Finding $iiint x^2,{rm d}x{rm d}y{rm d}z$ over the volume bounded by $frac{x^2}{a^2} + frac{y^2}{b^2} +...












3












$begingroup$



Finding the value of $$iiint x^2,mathrm dxmathrm dymathrm dz$$ over the volume bounded by the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2}= 1.$$




I am taking limits as follow:



$$-a leq x leq a$$
$$-frac{b}{a} (a^2-x^2)^{1/2} leq y leq frac{b}{a} (a^2-x^2)^{1/2}$$
$$-frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2} leq z leq frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2}$$
And solving the integral as $mathrm dz$ first then $mathrm dy$ and then $mathrm dx$. But it is just becoming harder and harder with each step.



I do not have a tutor. So, I highly need help of you masters. Please help me.






multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    With the use of generalized spherical coordinates it becomes really easy: $x=racos phi cos theta, ; y=rb sin phi cos theta, ; z=rc sin theta.$
    $endgroup$
    – user376343
    Dec 22 '18 at 22:41












  • $begingroup$
    @Naved THE Sheikh, study beta integrals, they can be of great use.
    $endgroup$
    – Awe Kumar Jha
    Dec 23 '18 at 3:55






  • 1




    $begingroup$
    @user376343 thank you so much for the advice. I'll definitely try this out.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:00










  • $begingroup$
    @Awe Kumar Jha I am yet learning beta and gamma functions. I have not practised its applications. I will learn them nicely and make my hand better on them. Thanks for the suggestion. 😊
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:03
















3












$begingroup$



Finding the value of $$iiint x^2,mathrm dxmathrm dymathrm dz$$ over the volume bounded by the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2}= 1.$$




I am taking limits as follow:



$$-a leq x leq a$$
$$-frac{b}{a} (a^2-x^2)^{1/2} leq y leq frac{b}{a} (a^2-x^2)^{1/2}$$
$$-frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2} leq z leq frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2}$$
And solving the integral as $mathrm dz$ first then $mathrm dy$ and then $mathrm dx$. But it is just becoming harder and harder with each step.



I do not have a tutor. So, I highly need help of you masters. Please help me.






multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    With the use of generalized spherical coordinates it becomes really easy: $x=racos phi cos theta, ; y=rb sin phi cos theta, ; z=rc sin theta.$
    $endgroup$
    – user376343
    Dec 22 '18 at 22:41












  • $begingroup$
    @Naved THE Sheikh, study beta integrals, they can be of great use.
    $endgroup$
    – Awe Kumar Jha
    Dec 23 '18 at 3:55






  • 1




    $begingroup$
    @user376343 thank you so much for the advice. I'll definitely try this out.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:00










  • $begingroup$
    @Awe Kumar Jha I am yet learning beta and gamma functions. I have not practised its applications. I will learn them nicely and make my hand better on them. Thanks for the suggestion. 😊
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:03














3












3








3


1



$begingroup$



Finding the value of $$iiint x^2,mathrm dxmathrm dymathrm dz$$ over the volume bounded by the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2}= 1.$$




I am taking limits as follow:



$$-a leq x leq a$$
$$-frac{b}{a} (a^2-x^2)^{1/2} leq y leq frac{b}{a} (a^2-x^2)^{1/2}$$
$$-frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2} leq z leq frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2}$$
And solving the integral as $mathrm dz$ first then $mathrm dy$ and then $mathrm dx$. But it is just becoming harder and harder with each step.



I do not have a tutor. So, I highly need help of you masters. Please help me.






multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$





Finding the value of $$iiint x^2,mathrm dxmathrm dymathrm dz$$ over the volume bounded by the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2}= 1.$$




I am taking limits as follow:



$$-a leq x leq a$$
$$-frac{b}{a} (a^2-x^2)^{1/2} leq y leq frac{b}{a} (a^2-x^2)^{1/2}$$
$$-frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2} leq z leq frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2}$$
And solving the integral as $mathrm dz$ first then $mathrm dy$ and then $mathrm dx$. But it is just becoming harder and harder with each step.



I do not have a tutor. So, I highly need help of you masters. Please help me.






multivariable-calculus definite-integrals




multivariable-calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 1:10









Saad

19.7k92352




19.7k92352










asked Dec 22 '18 at 22:25









Naved THE SheikhNaved THE Sheikh

186




186








  • 4




    $begingroup$
    With the use of generalized spherical coordinates it becomes really easy: $x=racos phi cos theta, ; y=rb sin phi cos theta, ; z=rc sin theta.$
    $endgroup$
    – user376343
    Dec 22 '18 at 22:41












  • $begingroup$
    @Naved THE Sheikh, study beta integrals, they can be of great use.
    $endgroup$
    – Awe Kumar Jha
    Dec 23 '18 at 3:55






  • 1




    $begingroup$
    @user376343 thank you so much for the advice. I'll definitely try this out.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:00










  • $begingroup$
    @Awe Kumar Jha I am yet learning beta and gamma functions. I have not practised its applications. I will learn them nicely and make my hand better on them. Thanks for the suggestion. 😊
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:03














  • 4




    $begingroup$
    With the use of generalized spherical coordinates it becomes really easy: $x=racos phi cos theta, ; y=rb sin phi cos theta, ; z=rc sin theta.$
    $endgroup$
    – user376343
    Dec 22 '18 at 22:41












  • $begingroup$
    @Naved THE Sheikh, study beta integrals, they can be of great use.
    $endgroup$
    – Awe Kumar Jha
    Dec 23 '18 at 3:55






  • 1




    $begingroup$
    @user376343 thank you so much for the advice. I'll definitely try this out.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:00










  • $begingroup$
    @Awe Kumar Jha I am yet learning beta and gamma functions. I have not practised its applications. I will learn them nicely and make my hand better on them. Thanks for the suggestion. 😊
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 20:03








4




4




$begingroup$
With the use of generalized spherical coordinates it becomes really easy: $x=racos phi cos theta, ; y=rb sin phi cos theta, ; z=rc sin theta.$
$endgroup$
– user376343
Dec 22 '18 at 22:41






$begingroup$
With the use of generalized spherical coordinates it becomes really easy: $x=racos phi cos theta, ; y=rb sin phi cos theta, ; z=rc sin theta.$
$endgroup$
– user376343
Dec 22 '18 at 22:41














$begingroup$
@Naved THE Sheikh, study beta integrals, they can be of great use.
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 3:55




$begingroup$
@Naved THE Sheikh, study beta integrals, they can be of great use.
$endgroup$
– Awe Kumar Jha
Dec 23 '18 at 3:55




1




1




$begingroup$
@user376343 thank you so much for the advice. I'll definitely try this out.
$endgroup$
– Naved THE Sheikh
Dec 24 '18 at 20:00




$begingroup$
@user376343 thank you so much for the advice. I'll definitely try this out.
$endgroup$
– Naved THE Sheikh
Dec 24 '18 at 20:00












$begingroup$
@Awe Kumar Jha I am yet learning beta and gamma functions. I have not practised its applications. I will learn them nicely and make my hand better on them. Thanks for the suggestion. 😊
$endgroup$
– Naved THE Sheikh
Dec 24 '18 at 20:03




$begingroup$
@Awe Kumar Jha I am yet learning beta and gamma functions. I have not practised its applications. I will learn them nicely and make my hand better on them. Thanks for the suggestion. 😊
$endgroup$
– Naved THE Sheikh
Dec 24 '18 at 20:03










1 Answer
1






active

oldest

votes


















3












$begingroup$

First, rewrite the integral as



$$int_{-a}^a x^2 intint_{E_x}1 dy dz dx $$



where $E_x$ is the ellipse in the $yz$-plane satisfying $frac{y^2}{b^2} + frac{z^2}{c^2} = (1-frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $frac{y^2}{b^2left(1-frac{x^2}{a^2}right)} + frac{z^2}{c^2left(1-frac{x^2}{a^2}right)}=1$, we see that the area of said ellipse is $pi bc(1-frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).



Thus, the triple integral simplifies to the single-variable integral



$$int_{-a}^a x^2 pi bc(1-frac{x^2}{a^2}) dx= frac{4}{15}pi a^3 bc$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank You So Much for your answer. I found it quite helpful.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 19:59











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%2f3049891%2ffinding-iiint-x2-rm-dx-rm-dy-rm-dz-over-the-volume-bounded-by-frac%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









3












$begingroup$

First, rewrite the integral as



$$int_{-a}^a x^2 intint_{E_x}1 dy dz dx $$



where $E_x$ is the ellipse in the $yz$-plane satisfying $frac{y^2}{b^2} + frac{z^2}{c^2} = (1-frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $frac{y^2}{b^2left(1-frac{x^2}{a^2}right)} + frac{z^2}{c^2left(1-frac{x^2}{a^2}right)}=1$, we see that the area of said ellipse is $pi bc(1-frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).



Thus, the triple integral simplifies to the single-variable integral



$$int_{-a}^a x^2 pi bc(1-frac{x^2}{a^2}) dx= frac{4}{15}pi a^3 bc$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank You So Much for your answer. I found it quite helpful.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 19:59
















3












$begingroup$

First, rewrite the integral as



$$int_{-a}^a x^2 intint_{E_x}1 dy dz dx $$



where $E_x$ is the ellipse in the $yz$-plane satisfying $frac{y^2}{b^2} + frac{z^2}{c^2} = (1-frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $frac{y^2}{b^2left(1-frac{x^2}{a^2}right)} + frac{z^2}{c^2left(1-frac{x^2}{a^2}right)}=1$, we see that the area of said ellipse is $pi bc(1-frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).



Thus, the triple integral simplifies to the single-variable integral



$$int_{-a}^a x^2 pi bc(1-frac{x^2}{a^2}) dx= frac{4}{15}pi a^3 bc$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank You So Much for your answer. I found it quite helpful.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 19:59














3












3








3





$begingroup$

First, rewrite the integral as



$$int_{-a}^a x^2 intint_{E_x}1 dy dz dx $$



where $E_x$ is the ellipse in the $yz$-plane satisfying $frac{y^2}{b^2} + frac{z^2}{c^2} = (1-frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $frac{y^2}{b^2left(1-frac{x^2}{a^2}right)} + frac{z^2}{c^2left(1-frac{x^2}{a^2}right)}=1$, we see that the area of said ellipse is $pi bc(1-frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).



Thus, the triple integral simplifies to the single-variable integral



$$int_{-a}^a x^2 pi bc(1-frac{x^2}{a^2}) dx= frac{4}{15}pi a^3 bc$$






share|cite|improve this answer









$endgroup$



First, rewrite the integral as



$$int_{-a}^a x^2 intint_{E_x}1 dy dz dx $$



where $E_x$ is the ellipse in the $yz$-plane satisfying $frac{y^2}{b^2} + frac{z^2}{c^2} = (1-frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $frac{y^2}{b^2left(1-frac{x^2}{a^2}right)} + frac{z^2}{c^2left(1-frac{x^2}{a^2}right)}=1$, we see that the area of said ellipse is $pi bc(1-frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).



Thus, the triple integral simplifies to the single-variable integral



$$int_{-a}^a x^2 pi bc(1-frac{x^2}{a^2}) dx= frac{4}{15}pi a^3 bc$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 3:52









suncup224suncup224

1,186716




1,186716












  • $begingroup$
    Thank You So Much for your answer. I found it quite helpful.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 19:59


















  • $begingroup$
    Thank You So Much for your answer. I found it quite helpful.
    $endgroup$
    – Naved THE Sheikh
    Dec 24 '18 at 19:59
















$begingroup$
Thank You So Much for your answer. I found it quite helpful.
$endgroup$
– Naved THE Sheikh
Dec 24 '18 at 19:59




$begingroup$
Thank You So Much for your answer. I found it quite helpful.
$endgroup$
– Naved THE Sheikh
Dec 24 '18 at 19:59


















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%2f3049891%2ffinding-iiint-x2-rm-dx-rm-dy-rm-dz-over-the-volume-bounded-by-frac%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more