Find the volume of ice cream cone using cylindrical/spherical coordinates












1












$begingroup$


I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone



Remember that $r^2=x^2+y^2$



I assumed for cylindrical coordinates the triple integral boundaries would be



$-sqrt{(9-r^2)}le z le -r$



$0le r le 3$



$0le theta le 2pi$



And for spherical coordinates the triple integral boundaries would be



$0le r le 3$



$pi/2le phi le pi$



$0le theta le 2pi$



However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.



So somethings wrong with my boundaries, as both integrals should equal the same volume.



This is my working out for cylindrical coordinate integral so far:



$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$



$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$



$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$



$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$



Let $u=9-r^2$



$du=-2r,dr$



$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$



$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$



$int_0^{2pi}[-9,+9],dtheta$



$int_0^{2pi}0,dtheta$



$=0$



So as you can see I can't proceed to the third integral since the second integral equals zero



Any help would be greatly appreciated :)






multivariable-calculus vector-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
    $endgroup$
    – Ian Miller
    Apr 7 '16 at 8:23


















1












$begingroup$


I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone



Remember that $r^2=x^2+y^2$



I assumed for cylindrical coordinates the triple integral boundaries would be



$-sqrt{(9-r^2)}le z le -r$



$0le r le 3$



$0le theta le 2pi$



And for spherical coordinates the triple integral boundaries would be



$0le r le 3$



$pi/2le phi le pi$



$0le theta le 2pi$



However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.



So somethings wrong with my boundaries, as both integrals should equal the same volume.



This is my working out for cylindrical coordinate integral so far:



$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$



$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$



$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$



$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$



Let $u=9-r^2$



$du=-2r,dr$



$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$



$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$



$int_0^{2pi}[-9,+9],dtheta$



$int_0^{2pi}0,dtheta$



$=0$



So as you can see I can't proceed to the third integral since the second integral equals zero



Any help would be greatly appreciated :)






multivariable-calculus vector-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
    $endgroup$
    – Ian Miller
    Apr 7 '16 at 8:23
















1












1








1


0



$begingroup$


I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone



Remember that $r^2=x^2+y^2$



I assumed for cylindrical coordinates the triple integral boundaries would be



$-sqrt{(9-r^2)}le z le -r$



$0le r le 3$



$0le theta le 2pi$



And for spherical coordinates the triple integral boundaries would be



$0le r le 3$



$pi/2le phi le pi$



$0le theta le 2pi$



However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.



So somethings wrong with my boundaries, as both integrals should equal the same volume.



This is my working out for cylindrical coordinate integral so far:



$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$



$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$



$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$



$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$



Let $u=9-r^2$



$du=-2r,dr$



$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$



$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$



$int_0^{2pi}[-9,+9],dtheta$



$int_0^{2pi}0,dtheta$



$=0$



So as you can see I can't proceed to the third integral since the second integral equals zero



Any help would be greatly appreciated :)






multivariable-calculus vector-analysis







share|cite|improve this question









$endgroup$




I'm stuck on what the boundaries are for the volume bounded by the cone $z=-sqrt{(x^2+y^2)}$ and the surface $z=-sqrt{(9-x^2-y^2)}$ $,,$-essentially an upside down ice cream cone



Remember that $r^2=x^2+y^2$



I assumed for cylindrical coordinates the triple integral boundaries would be



$-sqrt{(9-r^2)}le z le -r$



$0le r le 3$



$0le theta le 2pi$



And for spherical coordinates the triple integral boundaries would be



$0le r le 3$



$pi/2le phi le pi$



$0le theta le 2pi$



However upon entering these values into MATLAB, the cylindrical coordinates integral equals to zero, whilst the spherical coordinates integral equals to $18pi$.



So somethings wrong with my boundaries, as both integrals should equal the same volume.



This is my working out for cylindrical coordinate integral so far:



$int_0^{2pi}int_0^3int_{-sqrt{9-r^2}}^{-r} r,, dzdrdtheta$



$int_0^{2pi}int_0^3 [rz]_{-sqrt{9-r^2}}^{-r} ,, dzdrdtheta$



$int_0^{2pi}int_0^3 -r^2+{rsqrt{9-r^2}} ,, dzdrdtheta$



$int_0^{2pi}[(int_0^3 -r^2 ,dr)+(int_0^3 {rsqrt{9-r^2} ,dr)}]dtheta$



Let $u=9-r^2$



$du=-2r,dr$



$int_0^{2pi}[[frac{-r^3}{3}]_0^3,-frac{1}{2}int_0^3 u^{frac{1}{2}} ,du)],dtheta$



$int_0^{2pi}[-frac{27}{3},-frac{1}{2}[frac{2}{3}u^{frac{3}{2}}]_{r=0}^{r=3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-r^2)^{frac{3}{2}}]_{0}^{3} ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}[{frac{2}{3}}(9-3^2)^{frac{3}{2}},-{frac{2}{3}}(9-0^2)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{2}{frac{2}{3}}[(9-9)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[(0)^{frac{3}{2}},-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-(9)^{frac{3}{2}}] ],dtheta$



$int_0^{2pi}[-9,-frac{1}{3}[-27] ],dtheta$



$int_0^{2pi}[-9,+9],dtheta$



$int_0^{2pi}0,dtheta$



$=0$



So as you can see I can't proceed to the third integral since the second integral equals zero



Any help would be greatly appreciated :)






multivariable-calculus vector-analysis




multivariable-calculus vector-analysis






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share|cite|improve this question











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asked Apr 7 '16 at 8:15









AVeljAVelj

3610




3610












  • $begingroup$
    Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
    $endgroup$
    – Ian Miller
    Apr 7 '16 at 8:23




















  • $begingroup$
    Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
    $endgroup$
    – Ian Miller
    Apr 7 '16 at 8:23


















$begingroup$
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23






$begingroup$
Hi again. I don't think $18pi$ is correct either. Consider a sphere with radius of $3$. The volume would be $frac43pitimes3^3=36pi$. Your answer is half of this while your shape is clearly less than half a sphere. The reason for your spherical answer giving this is incorrect limits on $phi$. You should use $frac{3pi}{4}leqphileqpi$ as your cone has an angle of $frac{pi}{4}$. (I'll try to give you an actual answer once I get back to my computer.)
$endgroup$
– Ian Miller
Apr 7 '16 at 8:23












2 Answers
2






active

oldest

votes


















1












$begingroup$

By cylindrical coordinates the set up of the integral should be



$$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
    $endgroup$
    – manooooh
    Nov 26 '18 at 3:08






  • 1




    $begingroup$
    @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
    $endgroup$
    – gimusi
    Nov 26 '18 at 6:06





















0












$begingroup$

The surfaces intersect when
$$-r=-sqrt{9-r^2}$$
which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.



For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    By cylindrical coordinates the set up of the integral should be



    $$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
      $endgroup$
      – manooooh
      Nov 26 '18 at 3:08






    • 1




      $begingroup$
      @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
      $endgroup$
      – gimusi
      Nov 26 '18 at 6:06


















    1












    $begingroup$

    By cylindrical coordinates the set up of the integral should be



    $$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
      $endgroup$
      – manooooh
      Nov 26 '18 at 3:08






    • 1




      $begingroup$
      @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
      $endgroup$
      – gimusi
      Nov 26 '18 at 6:06
















    1












    1








    1





    $begingroup$

    By cylindrical coordinates the set up of the integral should be



    $$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$






    share|cite|improve this answer











    $endgroup$



    By cylindrical coordinates the set up of the integral should be



    $$int_0^{2pi}int_{-3}^{-frac3{sqrt2}}int_{0}^{sqrt{9-z^2}} r,, dzdrdtheta+int_0^{2pi}int_{-frac3{sqrt2}}^0int_{0}^{-z} r,, dzdrdtheta$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 '18 at 6:07

























    answered Feb 3 '18 at 3:20









    gimusigimusi

    92.9k94494




    92.9k94494












    • $begingroup$
      Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
      $endgroup$
      – manooooh
      Nov 26 '18 at 3:08






    • 1




      $begingroup$
      @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
      $endgroup$
      – gimusi
      Nov 26 '18 at 6:06




















    • $begingroup$
      Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
      $endgroup$
      – manooooh
      Nov 26 '18 at 3:08






    • 1




      $begingroup$
      @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
      $endgroup$
      – gimusi
      Nov 26 '18 at 6:06


















    $begingroup$
    Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
    $endgroup$
    – manooooh
    Nov 26 '18 at 3:08




    $begingroup$
    Wait, the polar coordinates are not for double integrals? I think it should be cylindrical coordinates, altough the first two coordinates are the same for the third integral we need a third variable.
    $endgroup$
    – manooooh
    Nov 26 '18 at 3:08




    1




    1




    $begingroup$
    @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
    $endgroup$
    – gimusi
    Nov 26 '18 at 6:06






    $begingroup$
    @manooooh Yes of course, I’ve used the uncorrect term here! I fix that, thanks.
    $endgroup$
    – gimusi
    Nov 26 '18 at 6:06













    0












    $begingroup$

    The surfaces intersect when
    $$-r=-sqrt{9-r^2}$$
    which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.



    For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The surfaces intersect when
      $$-r=-sqrt{9-r^2}$$
      which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.



      For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The surfaces intersect when
        $$-r=-sqrt{9-r^2}$$
        which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.



        For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)






        share|cite|improve this answer









        $endgroup$



        The surfaces intersect when
        $$-r=-sqrt{9-r^2}$$
        which is $r^2=frac92$, not $r^2=9$ as you have in your cylindrical attempt.



        For spherical coordinates $phi$ should be from $frac34pi$ to $pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call this $rho^2$ in order to distinguish it from $r^2=x^2+y^2$.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 7 '16 at 8:24









        DavidDavid

        68.1k664126




        68.1k664126






























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