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

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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











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1 Answer
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1 Answer
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active

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active

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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


















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