How to find the volume of ellipsoid using substitution in triple integral?












0












$begingroup$


The equation of ellipsoid is
$$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1$$
The hint for substitution is given as
$x=au$, $y=bv$ and $z=cv$.
I know that the projection of the ellipsoid on $xy-$plane is an ellipse with the equation
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$. But I don't know what will be the limits of $z$ in integration.
After transformation the Jacobian of transformation will be
$J=abc$. I don't know how to proceed this one and compute the volume of ellipsoid.
Please explain in detail about the limits before and after substitution with the help of the mathematica command that helps in plotting the ellipsoid along with its shadow.
THanks in advance!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$.
    $endgroup$
    – A.Γ.
    Jan 8 at 17:08










  • $begingroup$
    After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$
    $endgroup$
    – saulspatz
    Jan 8 at 17:08










  • $begingroup$
    @A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir!
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:11










  • $begingroup$
    The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult?
    $endgroup$
    – David K
    Jan 8 at 17:38


















0












$begingroup$


The equation of ellipsoid is
$$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1$$
The hint for substitution is given as
$x=au$, $y=bv$ and $z=cv$.
I know that the projection of the ellipsoid on $xy-$plane is an ellipse with the equation
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$. But I don't know what will be the limits of $z$ in integration.
After transformation the Jacobian of transformation will be
$J=abc$. I don't know how to proceed this one and compute the volume of ellipsoid.
Please explain in detail about the limits before and after substitution with the help of the mathematica command that helps in plotting the ellipsoid along with its shadow.
THanks in advance!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$.
    $endgroup$
    – A.Γ.
    Jan 8 at 17:08










  • $begingroup$
    After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$
    $endgroup$
    – saulspatz
    Jan 8 at 17:08










  • $begingroup$
    @A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir!
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:11










  • $begingroup$
    The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult?
    $endgroup$
    – David K
    Jan 8 at 17:38
















0












0








0





$begingroup$


The equation of ellipsoid is
$$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1$$
The hint for substitution is given as
$x=au$, $y=bv$ and $z=cv$.
I know that the projection of the ellipsoid on $xy-$plane is an ellipse with the equation
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$. But I don't know what will be the limits of $z$ in integration.
After transformation the Jacobian of transformation will be
$J=abc$. I don't know how to proceed this one and compute the volume of ellipsoid.
Please explain in detail about the limits before and after substitution with the help of the mathematica command that helps in plotting the ellipsoid along with its shadow.
THanks in advance!










share|cite|improve this question









$endgroup$




The equation of ellipsoid is
$$frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1$$
The hint for substitution is given as
$x=au$, $y=bv$ and $z=cv$.
I know that the projection of the ellipsoid on $xy-$plane is an ellipse with the equation
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$. But I don't know what will be the limits of $z$ in integration.
After transformation the Jacobian of transformation will be
$J=abc$. I don't know how to proceed this one and compute the volume of ellipsoid.
Please explain in detail about the limits before and after substitution with the help of the mathematica command that helps in plotting the ellipsoid along with its shadow.
THanks in advance!







calculus multivariable-calculus multiple-integral






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share|cite|improve this question











share|cite|improve this question




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asked Jan 8 at 17:02









Noor AslamNoor Aslam

15112




15112








  • 1




    $begingroup$
    If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$.
    $endgroup$
    – A.Γ.
    Jan 8 at 17:08










  • $begingroup$
    After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$
    $endgroup$
    – saulspatz
    Jan 8 at 17:08










  • $begingroup$
    @A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir!
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:11










  • $begingroup$
    The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult?
    $endgroup$
    – David K
    Jan 8 at 17:38
















  • 1




    $begingroup$
    If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$.
    $endgroup$
    – A.Γ.
    Jan 8 at 17:08










  • $begingroup$
    After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$
    $endgroup$
    – saulspatz
    Jan 8 at 17:08










  • $begingroup$
    @A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir!
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:11










  • $begingroup$
    The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult?
    $endgroup$
    – David K
    Jan 8 at 17:38










1




1




$begingroup$
If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$.
$endgroup$
– A.Γ.
Jan 8 at 17:08




$begingroup$
If you mean (how it should be) $z=cw$ then after the substitution the ellipsoid becomes the famous $u^2+v^2+w^2=1$.
$endgroup$
– A.Γ.
Jan 8 at 17:08












$begingroup$
After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$
$endgroup$
– saulspatz
Jan 8 at 17:08




$begingroup$
After making the substitution, you have $u^2+v^2+w^2=1$ so the range of all variables is $[-1,1]$ I assume you meant $z=cw,$ not $z=cv.$
$endgroup$
– saulspatz
Jan 8 at 17:08












$begingroup$
@A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir!
$endgroup$
– Noor Aslam
Jan 8 at 17:11




$begingroup$
@A.Γ. I know after transformation it becomes sphere with centre at origin, I am interested how to find the limits of integration before and after transformation with a proper procedure Sir!
$endgroup$
– Noor Aslam
Jan 8 at 17:11












$begingroup$
The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult?
$endgroup$
– David K
Jan 8 at 17:38






$begingroup$
The whole point of the substitution is that you don't need to find integration limits before the substitution, and for that matter you should know the value of the integral after substitution without writing formulas for the limits of the new variables $u,v,w.$ This should be an easy exercise; why make it difficult?
$endgroup$
– David K
Jan 8 at 17:38












1 Answer
1






active

oldest

votes


















3












$begingroup$

I think you're mixing two things: setting up the integration limits if you don't use the suggested substitution and how to proceed if you want to use that substitution. I'll comment on both.





You can project the ellipsoid onto the $xy$-plane to get an ellipse, but then you're not using the suggested substitution.




$$frac{x^2}{a^2}+frac{y^2}{b^2}=1 tag{$color{blue}{*}$}$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$.




If you want to do it like this, you need to be careful with the limits of integration. The limits you suggest don't bound an ellipse in the $xy$-plane, that would be a rectangle!



You can fix $x$ from $-a$ to $a$, but then $y$ follows from $(color{blue}{*})$ above to get:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} ldots ,mbox{d}y ,mbox{d}x$$
Now you either directly integrate a suited function to get the desired volume, or you set up the bounds for $z$ going back to the original equation of the ellipsoid and integrate the function $1$ to get the volume, which gives:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} int_{-csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}}^{csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}} 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
The first step of integration (with respect to $z$) is easy and using symmetry allows to simplify this a bit further to:
$$8int_{0}^{a} int_{0}^{bsqrt{1-tfrac{x^2}{a^2}}} csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}} ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
And you could take it from there.





Or you would go for the suggested substitution which turns the ellipsoid into a sphere of radius $1$, centered at the origin. Don't forgot to include the Jacobian (as a consequence of the substitution). For the suggested substitution:




The hint for substitution is given as
$x=au$, $y=bv$ and $z=ccolor{red}{w}$.




the Jacobian is simply $abc$ so the integral over the ellipsoid $E$ is turned into an integral over the (simple) sphere $S$:
$$iiint_E 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x = abc iiint_S 1 ,mbox{d}w ,mbox{d}v ,mbox{d}u$$and you're done if you know the volume of a sphere with radius $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:57










  • $begingroup$
    You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
    $endgroup$
    – StackTD
    Jan 8 at 18:03












  • $begingroup$
    Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:11












  • $begingroup$
    It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
    $endgroup$
    – StackTD
    Jan 8 at 18:15












  • $begingroup$
    sir I have multiplied 8 with $frac{4pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:45











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

I think you're mixing two things: setting up the integration limits if you don't use the suggested substitution and how to proceed if you want to use that substitution. I'll comment on both.





You can project the ellipsoid onto the $xy$-plane to get an ellipse, but then you're not using the suggested substitution.




$$frac{x^2}{a^2}+frac{y^2}{b^2}=1 tag{$color{blue}{*}$}$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$.




If you want to do it like this, you need to be careful with the limits of integration. The limits you suggest don't bound an ellipse in the $xy$-plane, that would be a rectangle!



You can fix $x$ from $-a$ to $a$, but then $y$ follows from $(color{blue}{*})$ above to get:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} ldots ,mbox{d}y ,mbox{d}x$$
Now you either directly integrate a suited function to get the desired volume, or you set up the bounds for $z$ going back to the original equation of the ellipsoid and integrate the function $1$ to get the volume, which gives:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} int_{-csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}}^{csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}} 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
The first step of integration (with respect to $z$) is easy and using symmetry allows to simplify this a bit further to:
$$8int_{0}^{a} int_{0}^{bsqrt{1-tfrac{x^2}{a^2}}} csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}} ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
And you could take it from there.





Or you would go for the suggested substitution which turns the ellipsoid into a sphere of radius $1$, centered at the origin. Don't forgot to include the Jacobian (as a consequence of the substitution). For the suggested substitution:




The hint for substitution is given as
$x=au$, $y=bv$ and $z=ccolor{red}{w}$.




the Jacobian is simply $abc$ so the integral over the ellipsoid $E$ is turned into an integral over the (simple) sphere $S$:
$$iiint_E 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x = abc iiint_S 1 ,mbox{d}w ,mbox{d}v ,mbox{d}u$$and you're done if you know the volume of a sphere with radius $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:57










  • $begingroup$
    You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
    $endgroup$
    – StackTD
    Jan 8 at 18:03












  • $begingroup$
    Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:11












  • $begingroup$
    It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
    $endgroup$
    – StackTD
    Jan 8 at 18:15












  • $begingroup$
    sir I have multiplied 8 with $frac{4pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:45
















3












$begingroup$

I think you're mixing two things: setting up the integration limits if you don't use the suggested substitution and how to proceed if you want to use that substitution. I'll comment on both.





You can project the ellipsoid onto the $xy$-plane to get an ellipse, but then you're not using the suggested substitution.




$$frac{x^2}{a^2}+frac{y^2}{b^2}=1 tag{$color{blue}{*}$}$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$.




If you want to do it like this, you need to be careful with the limits of integration. The limits you suggest don't bound an ellipse in the $xy$-plane, that would be a rectangle!



You can fix $x$ from $-a$ to $a$, but then $y$ follows from $(color{blue}{*})$ above to get:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} ldots ,mbox{d}y ,mbox{d}x$$
Now you either directly integrate a suited function to get the desired volume, or you set up the bounds for $z$ going back to the original equation of the ellipsoid and integrate the function $1$ to get the volume, which gives:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} int_{-csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}}^{csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}} 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
The first step of integration (with respect to $z$) is easy and using symmetry allows to simplify this a bit further to:
$$8int_{0}^{a} int_{0}^{bsqrt{1-tfrac{x^2}{a^2}}} csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}} ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
And you could take it from there.





Or you would go for the suggested substitution which turns the ellipsoid into a sphere of radius $1$, centered at the origin. Don't forgot to include the Jacobian (as a consequence of the substitution). For the suggested substitution:




The hint for substitution is given as
$x=au$, $y=bv$ and $z=ccolor{red}{w}$.




the Jacobian is simply $abc$ so the integral over the ellipsoid $E$ is turned into an integral over the (simple) sphere $S$:
$$iiint_E 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x = abc iiint_S 1 ,mbox{d}w ,mbox{d}v ,mbox{d}u$$and you're done if you know the volume of a sphere with radius $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:57










  • $begingroup$
    You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
    $endgroup$
    – StackTD
    Jan 8 at 18:03












  • $begingroup$
    Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:11












  • $begingroup$
    It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
    $endgroup$
    – StackTD
    Jan 8 at 18:15












  • $begingroup$
    sir I have multiplied 8 with $frac{4pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:45














3












3








3





$begingroup$

I think you're mixing two things: setting up the integration limits if you don't use the suggested substitution and how to proceed if you want to use that substitution. I'll comment on both.





You can project the ellipsoid onto the $xy$-plane to get an ellipse, but then you're not using the suggested substitution.




$$frac{x^2}{a^2}+frac{y^2}{b^2}=1 tag{$color{blue}{*}$}$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$.




If you want to do it like this, you need to be careful with the limits of integration. The limits you suggest don't bound an ellipse in the $xy$-plane, that would be a rectangle!



You can fix $x$ from $-a$ to $a$, but then $y$ follows from $(color{blue}{*})$ above to get:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} ldots ,mbox{d}y ,mbox{d}x$$
Now you either directly integrate a suited function to get the desired volume, or you set up the bounds for $z$ going back to the original equation of the ellipsoid and integrate the function $1$ to get the volume, which gives:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} int_{-csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}}^{csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}} 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
The first step of integration (with respect to $z$) is easy and using symmetry allows to simplify this a bit further to:
$$8int_{0}^{a} int_{0}^{bsqrt{1-tfrac{x^2}{a^2}}} csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}} ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
And you could take it from there.





Or you would go for the suggested substitution which turns the ellipsoid into a sphere of radius $1$, centered at the origin. Don't forgot to include the Jacobian (as a consequence of the substitution). For the suggested substitution:




The hint for substitution is given as
$x=au$, $y=bv$ and $z=ccolor{red}{w}$.




the Jacobian is simply $abc$ so the integral over the ellipsoid $E$ is turned into an integral over the (simple) sphere $S$:
$$iiint_E 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x = abc iiint_S 1 ,mbox{d}w ,mbox{d}v ,mbox{d}u$$and you're done if you know the volume of a sphere with radius $1$.






share|cite|improve this answer









$endgroup$



I think you're mixing two things: setting up the integration limits if you don't use the suggested substitution and how to proceed if you want to use that substitution. I'll comment on both.





You can project the ellipsoid onto the $xy$-plane to get an ellipse, but then you're not using the suggested substitution.




$$frac{x^2}{a^2}+frac{y^2}{b^2}=1 tag{$color{blue}{*}$}$$
So before transformation or substitution the curves that bounds the projection of the ellipsoid in $xy-$plane will be $x=-a$, $x=a$ and $y=-b$, $y=b$.




If you want to do it like this, you need to be careful with the limits of integration. The limits you suggest don't bound an ellipse in the $xy$-plane, that would be a rectangle!



You can fix $x$ from $-a$ to $a$, but then $y$ follows from $(color{blue}{*})$ above to get:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} ldots ,mbox{d}y ,mbox{d}x$$
Now you either directly integrate a suited function to get the desired volume, or you set up the bounds for $z$ going back to the original equation of the ellipsoid and integrate the function $1$ to get the volume, which gives:
$$int_{-a}^{a} int_{-bsqrt{1-tfrac{x^2}{a^2}}}^{bsqrt{1-tfrac{x^2}{a^2}}} int_{-csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}}^{csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}}} 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
The first step of integration (with respect to $z$) is easy and using symmetry allows to simplify this a bit further to:
$$8int_{0}^{a} int_{0}^{bsqrt{1-tfrac{x^2}{a^2}}} csqrt{1-tfrac{x^2}{a^2}-tfrac{y^2}{b^2}} ,mbox{d}z ,mbox{d}y ,mbox{d}x$$
And you could take it from there.





Or you would go for the suggested substitution which turns the ellipsoid into a sphere of radius $1$, centered at the origin. Don't forgot to include the Jacobian (as a consequence of the substitution). For the suggested substitution:




The hint for substitution is given as
$x=au$, $y=bv$ and $z=ccolor{red}{w}$.




the Jacobian is simply $abc$ so the integral over the ellipsoid $E$ is turned into an integral over the (simple) sphere $S$:
$$iiint_E 1 ,mbox{d}z ,mbox{d}y ,mbox{d}x = abc iiint_S 1 ,mbox{d}w ,mbox{d}v ,mbox{d}u$$and you're done if you know the volume of a sphere with radius $1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 17:29









StackTDStackTD

22.6k2049




22.6k2049












  • $begingroup$
    I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:57










  • $begingroup$
    You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
    $endgroup$
    – StackTD
    Jan 8 at 18:03












  • $begingroup$
    Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:11












  • $begingroup$
    It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
    $endgroup$
    – StackTD
    Jan 8 at 18:15












  • $begingroup$
    sir I have multiplied 8 with $frac{4pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:45


















  • $begingroup$
    I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
    $endgroup$
    – Noor Aslam
    Jan 8 at 17:57










  • $begingroup$
    You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
    $endgroup$
    – StackTD
    Jan 8 at 18:03












  • $begingroup$
    Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:11












  • $begingroup$
    It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
    $endgroup$
    – StackTD
    Jan 8 at 18:15












  • $begingroup$
    sir I have multiplied 8 with $frac{4pi}{3}$
    $endgroup$
    – Noor Aslam
    Jan 8 at 18:45
















$begingroup$
I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
$endgroup$
– Noor Aslam
Jan 8 at 17:57




$begingroup$
I know the volume of the unit sphere is $frac{4pi}{3}$, can I compute this $frac{4pi}{3}$, from the integral putting limits on it $intintint_{S}dwdvdu$. I shall be very thankfull! Sir
$endgroup$
– Noor Aslam
Jan 8 at 17:57












$begingroup$
You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
$endgroup$
– StackTD
Jan 8 at 18:03






$begingroup$
You either directly use the fact that you know the volume of the unit sphere at this point (and the volume of the ellipsoid follows immediately), or you can compute it via integration if you can't use it - but then the advantage of the substitution is limited to simpler calculations rather than almost no calculations.
$endgroup$
– StackTD
Jan 8 at 18:03














$begingroup$
Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
$endgroup$
– Noor Aslam
Jan 8 at 18:11






$begingroup$
Sir if want to calculate it through finding the integration after substitution the integral is not still simple, How can I calculate? The integral after substitution is of the form $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu$, I don't know how to show that $8int_{0}^{1}int_{0}^{sqrt{1-u^2}}int_{0}^{sqrt{1-u^2-v^2}}dwdvdu=frac{32pi}{3}$
$endgroup$
– Noor Aslam
Jan 8 at 18:11














$begingroup$
It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
$endgroup$
– StackTD
Jan 8 at 18:15






$begingroup$
It's 4 in the numerator, not 32. How annoying that integral is, depends on what you can use: I would do it with spherical coordinates or avoid a triple integral by calculating it as the volume of a solid of revolution. But that all depends on your context; if the issue now is how to find the volume of a sphere through integration: searching here on this site (or on google) will give you a lot of information and different approaches.
$endgroup$
– StackTD
Jan 8 at 18:15














$begingroup$
sir I have multiplied 8 with $frac{4pi}{3}$
$endgroup$
– Noor Aslam
Jan 8 at 18:45




$begingroup$
sir I have multiplied 8 with $frac{4pi}{3}$
$endgroup$
– Noor Aslam
Jan 8 at 18:45


















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