Moment of inertia of a solid right circular cone












1












$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 '18 at 21:57









      Km356Km356

      887




      887






















          1 Answer
          1






          active

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          1












          $begingroup$

          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 $0 le r le r_{MAX}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
            $endgroup$
            – Km356
            Nov 24 '18 at 22:19










          • $begingroup$
            @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:22










          • $begingroup$
            @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:23










          • $begingroup$
            yes that is indeed what I missed,, thanks
            $endgroup$
            – Km356
            Nov 24 '18 at 22:25











          Your Answer





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          1 Answer
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          active

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          1 Answer
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          active

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          active

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          1












          $begingroup$

          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 $0 le r le r_{MAX}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
            $endgroup$
            – Km356
            Nov 24 '18 at 22:19










          • $begingroup$
            @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:22










          • $begingroup$
            @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:23










          • $begingroup$
            yes that is indeed what I missed,, thanks
            $endgroup$
            – Km356
            Nov 24 '18 at 22:25
















          1












          $begingroup$

          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 $0 le r le r_{MAX}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
            $endgroup$
            – Km356
            Nov 24 '18 at 22:19










          • $begingroup$
            @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:22










          • $begingroup$
            @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:23










          • $begingroup$
            yes that is indeed what I missed,, thanks
            $endgroup$
            – Km356
            Nov 24 '18 at 22:25














          1












          1








          1





          $begingroup$

          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 $0 le r le r_{MAX}$.






          share|cite|improve this answer











          $endgroup$



          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 $0 le r le r_{MAX}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 '18 at 22:22

























          answered Nov 24 '18 at 22:03









          gimusigimusi

          1




          1












          • $begingroup$
            isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
            $endgroup$
            – Km356
            Nov 24 '18 at 22:19










          • $begingroup$
            @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:22










          • $begingroup$
            @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:23










          • $begingroup$
            yes that is indeed what I missed,, thanks
            $endgroup$
            – Km356
            Nov 24 '18 at 22:25


















          • $begingroup$
            isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
            $endgroup$
            – Km356
            Nov 24 '18 at 22:19










          • $begingroup$
            @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:22










          • $begingroup$
            @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
            $endgroup$
            – gimusi
            Nov 24 '18 at 22:23










          • $begingroup$
            yes that is indeed what I missed,, thanks
            $endgroup$
            – Km356
            Nov 24 '18 at 22:25
















          $begingroup$
          isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
          $endgroup$
          – Km356
          Nov 24 '18 at 22:19




          $begingroup$
          isn't the lower limit for z a function of r? the integral you wrote doesn't lead to the correct answer
          $endgroup$
          – Km356
          Nov 24 '18 at 22:19












          $begingroup$
          @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
          $endgroup$
          – gimusi
          Nov 24 '18 at 22:22




          $begingroup$
          @Km356 Yes of course sorry, of course we need $r(z)$! I fix that
          $endgroup$
          – gimusi
          Nov 24 '18 at 22:22












          $begingroup$
          @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
          $endgroup$
          – gimusi
          Nov 24 '18 at 22:23




          $begingroup$
          @Km356 That value is indeed the $r_{MAX}$ and that the reason why we can't substitute that with $r$ inside the integral.
          $endgroup$
          – gimusi
          Nov 24 '18 at 22:23












          $begingroup$
          yes that is indeed what I missed,, thanks
          $endgroup$
          – Km356
          Nov 24 '18 at 22:25




          $begingroup$
          yes that is indeed what I missed,, thanks
          $endgroup$
          – Km356
          Nov 24 '18 at 22:25


















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