Volume of Solid of Revolution (Glass)












0












$begingroup$


We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
glass



What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?



Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$






integration volume solid-of-revolution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you show your attempt? Otherwise we are not able to help you.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:28










  • $begingroup$
    See my hint below. Now you should know how to start.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:35










  • $begingroup$
    Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:21










  • $begingroup$
    I have done this using geogebra, however I am not sure what my integral should be
    $endgroup$
    – M Papas
    Dec 19 '18 at 16:22










  • $begingroup$
    Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:30
















0












$begingroup$


We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
glass



What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?



Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$






integration volume solid-of-revolution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you show your attempt? Otherwise we are not able to help you.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:28










  • $begingroup$
    See my hint below. Now you should know how to start.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:35










  • $begingroup$
    Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:21










  • $begingroup$
    I have done this using geogebra, however I am not sure what my integral should be
    $endgroup$
    – M Papas
    Dec 19 '18 at 16:22










  • $begingroup$
    Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:30














0












0








0





$begingroup$


We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
glass



What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?



Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$






integration volume solid-of-revolution







share|cite|improve this question











$endgroup$




We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
glass



What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?



Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$






integration volume solid-of-revolution




integration volume solid-of-revolution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 13:06







M Papas

















asked Dec 19 '18 at 15:22









M PapasM Papas

76




76












  • $begingroup$
    Why don't you show your attempt? Otherwise we are not able to help you.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:28










  • $begingroup$
    See my hint below. Now you should know how to start.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:35










  • $begingroup$
    Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:21










  • $begingroup$
    I have done this using geogebra, however I am not sure what my integral should be
    $endgroup$
    – M Papas
    Dec 19 '18 at 16:22










  • $begingroup$
    Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:30


















  • $begingroup$
    Why don't you show your attempt? Otherwise we are not able to help you.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:28










  • $begingroup$
    See my hint below. Now you should know how to start.
    $endgroup$
    – Robert Z
    Dec 19 '18 at 15:35










  • $begingroup$
    Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:21










  • $begingroup$
    I have done this using geogebra, however I am not sure what my integral should be
    $endgroup$
    – M Papas
    Dec 19 '18 at 16:22










  • $begingroup$
    Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:30
















$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28




$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28












$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35




$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35












$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21




$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21












$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22




$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22












$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30




$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30










3 Answers
3






active

oldest

votes


















0












$begingroup$

The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.



The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.



The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
$$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
You can now get the upper limit of $x$ from the above equation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    which integral should I use to calculate the volume?
    $endgroup$
    – M Papas
    Dec 19 '18 at 16:53










  • $begingroup$
    Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
    $endgroup$
    – Andrei
    Dec 19 '18 at 16:57












  • $begingroup$
    In the disk method what should my integral be? I didint really get it
    $endgroup$
    – M Papas
    Dec 19 '18 at 17:02










  • $begingroup$
    You integrate over $y$, between $a$ and $A$ the volume of the disk
    $endgroup$
    – Andrei
    Dec 19 '18 at 17:07



















1












$begingroup$

I do the problem in cylindrical coordinates as follows:
Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
(I figured this by reconciling $r$ with $x$ in the $x-z$ plane)



Now the integral becomes terribly simple:
$$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    So in your drawing you show that you are almost near to find the good start.



    Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.



    Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.



    When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.



    Then just integrate the above along $y$, between the bounds you need.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.



      The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.



      The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
      $$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
      You can now get the upper limit of $x$ from the above equation.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        which integral should I use to calculate the volume?
        $endgroup$
        – M Papas
        Dec 19 '18 at 16:53










      • $begingroup$
        Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
        $endgroup$
        – Andrei
        Dec 19 '18 at 16:57












      • $begingroup$
        In the disk method what should my integral be? I didint really get it
        $endgroup$
        – M Papas
        Dec 19 '18 at 17:02










      • $begingroup$
        You integrate over $y$, between $a$ and $A$ the volume of the disk
        $endgroup$
        – Andrei
        Dec 19 '18 at 17:07
















      0












      $begingroup$

      The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.



      The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.



      The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
      $$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
      You can now get the upper limit of $x$ from the above equation.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        which integral should I use to calculate the volume?
        $endgroup$
        – M Papas
        Dec 19 '18 at 16:53










      • $begingroup$
        Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
        $endgroup$
        – Andrei
        Dec 19 '18 at 16:57












      • $begingroup$
        In the disk method what should my integral be? I didint really get it
        $endgroup$
        – M Papas
        Dec 19 '18 at 17:02










      • $begingroup$
        You integrate over $y$, between $a$ and $A$ the volume of the disk
        $endgroup$
        – Andrei
        Dec 19 '18 at 17:07














      0












      0








      0





      $begingroup$

      The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.



      The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.



      The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
      $$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
      You can now get the upper limit of $x$ from the above equation.






      share|cite|improve this answer









      $endgroup$



      The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.



      The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.



      The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
      $$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
      You can now get the upper limit of $x$ from the above equation.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 19 '18 at 16:50









      AndreiAndrei

      13.3k21230




      13.3k21230












      • $begingroup$
        which integral should I use to calculate the volume?
        $endgroup$
        – M Papas
        Dec 19 '18 at 16:53










      • $begingroup$
        Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
        $endgroup$
        – Andrei
        Dec 19 '18 at 16:57












      • $begingroup$
        In the disk method what should my integral be? I didint really get it
        $endgroup$
        – M Papas
        Dec 19 '18 at 17:02










      • $begingroup$
        You integrate over $y$, between $a$ and $A$ the volume of the disk
        $endgroup$
        – Andrei
        Dec 19 '18 at 17:07


















      • $begingroup$
        which integral should I use to calculate the volume?
        $endgroup$
        – M Papas
        Dec 19 '18 at 16:53










      • $begingroup$
        Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
        $endgroup$
        – Andrei
        Dec 19 '18 at 16:57












      • $begingroup$
        In the disk method what should my integral be? I didint really get it
        $endgroup$
        – M Papas
        Dec 19 '18 at 17:02










      • $begingroup$
        You integrate over $y$, between $a$ and $A$ the volume of the disk
        $endgroup$
        – Andrei
        Dec 19 '18 at 17:07
















      $begingroup$
      which integral should I use to calculate the volume?
      $endgroup$
      – M Papas
      Dec 19 '18 at 16:53




      $begingroup$
      which integral should I use to calculate the volume?
      $endgroup$
      – M Papas
      Dec 19 '18 at 16:53












      $begingroup$
      Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
      $endgroup$
      – Andrei
      Dec 19 '18 at 16:57






      $begingroup$
      Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
      $endgroup$
      – Andrei
      Dec 19 '18 at 16:57














      $begingroup$
      In the disk method what should my integral be? I didint really get it
      $endgroup$
      – M Papas
      Dec 19 '18 at 17:02




      $begingroup$
      In the disk method what should my integral be? I didint really get it
      $endgroup$
      – M Papas
      Dec 19 '18 at 17:02












      $begingroup$
      You integrate over $y$, between $a$ and $A$ the volume of the disk
      $endgroup$
      – Andrei
      Dec 19 '18 at 17:07




      $begingroup$
      You integrate over $y$, between $a$ and $A$ the volume of the disk
      $endgroup$
      – Andrei
      Dec 19 '18 at 17:07











      1












      $begingroup$

      I do the problem in cylindrical coordinates as follows:
      Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
      (I figured this by reconciling $r$ with $x$ in the $x-z$ plane)



      Now the integral becomes terribly simple:
      $$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        I do the problem in cylindrical coordinates as follows:
        Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
        (I figured this by reconciling $r$ with $x$ in the $x-z$ plane)



        Now the integral becomes terribly simple:
        $$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          I do the problem in cylindrical coordinates as follows:
          Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
          (I figured this by reconciling $r$ with $x$ in the $x-z$ plane)



          Now the integral becomes terribly simple:
          $$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$






          share|cite|improve this answer











          $endgroup$



          I do the problem in cylindrical coordinates as follows:
          Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
          (I figured this by reconciling $r$ with $x$ in the $x-z$ plane)



          Now the integral becomes terribly simple:
          $$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 17:17

























          answered Dec 19 '18 at 17:11









          Sameer BahetiSameer Baheti

          5618




          5618























              0












              $begingroup$

              So in your drawing you show that you are almost near to find the good start.



              Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.



              Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.



              When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.



              Then just integrate the above along $y$, between the bounds you need.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                So in your drawing you show that you are almost near to find the good start.



                Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.



                Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.



                When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.



                Then just integrate the above along $y$, between the bounds you need.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  So in your drawing you show that you are almost near to find the good start.



                  Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.



                  Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.



                  When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.



                  Then just integrate the above along $y$, between the bounds you need.






                  share|cite|improve this answer









                  $endgroup$



                  So in your drawing you show that you are almost near to find the good start.



                  Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.



                  Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.



                  When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.



                  Then just integrate the above along $y$, between the bounds you need.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 16:55









                  G CabG Cab

                  20.4k31341




                  20.4k31341






























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