evaluate this elliptic hyperbloid volume?












1












$begingroup$


How can I calculate the volume of this region in cylindrical coordinates?



$D={2x^2+y^2=z^2+4,|z| le 2}$



I think I got this wrong :



$$operatorname{Volume} = 2int_{0}^{2pi}int_{0}^{2} int_{0}^{sqrt{2r^2cos^2theta+r^2sin^2theta-4}}rdzdrdtheta$$



The problem is that the region is an ellipse and not a circle. I can integrate for $x^2+y^2=z^2+4$. Could you explain to me how to do it?



I think, what's wrong is also the Jacobian :



$x=2rcostheta$



$y=frac{1}{sqrt{2}}rcostheta$



$|J|=2sqrt{2}$ ?



so maybe z range like this : $z={0,sqrt{8r^2cos^2theta+frac{1}{2}r^2sin^2theta-4}}$






calculus integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=rcostheta, y=rsintheta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a ucos v, y=b usin v, z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2pi$ but is not the polar angle, and $u$ runs from $0$ to $1$.
    $endgroup$
    – Klaas van Aarsen
    Jan 16 at 20:22


















1












$begingroup$


How can I calculate the volume of this region in cylindrical coordinates?



$D={2x^2+y^2=z^2+4,|z| le 2}$



I think I got this wrong :



$$operatorname{Volume} = 2int_{0}^{2pi}int_{0}^{2} int_{0}^{sqrt{2r^2cos^2theta+r^2sin^2theta-4}}rdzdrdtheta$$



The problem is that the region is an ellipse and not a circle. I can integrate for $x^2+y^2=z^2+4$. Could you explain to me how to do it?



I think, what's wrong is also the Jacobian :



$x=2rcostheta$



$y=frac{1}{sqrt{2}}rcostheta$



$|J|=2sqrt{2}$ ?



so maybe z range like this : $z={0,sqrt{8r^2cos^2theta+frac{1}{2}r^2sin^2theta-4}}$






calculus integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=rcostheta, y=rsintheta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a ucos v, y=b usin v, z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2pi$ but is not the polar angle, and $u$ runs from $0$ to $1$.
    $endgroup$
    – Klaas van Aarsen
    Jan 16 at 20:22
















1












1








1





$begingroup$


How can I calculate the volume of this region in cylindrical coordinates?



$D={2x^2+y^2=z^2+4,|z| le 2}$



I think I got this wrong :



$$operatorname{Volume} = 2int_{0}^{2pi}int_{0}^{2} int_{0}^{sqrt{2r^2cos^2theta+r^2sin^2theta-4}}rdzdrdtheta$$



The problem is that the region is an ellipse and not a circle. I can integrate for $x^2+y^2=z^2+4$. Could you explain to me how to do it?



I think, what's wrong is also the Jacobian :



$x=2rcostheta$



$y=frac{1}{sqrt{2}}rcostheta$



$|J|=2sqrt{2}$ ?



so maybe z range like this : $z={0,sqrt{8r^2cos^2theta+frac{1}{2}r^2sin^2theta-4}}$






calculus integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$




How can I calculate the volume of this region in cylindrical coordinates?



$D={2x^2+y^2=z^2+4,|z| le 2}$



I think I got this wrong :



$$operatorname{Volume} = 2int_{0}^{2pi}int_{0}^{2} int_{0}^{sqrt{2r^2cos^2theta+r^2sin^2theta-4}}rdzdrdtheta$$



The problem is that the region is an ellipse and not a circle. I can integrate for $x^2+y^2=z^2+4$. Could you explain to me how to do it?



I think, what's wrong is also the Jacobian :



$x=2rcostheta$



$y=frac{1}{sqrt{2}}rcostheta$



$|J|=2sqrt{2}$ ?



so maybe z range like this : $z={0,sqrt{8r^2cos^2theta+frac{1}{2}r^2sin^2theta-4}}$






calculus integration multivariable-calculus definite-integrals




calculus integration multivariable-calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 20:24









InsideOut

5,14431034




5,14431034










asked Jan 16 at 19:23









NPLSNPLS

7812




7812












  • $begingroup$
    The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=rcostheta, y=rsintheta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a ucos v, y=b usin v, z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2pi$ but is not the polar angle, and $u$ runs from $0$ to $1$.
    $endgroup$
    – Klaas van Aarsen
    Jan 16 at 20:22




















  • $begingroup$
    The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=rcostheta, y=rsintheta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a ucos v, y=b usin v, z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2pi$ but is not the polar angle, and $u$ runs from $0$ to $1$.
    $endgroup$
    – Klaas van Aarsen
    Jan 16 at 20:22


















$begingroup$
The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=rcostheta, y=rsintheta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a ucos v, y=b usin v, z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2pi$ but is not the polar angle, and $u$ runs from $0$ to $1$.
$endgroup$
– Klaas van Aarsen
Jan 16 at 20:22






$begingroup$
The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=rcostheta, y=rsintheta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a ucos v, y=b usin v, z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2pi$ but is not the polar angle, and $u$ runs from $0$ to $1$.
$endgroup$
– Klaas van Aarsen
Jan 16 at 20:22












1 Answer
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$begingroup$

It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.



Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.



For a constant $z$ we have the ellipse:
$$frac{x^2}{frac{z^2+4}{2}}+frac{y^2}{z^2+4}=1$$
It has semi axes $a=sqrt{frac{z^2+4}{2}}$ and $b=sqrt{z^2+4}$.



Since the area of an ellipse is $pi a b$, the volume is:
$$text{Volume}=int_{-2}^2 pi a b,dz = int_{-2}^2 picdotfrac{z^2+4}{sqrt 2},dz$$






share|cite|improve this answer









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    $begingroup$

    It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.



    Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.



    For a constant $z$ we have the ellipse:
    $$frac{x^2}{frac{z^2+4}{2}}+frac{y^2}{z^2+4}=1$$
    It has semi axes $a=sqrt{frac{z^2+4}{2}}$ and $b=sqrt{z^2+4}$.



    Since the area of an ellipse is $pi a b$, the volume is:
    $$text{Volume}=int_{-2}^2 pi a b,dz = int_{-2}^2 picdotfrac{z^2+4}{sqrt 2},dz$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.



      Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.



      For a constant $z$ we have the ellipse:
      $$frac{x^2}{frac{z^2+4}{2}}+frac{y^2}{z^2+4}=1$$
      It has semi axes $a=sqrt{frac{z^2+4}{2}}$ and $b=sqrt{z^2+4}$.



      Since the area of an ellipse is $pi a b$, the volume is:
      $$text{Volume}=int_{-2}^2 pi a b,dz = int_{-2}^2 picdotfrac{z^2+4}{sqrt 2},dz$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.



        Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.



        For a constant $z$ we have the ellipse:
        $$frac{x^2}{frac{z^2+4}{2}}+frac{y^2}{z^2+4}=1$$
        It has semi axes $a=sqrt{frac{z^2+4}{2}}$ and $b=sqrt{z^2+4}$.



        Since the area of an ellipse is $pi a b$, the volume is:
        $$text{Volume}=int_{-2}^2 pi a b,dz = int_{-2}^2 picdotfrac{z^2+4}{sqrt 2},dz$$






        share|cite|improve this answer









        $endgroup$



        It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.



        Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.



        For a constant $z$ we have the ellipse:
        $$frac{x^2}{frac{z^2+4}{2}}+frac{y^2}{z^2+4}=1$$
        It has semi axes $a=sqrt{frac{z^2+4}{2}}$ and $b=sqrt{z^2+4}$.



        Since the area of an ellipse is $pi a b$, the volume is:
        $$text{Volume}=int_{-2}^2 pi a b,dz = int_{-2}^2 picdotfrac{z^2+4}{sqrt 2},dz$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 20:05









        Klaas van AarsenKlaas van Aarsen

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