Find the center of mass of the body bounded by $z=0$ and $z=H-x^2-y^2$
$begingroup$
First I switched to cylindrical coordinate system: $0 leq z leq H - r^2$.
This is the mass of the body:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r,dphi ,dr ,dz = frac43frac25pi H^{frac52}$$
Now the other thing:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r z ,dphi ,dr ,dz = frac43frac25frac27pi H^{frac72}$$
And $$C_z = frac{frac43frac25frac27pi H^{frac72}}{frac43frac25pi H^{frac52}} = frac27 H$$
However my teacher says the asnwer is $frac{H}{3}$. Where is the mistake? Is there an easier way to calculate this (maybe simpler integrals)?
integration definite-integrals
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
$endgroup$
add a comment |
$begingroup$
First I switched to cylindrical coordinate system: $0 leq z leq H - r^2$.
This is the mass of the body:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r,dphi ,dr ,dz = frac43frac25pi H^{frac52}$$
Now the other thing:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r z ,dphi ,dr ,dz = frac43frac25frac27pi H^{frac72}$$
And $$C_z = frac{frac43frac25frac27pi H^{frac72}}{frac43frac25pi H^{frac52}} = frac27 H$$
However my teacher says the asnwer is $frac{H}{3}$. Where is the mistake? Is there an easier way to calculate this (maybe simpler integrals)?
integration definite-integrals
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
$endgroup$
add a comment |
$begingroup$
First I switched to cylindrical coordinate system: $0 leq z leq H - r^2$.
This is the mass of the body:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r,dphi ,dr ,dz = frac43frac25pi H^{frac52}$$
Now the other thing:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r z ,dphi ,dr ,dz = frac43frac25frac27pi H^{frac72}$$
And $$C_z = frac{frac43frac25frac27pi H^{frac72}}{frac43frac25pi H^{frac52}} = frac27 H$$
However my teacher says the asnwer is $frac{H}{3}$. Where is the mistake? Is there an easier way to calculate this (maybe simpler integrals)?
integration definite-integrals
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
$endgroup$
First I switched to cylindrical coordinate system: $0 leq z leq H - r^2$.
This is the mass of the body:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r,dphi ,dr ,dz = frac43frac25pi H^{frac52}$$
Now the other thing:
$$int_0^Hint_{-sqrt{H-z}}^{sqrt{H-z}}int_0^{2pi}r z ,dphi ,dr ,dz = frac43frac25frac27pi H^{frac72}$$
And $$C_z = frac{frac43frac25frac27pi H^{frac72}}{frac43frac25pi H^{frac52}} = frac27 H$$
However my teacher says the asnwer is $frac{H}{3}$. Where is the mistake? Is there an easier way to calculate this (maybe simpler integrals)?
integration definite-integrals
integration definite-integrals
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
asked Dec 10 '18 at 3:12
SlowerPhotonSlowerPhoton
408111
408111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The lower limits of the $r$-integrals are incorrect: they should just be zero. Putting that aside, you've made errors computing the integrals you presented. They should be proportional to $H^2$ and $H^3$, not $H^{5/2}$ and $H^{7/2}$. The mass of the solid is
$$int_0^H int_0^{sqrt{H - z}} int_0^{2pi} r, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi r, dr, dz = int_0^H pi(H - z), dz = frac{pi H^2}{2}$$
and the moment integral is
$$int_0^H int_0^{sqrt{H-z}} int_0^{2pi} zr, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi rz, dr, dz = int_0^H pi z(H-z), dz = frac{pi H^3}{6}$$ Therefore,
$$C_z =frac{pi H^3/6}{pi H^2/2} = frac{H}{3}$$ as your teacher claimed.
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
$endgroup$
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%2f3033389%2ffind-the-center-of-mass-of-the-body-bounded-by-z-0-and-z-h-x2-y2%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 lower limits of the $r$-integrals are incorrect: they should just be zero. Putting that aside, you've made errors computing the integrals you presented. They should be proportional to $H^2$ and $H^3$, not $H^{5/2}$ and $H^{7/2}$. The mass of the solid is
$$int_0^H int_0^{sqrt{H - z}} int_0^{2pi} r, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi r, dr, dz = int_0^H pi(H - z), dz = frac{pi H^2}{2}$$
and the moment integral is
$$int_0^H int_0^{sqrt{H-z}} int_0^{2pi} zr, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi rz, dr, dz = int_0^H pi z(H-z), dz = frac{pi H^3}{6}$$ Therefore,
$$C_z =frac{pi H^3/6}{pi H^2/2} = frac{H}{3}$$ as your teacher claimed.
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
$endgroup$
add a comment |
$begingroup$
The lower limits of the $r$-integrals are incorrect: they should just be zero. Putting that aside, you've made errors computing the integrals you presented. They should be proportional to $H^2$ and $H^3$, not $H^{5/2}$ and $H^{7/2}$. The mass of the solid is
$$int_0^H int_0^{sqrt{H - z}} int_0^{2pi} r, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi r, dr, dz = int_0^H pi(H - z), dz = frac{pi H^2}{2}$$
and the moment integral is
$$int_0^H int_0^{sqrt{H-z}} int_0^{2pi} zr, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi rz, dr, dz = int_0^H pi z(H-z), dz = frac{pi H^3}{6}$$ Therefore,
$$C_z =frac{pi H^3/6}{pi H^2/2} = frac{H}{3}$$ as your teacher claimed.
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
$endgroup$
add a comment |
$begingroup$
The lower limits of the $r$-integrals are incorrect: they should just be zero. Putting that aside, you've made errors computing the integrals you presented. They should be proportional to $H^2$ and $H^3$, not $H^{5/2}$ and $H^{7/2}$. The mass of the solid is
$$int_0^H int_0^{sqrt{H - z}} int_0^{2pi} r, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi r, dr, dz = int_0^H pi(H - z), dz = frac{pi H^2}{2}$$
and the moment integral is
$$int_0^H int_0^{sqrt{H-z}} int_0^{2pi} zr, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi rz, dr, dz = int_0^H pi z(H-z), dz = frac{pi H^3}{6}$$ Therefore,
$$C_z =frac{pi H^3/6}{pi H^2/2} = frac{H}{3}$$ as your teacher claimed.
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
$endgroup$
The lower limits of the $r$-integrals are incorrect: they should just be zero. Putting that aside, you've made errors computing the integrals you presented. They should be proportional to $H^2$ and $H^3$, not $H^{5/2}$ and $H^{7/2}$. The mass of the solid is
$$int_0^H int_0^{sqrt{H - z}} int_0^{2pi} r, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi r, dr, dz = int_0^H pi(H - z), dz = frac{pi H^2}{2}$$
and the moment integral is
$$int_0^H int_0^{sqrt{H-z}} int_0^{2pi} zr, dphi, dr, dz = int_0^H int_0^{sqrt{H-z}} 2pi rz, dr, dz = int_0^H pi z(H-z), dz = frac{pi H^3}{6}$$ Therefore,
$$C_z =frac{pi H^3/6}{pi H^2/2} = frac{H}{3}$$ as your teacher claimed.
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
answered Dec 10 '18 at 4:27
kobekobe
34.9k22248
34.9k22248
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%2f3033389%2ffind-the-center-of-mass-of-the-body-bounded-by-z-0-and-z-h-x2-y2%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
