Moment of inertia of a solid right circular cone
So I was trying to do this using triple integrals (I know there are simpler ways) when I encountered something I can't really justify..here it is
the mass is uniformly distributed so $dm=rho dv$ and the equation of the cone is $$ x^2 + y^2 = frac{R^2}{h^2}z^2$$
where R is the radius of circular top of the cone and h is it's height , now the square of the distance from the z axis is $x^2 + y^2 $ and I evaluate the integral using cylindrical coordinates and the equation becomes $r=frac{R}{h}z$ (taking the positive root)
$$iiint_v r^3rho,dz,dr,dtheta$$ where the limits of z are $z=frac{h}{R}r$ and h , r from zero to R and $theta$ from zero to $2pi$
I do the integral and get the correct answer which $frac{3MR^2}{10}$
here's my question..when I replace the $r^3$ term in the integrand with $r^3=frac{R^3}{h^3}z^3$ and do the integral (using the same limits and order of integration) I get an incorrect result and I can't explain why this happens
multivariable-calculus
asked Nov 24 at 21:57
Km356
887
add a comment |
So I was trying to do this using triple integrals (I know there are simpler ways) when I encountered something I can't really justify..here it is
the mass is uniformly distributed so $dm=rho dv$ and the equation of the cone is $$ x^2 + y^2 = frac{R^2}{h^2}z^2$$
where R is the radius of circular top of the cone and h is it's height , now the square of the distance from the z axis is $x^2 + y^2 $ and I evaluate the integral using cylindrical coordinates and the equation becomes $r=frac{R}{h}z$ (taking the positive root)
$$iiint_v r^3rho,dz,dr,dtheta$$ where the limits of z are $z=frac{h}{R}r$ and h , r from zero to R and $theta$ from zero to $2pi$
I do the integral and get the correct answer which $frac{3MR^2}{10}$
here's my question..when I replace the $r^3$ term in the integrand with $r^3=frac{R^3}{h^3}z^3$ and do the integral (using the same limits and order of integration) I get an incorrect result and I can't explain why this happens
multivariable-calculus
asked Nov 24 at 21:57
Km356
887
add a comment |
So I was trying to do this using triple integrals (I know there are simpler ways) when I encountered something I can't really justify..here it is
the mass is uniformly distributed so $dm=rho dv$ and the equation of the cone is $$ x^2 + y^2 = frac{R^2}{h^2}z^2$$
where R is the radius of circular top of the cone and h is it's height , now the square of the distance from the z axis is $x^2 + y^2 $ and I evaluate the integral using cylindrical coordinates and the equation becomes $r=frac{R}{h}z$ (taking the positive root)
$$iiint_v r^3rho,dz,dr,dtheta$$ where the limits of z are $z=frac{h}{R}r$ and h , r from zero to R and $theta$ from zero to $2pi$
I do the integral and get the correct answer which $frac{3MR^2}{10}$
here's my question..when I replace the $r^3$ term in the integrand with $r^3=frac{R^3}{h^3}z^3$ and do the integral (using the same limits and order of integration) I get an incorrect result and I can't explain why this happens
multivariable-calculus
asked Nov 24 at 21:57
Km356
887
So I was trying to do this using triple integrals (I know there are simpler ways) when I encountered something I can't really justify..here it is
the mass is uniformly distributed so $dm=rho dv$ and the equation of the cone is $$ x^2 + y^2 = frac{R^2}{h^2}z^2$$
where R is the radius of circular top of the cone and h is it's height , now the square of the distance from the z axis is $x^2 + y^2 $ and I evaluate the integral using cylindrical coordinates and the equation becomes $r=frac{R}{h}z$ (taking the positive root)
$$iiint_v r^3rho,dz,dr,dtheta$$ where the limits of z are $z=frac{h}{R}r$ and h , r from zero to R and $theta$ from zero to $2pi$
I do the integral and get the correct answer which $frac{3MR^2}{10}$
here's my question..when I replace the $r^3$ term in the integrand with $r^3=frac{R^3}{h^3}z^3$ and do the integral (using the same limits and order of integration) I get an incorrect result and I can't explain why this happens
multivariable-calculus
multivariable-calculus
asked Nov 24 at 21:57
Km356
887
asked Nov 24 at 21:57
Km356
887
asked Nov 24 at 21:57
Km356
887
asked Nov 24 at 21:57
Km356
887
asked Nov 24 at 21:57
Km356
887
887
add a comment |
add a comment |
1 Answer
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Yes the correct set up is as follows
$$int_0^{2pi},dthetaint_0^h dz int_0^{frac{R}{h}z} rho r^3,dr$$
we can't substitute $r^3=frac{R^3}{h^3}z^3$ because the $r$ indicated here represent the maximum radius to be considered for any fixed $z$, that is $$r_{MAX}^3=frac{R^3}{h^3}z^3$$ but for any $z$ we have $0 le r le r_{MAX}$.
answered Nov 24 at 22:03
gimusi
1
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Yes the correct set up is as follows
$$int_0^{2pi},dthetaint_0^h dz int_0^{frac{R}{h}z} rho r^3,dr$$
we can't substitute $r^3=frac{R^3}{h^3}z^3$ because the $r$ indicated here represent the maximum radius to be considered for any fixed $z$, that is $$r_{MAX}^3=frac{R^3}{h^3}z^3$$ but for any $z$ we have $0 le r le r_{MAX}$.
answered Nov 24 at 22:03
gimusi
1
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
add a comment |
Yes the correct set up is as follows
$$int_0^{2pi},dthetaint_0^h dz int_0^{frac{R}{h}z} rho r^3,dr$$
we can't substitute $r^3=frac{R^3}{h^3}z^3$ because the $r$ indicated here represent the maximum radius to be considered for any fixed $z$, that is $$r_{MAX}^3=frac{R^3}{h^3}z^3$$ but for any $z$ we have $0 le r le r_{MAX}$.
answered Nov 24 at 22:03
gimusi
1
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
add a comment |
Yes the correct set up is as follows
$$int_0^{2pi},dthetaint_0^h dz int_0^{frac{R}{h}z} rho r^3,dr$$
we can't substitute $r^3=frac{R^3}{h^3}z^3$ because the $r$ indicated here represent the maximum radius to be considered for any fixed $z$, that is $$r_{MAX}^3=frac{R^3}{h^3}z^3$$ but for any $z$ we have $0 le r le r_{MAX}$.
answered Nov 24 at 22:03
gimusi
1
Yes the correct set up is as follows
$$int_0^{2pi},dthetaint_0^h dz int_0^{frac{R}{h}z} rho r^3,dr$$
we can't substitute $r^3=frac{R^3}{h^3}z^3$ because the $r$ indicated here represent the maximum radius to be considered for any fixed $z$, that is $$r_{MAX}^3=frac{R^3}{h^3}z^3$$ but for any $z$ we have $0 le r le r_{MAX}$.
answered Nov 24 at 22:03
gimusi
1
edited Nov 24 at 22:22
answered Nov 24 at 22:03
gimusi
1
answered Nov 24 at 22:03
gimusi
1
answered Nov 24 at 22:03
gimusi
1
1
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
add a comment |
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
– Km356
Nov 24 at 22:19
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 Yes of course sorry, of course we need $r(z)$! I fix that
– gimusi
Nov 24 at 22:22
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
@Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
– gimusi
Nov 24 at 22:23
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
yes that is indeed what I missed,, thanks
– Km356
Nov 24 at 22:25
add a comment |
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