To find the moment of inertia about co-ordinate axes of a tetrahedron in the first octant bounded by x+y+z=1
$begingroup$
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
asked Dec 10 '18 at 13:54
tNotrtNotr
142
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 13:54
tNotrtNotr
142
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 10 '18 at 13:54
tNotrtNotr
142
$endgroup$
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
asked Dec 10 '18 at 13:54
tNotrtNotr
142
asked Dec 10 '18 at 13:54
tNotrtNotr
142
edited Dec 10 '18 at 14:04
tNotr
asked Dec 10 '18 at 13:54
tNotrtNotr
142
asked Dec 10 '18 at 13:54
tNotrtNotr
142
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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