|4x^2+4y^2+4f(z)^2f'(z)^2|}}{|-2f(z)f'(z)|}dxdydz$
But I'm really not sure how to continue from here... Any hints?
Or perhaps, is there any way doing it using some parameterization?
calculus integration multivariable-calculus manifolds multiple-integral
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
$endgroup$
add a comment |
$begingroup$
Describe the area of the following manifold: $M={(x,y,z)|x^2+y^2=f(z)^2, 0lt z lt 1}$, as a one dimensional integral.
My attempt -
I define $F(x,y,z) = x^2+y^2-f(z)^2$ and therefore $M={(x,y,z)| F(x,y,z) = 0}$.
The formula(as much as I know) for such defintion of manifolds over $mathbb{R}^3$ is: $int_M dA = int F(x,y,z)frac{sqrt{|nabla F|}}{|F_z|}dxdydz$.
In our case: $F_z = -2f(z)f'(z)$, and $|nabla F|= |4x^2+4y^2+4f(z)^2f'(z)^2|$.
Putting it all together: $int_M dA = int_0^1int_0^{f(z)^2}int_0^{f(z)^2} (x^2+y^2-f(z)^2)frac{sqrt{|4x^2+4y^2+4f(z)^2f'(z)^2|}}{|-2f(z)f'(z)|}dxdydz$
But I'm really not sure how to continue from here... Any hints?
Or perhaps, is there any way doing it using some parameterization?
calculus integration multivariable-calculus manifolds multiple-integral
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
$endgroup$
add a comment |
$begingroup$
Describe the area of the following manifold: $M={(x,y,z)|x^2+y^2=f(z)^2, 0lt z lt 1}$, as a one dimensional integral.
My attempt -
I define $F(x,y,z) = x^2+y^2-f(z)^2$ and therefore $M={(x,y,z)| F(x,y,z) = 0}$.
The formula(as much as I know) for such defintion of manifolds over $mathbb{R}^3$ is: $int_M dA = int F(x,y,z)frac{sqrt{|nabla F|}}{|F_z|}dxdydz$.
In our case: $F_z = -2f(z)f'(z)$, and $|nabla F|= |4x^2+4y^2+4f(z)^2f'(z)^2|$.
Putting it all together: $int_M dA = int_0^1int_0^{f(z)^2}int_0^{f(z)^2} (x^2+y^2-f(z)^2)frac{sqrt{|4x^2+4y^2+4f(z)^2f'(z)^2|}}{|-2f(z)f'(z)|}dxdydz$
But I'm really not sure how to continue from here... Any hints?
Or perhaps, is there any way doing it using some parameterization?
calculus integration multivariable-calculus manifolds multiple-integral
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
$endgroup$
Describe the area of the following manifold: $M={(x,y,z)|x^2+y^2=f(z)^2, 0lt z lt 1}$, as a one dimensional integral.
My attempt -
I define $F(x,y,z) = x^2+y^2-f(z)^2$ and therefore $M={(x,y,z)| F(x,y,z) = 0}$.
The formula(as much as I know) for such defintion of manifolds over $mathbb{R}^3$ is: $int_M dA = int F(x,y,z)frac{sqrt{|nabla F|}}{|F_z|}dxdydz$.
In our case: $F_z = -2f(z)f'(z)$, and $|nabla F|= |4x^2+4y^2+4f(z)^2f'(z)^2|$.
Putting it all together: $int_M dA = int_0^1int_0^{f(z)^2}int_0^{f(z)^2} (x^2+y^2-f(z)^2)frac{sqrt{|4x^2+4y^2+4f(z)^2f'(z)^2|}}{|-2f(z)f'(z)|}dxdydz$
But I'm really not sure how to continue from here... Any hints?
Or perhaps, is there any way doing it using some parameterization?
calculus integration multivariable-calculus manifolds multiple-integral
calculus integration multivariable-calculus manifolds multiple-integral
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
edited Dec 14 '18 at 20:18
ChikChak
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
asked Dec 14 '18 at 20:10
ChikChakChikChak
776418
776418
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Switching to cylinder coordinates gives $r^2 = f(z)^2$, so $r = |f(z)|, zin(0,1), theta in [0,2pi]$. Your surface is then composed of horizontal strips (perpendicular to the $z$-axis) of thickness $dz$, radius $r(z) = |f(z)|$, and hence area $2pi |f(z)| dz$. So the total area would be
$$
A = 2 pi int_0^1 |f(z)| dz
$$
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
$endgroup$
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
1
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
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%2f3039831%2fcalculating-integral-over-manifold-given-by-equations%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
$begingroup$
Switching to cylinder coordinates gives $r^2 = f(z)^2$, so $r = |f(z)|, zin(0,1), theta in [0,2pi]$. Your surface is then composed of horizontal strips (perpendicular to the $z$-axis) of thickness $dz$, radius $r(z) = |f(z)|$, and hence area $2pi |f(z)| dz$. So the total area would be
$$
A = 2 pi int_0^1 |f(z)| dz
$$
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
$endgroup$
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
1
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
add a comment |
$begingroup$
Switching to cylinder coordinates gives $r^2 = f(z)^2$, so $r = |f(z)|, zin(0,1), theta in [0,2pi]$. Your surface is then composed of horizontal strips (perpendicular to the $z$-axis) of thickness $dz$, radius $r(z) = |f(z)|$, and hence area $2pi |f(z)| dz$. So the total area would be
$$
A = 2 pi int_0^1 |f(z)| dz
$$
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
$endgroup$
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
1
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
add a comment |
$begingroup$
Switching to cylinder coordinates gives $r^2 = f(z)^2$, so $r = |f(z)|, zin(0,1), theta in [0,2pi]$. Your surface is then composed of horizontal strips (perpendicular to the $z$-axis) of thickness $dz$, radius $r(z) = |f(z)|$, and hence area $2pi |f(z)| dz$. So the total area would be
$$
A = 2 pi int_0^1 |f(z)| dz
$$
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
$endgroup$
Switching to cylinder coordinates gives $r^2 = f(z)^2$, so $r = |f(z)|, zin(0,1), theta in [0,2pi]$. Your surface is then composed of horizontal strips (perpendicular to the $z$-axis) of thickness $dz$, radius $r(z) = |f(z)|$, and hence area $2pi |f(z)| dz$. So the total area would be
$$
A = 2 pi int_0^1 |f(z)| dz
$$
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
answered Dec 14 '18 at 20:27
MatthiasMatthias
3287
3287
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
1
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
add a comment |
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
1
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
$begingroup$
But how can we just switch to cylinder coordinates, without knowing how will it affect $z$?
$endgroup$
– ChikChak
Dec 14 '18 at 20:35
1
1
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
$begingroup$
$z$ doesn't get affected: $x=rcostheta,; y=rsintheta,; z=z$.
$endgroup$
– Matthias
Dec 14 '18 at 20:36
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%2f3039831%2fcalculating-integral-over-manifold-given-by-equations%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
