To find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by x+y+z=1












0












$begingroup$


I need to find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by $x+y+z=1$.



My attempt:



$x+y+z = 1$ forms the base of the tetrahedron and my limits for integration based on integrating first with respect to z are:



limits for $z: [0,1-x-y]$



limits for $y:[0,1-x]$



limits for $x:[0,1]$



The formula for moment of inertia in 3 dimensions is:



$ I_x = iiint( y^2+ z^2 ) delta dV $ where $delta$ is the density function



$ I_y = iiint( x^2+ z^2 ) delta dV $



$ I_z = iiint( x^2+ y^2 ) delta dV $



then I found the moment of inertia about x axis,



$I_x = int_0^1int_0^{1-x}int_0^{1-x-y}( y^2+ z^2 ) dzdydz$



$I_x = int_0^1int_0^{1-x}( y^2z+ frac{z^3}{3} )|_0^{1-x-y} dydz$



.



.



.
$I_x = frac{1}{30}$

Similarly for y and z.
For now, I just want to confirm whether I'm moving in the right direction and whether I've missed something or not. Thank you.



Edit: As far as I know, tetrahedron has to do with triple integration. However, the official method for this question is double integration for x and y moments and triple for z. Why is this so?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    I see it right. But I did not know about that official method (taken apart the very same definition of momentum of inertia). Can you give a link?
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:09










  • $begingroup$
    finally someone answered! That was just a hint from my professor but my question is this, I would find Ix and Iy using triple integration just like I did for Iz. But I'm not sure why? Why can't it be done by double integration alone? Thank you
    $endgroup$
    – tNotr
    Dec 10 '18 at 15:13










  • $begingroup$
    As far as the definition is concerned, the moment of inertia is the weighted sum of the function $rho(vec r)$, the density, over the entire space with weight $mathbb a^2$, with $mathbb a$ the distance to some line, so is, is the integral over the space of $rho(vec r)mathbb a^2$. I'd say the triple integral is the natural way for the calculation of the moment of inertia. Sometimes the symmetry of $rho$ makes possible to do with a double o single integral (e.g. a cone), but to my understanding even in this case it needs some previous use of integrals.
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:26


















0












$begingroup$


I need to find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by $x+y+z=1$.



My attempt:



$x+y+z = 1$ forms the base of the tetrahedron and my limits for integration based on integrating first with respect to z are:



limits for $z: [0,1-x-y]$



limits for $y:[0,1-x]$



limits for $x:[0,1]$



The formula for moment of inertia in 3 dimensions is:



$ I_x = iiint( y^2+ z^2 ) delta dV $ where $delta$ is the density function



$ I_y = iiint( x^2+ z^2 ) delta dV $



$ I_z = iiint( x^2+ y^2 ) delta dV $



then I found the moment of inertia about x axis,



$I_x = int_0^1int_0^{1-x}int_0^{1-x-y}( y^2+ z^2 ) dzdydz$



$I_x = int_0^1int_0^{1-x}( y^2z+ frac{z^3}{3} )|_0^{1-x-y} dydz$



.



.



.
$I_x = frac{1}{30}$

Similarly for y and z.
For now, I just want to confirm whether I'm moving in the right direction and whether I've missed something or not. Thank you.



Edit: As far as I know, tetrahedron has to do with triple integration. However, the official method for this question is double integration for x and y moments and triple for z. Why is this so?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    I see it right. But I did not know about that official method (taken apart the very same definition of momentum of inertia). Can you give a link?
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:09










  • $begingroup$
    finally someone answered! That was just a hint from my professor but my question is this, I would find Ix and Iy using triple integration just like I did for Iz. But I'm not sure why? Why can't it be done by double integration alone? Thank you
    $endgroup$
    – tNotr
    Dec 10 '18 at 15:13










  • $begingroup$
    As far as the definition is concerned, the moment of inertia is the weighted sum of the function $rho(vec r)$, the density, over the entire space with weight $mathbb a^2$, with $mathbb a$ the distance to some line, so is, is the integral over the space of $rho(vec r)mathbb a^2$. I'd say the triple integral is the natural way for the calculation of the moment of inertia. Sometimes the symmetry of $rho$ makes possible to do with a double o single integral (e.g. a cone), but to my understanding even in this case it needs some previous use of integrals.
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:26
















0












0








0





$begingroup$


I need to find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by $x+y+z=1$.



My attempt:



$x+y+z = 1$ forms the base of the tetrahedron and my limits for integration based on integrating first with respect to z are:



limits for $z: [0,1-x-y]$



limits for $y:[0,1-x]$



limits for $x:[0,1]$



The formula for moment of inertia in 3 dimensions is:



$ I_x = iiint( y^2+ z^2 ) delta dV $ where $delta$ is the density function



$ I_y = iiint( x^2+ z^2 ) delta dV $



$ I_z = iiint( x^2+ y^2 ) delta dV $



then I found the moment of inertia about x axis,



$I_x = int_0^1int_0^{1-x}int_0^{1-x-y}( y^2+ z^2 ) dzdydz$



$I_x = int_0^1int_0^{1-x}( y^2z+ frac{z^3}{3} )|_0^{1-x-y} dydz$



.



.



.
$I_x = frac{1}{30}$

Similarly for y and z.
For now, I just want to confirm whether I'm moving in the right direction and whether I've missed something or not. Thank you.



Edit: As far as I know, tetrahedron has to do with triple integration. However, the official method for this question is double integration for x and y moments and triple for z. Why is this so?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$




I need to find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by $x+y+z=1$.



My attempt:



$x+y+z = 1$ forms the base of the tetrahedron and my limits for integration based on integrating first with respect to z are:



limits for $z: [0,1-x-y]$



limits for $y:[0,1-x]$



limits for $x:[0,1]$



The formula for moment of inertia in 3 dimensions is:



$ I_x = iiint( y^2+ z^2 ) delta dV $ where $delta$ is the density function



$ I_y = iiint( x^2+ z^2 ) delta dV $



$ I_z = iiint( x^2+ y^2 ) delta dV $



then I found the moment of inertia about x axis,



$I_x = int_0^1int_0^{1-x}int_0^{1-x-y}( y^2+ z^2 ) dzdydz$



$I_x = int_0^1int_0^{1-x}( y^2z+ frac{z^3}{3} )|_0^{1-x-y} dydz$



.



.



.
$I_x = frac{1}{30}$

Similarly for y and z.
For now, I just want to confirm whether I'm moving in the right direction and whether I've missed something or not. Thank you.



Edit: As far as I know, tetrahedron has to do with triple integration. However, the official method for this question is double integration for x and y moments and triple for z. Why is this so?






calculus integration multivariable-calculus




calculus integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 14:04







tNotr

















asked Dec 10 '18 at 13:54









tNotrtNotr

142




142












  • $begingroup$
    I see it right. But I did not know about that official method (taken apart the very same definition of momentum of inertia). Can you give a link?
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:09










  • $begingroup$
    finally someone answered! That was just a hint from my professor but my question is this, I would find Ix and Iy using triple integration just like I did for Iz. But I'm not sure why? Why can't it be done by double integration alone? Thank you
    $endgroup$
    – tNotr
    Dec 10 '18 at 15:13










  • $begingroup$
    As far as the definition is concerned, the moment of inertia is the weighted sum of the function $rho(vec r)$, the density, over the entire space with weight $mathbb a^2$, with $mathbb a$ the distance to some line, so is, is the integral over the space of $rho(vec r)mathbb a^2$. I'd say the triple integral is the natural way for the calculation of the moment of inertia. Sometimes the symmetry of $rho$ makes possible to do with a double o single integral (e.g. a cone), but to my understanding even in this case it needs some previous use of integrals.
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:26




















  • $begingroup$
    I see it right. But I did not know about that official method (taken apart the very same definition of momentum of inertia). Can you give a link?
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:09










  • $begingroup$
    finally someone answered! That was just a hint from my professor but my question is this, I would find Ix and Iy using triple integration just like I did for Iz. But I'm not sure why? Why can't it be done by double integration alone? Thank you
    $endgroup$
    – tNotr
    Dec 10 '18 at 15:13










  • $begingroup$
    As far as the definition is concerned, the moment of inertia is the weighted sum of the function $rho(vec r)$, the density, over the entire space with weight $mathbb a^2$, with $mathbb a$ the distance to some line, so is, is the integral over the space of $rho(vec r)mathbb a^2$. I'd say the triple integral is the natural way for the calculation of the moment of inertia. Sometimes the symmetry of $rho$ makes possible to do with a double o single integral (e.g. a cone), but to my understanding even in this case it needs some previous use of integrals.
    $endgroup$
    – Rafa Budría
    Dec 10 '18 at 15:26


















$begingroup$
I see it right. But I did not know about that official method (taken apart the very same definition of momentum of inertia). Can you give a link?
$endgroup$
– Rafa Budría
Dec 10 '18 at 15:09




$begingroup$
I see it right. But I did not know about that official method (taken apart the very same definition of momentum of inertia). Can you give a link?
$endgroup$
– Rafa Budría
Dec 10 '18 at 15:09












$begingroup$
finally someone answered! That was just a hint from my professor but my question is this, I would find Ix and Iy using triple integration just like I did for Iz. But I'm not sure why? Why can't it be done by double integration alone? Thank you
$endgroup$
– tNotr
Dec 10 '18 at 15:13




$begingroup$
finally someone answered! That was just a hint from my professor but my question is this, I would find Ix and Iy using triple integration just like I did for Iz. But I'm not sure why? Why can't it be done by double integration alone? Thank you
$endgroup$
– tNotr
Dec 10 '18 at 15:13












$begingroup$
As far as the definition is concerned, the moment of inertia is the weighted sum of the function $rho(vec r)$, the density, over the entire space with weight $mathbb a^2$, with $mathbb a$ the distance to some line, so is, is the integral over the space of $rho(vec r)mathbb a^2$. I'd say the triple integral is the natural way for the calculation of the moment of inertia. Sometimes the symmetry of $rho$ makes possible to do with a double o single integral (e.g. a cone), but to my understanding even in this case it needs some previous use of integrals.
$endgroup$
– Rafa Budría
Dec 10 '18 at 15:26






$begingroup$
As far as the definition is concerned, the moment of inertia is the weighted sum of the function $rho(vec r)$, the density, over the entire space with weight $mathbb a^2$, with $mathbb a$ the distance to some line, so is, is the integral over the space of $rho(vec r)mathbb a^2$. I'd say the triple integral is the natural way for the calculation of the moment of inertia. Sometimes the symmetry of $rho$ makes possible to do with a double o single integral (e.g. a cone), but to my understanding even in this case it needs some previous use of integrals.
$endgroup$
– Rafa Budría
Dec 10 '18 at 15:26












0






active

oldest

votes











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033937%2fto-find-the-moment-of-inertia-about-co-ordinate-axes-of-a-tetrahedron-in-the-fir%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033937%2fto-find-the-moment-of-inertia-about-co-ordinate-axes-of-a-tetrahedron-in-the-fir%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
Read more