volume of surface given by $(x^2+y^2+z^2)^2=x$












1












$begingroup$


A question asks me to find the volume of the surface $(x^2+y^2+z^2)^2=x$ this looks like a very difficult triple integral to evaluate using cartesian coordinates, so I though I would describe the set in spherical coordinates.
Doing so I got $p^2=sin(phi)cos(theta)$. But this still seems like a very difficult integral to evaluate. Am I doing something wrong, or am I completely missing something?






integration multivariable-calculus volume spherical-coordinates







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute.
    $endgroup$
    – Aaron
    Dec 11 '18 at 1:48












  • $begingroup$
    Can you explain how that is?
    $endgroup$
    – Skrrrrrtttt
    Dec 11 '18 at 2:10
















1












$begingroup$


A question asks me to find the volume of the surface $(x^2+y^2+z^2)^2=x$ this looks like a very difficult triple integral to evaluate using cartesian coordinates, so I though I would describe the set in spherical coordinates.
Doing so I got $p^2=sin(phi)cos(theta)$. But this still seems like a very difficult integral to evaluate. Am I doing something wrong, or am I completely missing something?






integration multivariable-calculus volume spherical-coordinates







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute.
    $endgroup$
    – Aaron
    Dec 11 '18 at 1:48












  • $begingroup$
    Can you explain how that is?
    $endgroup$
    – Skrrrrrtttt
    Dec 11 '18 at 2:10














1












1








1


1



$begingroup$


A question asks me to find the volume of the surface $(x^2+y^2+z^2)^2=x$ this looks like a very difficult triple integral to evaluate using cartesian coordinates, so I though I would describe the set in spherical coordinates.
Doing so I got $p^2=sin(phi)cos(theta)$. But this still seems like a very difficult integral to evaluate. Am I doing something wrong, or am I completely missing something?






integration multivariable-calculus volume spherical-coordinates







share|cite|improve this question









$endgroup$




A question asks me to find the volume of the surface $(x^2+y^2+z^2)^2=x$ this looks like a very difficult triple integral to evaluate using cartesian coordinates, so I though I would describe the set in spherical coordinates.
Doing so I got $p^2=sin(phi)cos(theta)$. But this still seems like a very difficult integral to evaluate. Am I doing something wrong, or am I completely missing something?






integration multivariable-calculus volume spherical-coordinates




integration multivariable-calculus volume spherical-coordinates






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 1:45









SkrrrrrttttSkrrrrrtttt

370110




370110












  • $begingroup$
    Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute.
    $endgroup$
    – Aaron
    Dec 11 '18 at 1:48












  • $begingroup$
    Can you explain how that is?
    $endgroup$
    – Skrrrrrtttt
    Dec 11 '18 at 2:10


















  • $begingroup$
    Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute.
    $endgroup$
    – Aaron
    Dec 11 '18 at 1:48












  • $begingroup$
    Can you explain how that is?
    $endgroup$
    – Skrrrrrtttt
    Dec 11 '18 at 2:10
















$begingroup$
Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute.
$endgroup$
– Aaron
Dec 11 '18 at 1:48






$begingroup$
Are you looking for the volume bounded by the surface, or the surface area? If you are looking for the volume, note that for a fixed value of $x$, $y^2+z^2$ will be a constant, and so you have cross sections which are circles whose radius you can compute.
$endgroup$
– Aaron
Dec 11 '18 at 1:48














$begingroup$
Can you explain how that is?
$endgroup$
– Skrrrrrtttt
Dec 11 '18 at 2:10




$begingroup$
Can you explain how that is?
$endgroup$
– Skrrrrrtttt
Dec 11 '18 at 2:10










3 Answers
3






active

oldest

votes


















2












$begingroup$

You can use polar coordinates to evaluate the integral. However, you should swap
the roles of $x,y,z$ directions to simplify the expressions.



In polar coordinates $(r,theta,phi) mapsto ( rcostheta, rsinthetacosphi, rsinthetasinphi)$ where $r ge 0$, $theta in [0,pi]$ and $phi in [0,2pi)$. The surface becomes
$$(x^2+y^2+z^2)^2 = x iff r^4 = rcostheta$$
Since $r ge 0$, the relevant range of $theta$ is $[0,pi/2]$ where $costheta ge 0$.
The volume we seek becomes



$$int_0^{2pi} int_0^{pi/2} left(int_0^{sqrt[3]{costheta}} r^2 drright) sintheta dtheta dphi
=int_0^{2pi}int_0^{pi/2} left[ frac13 r^3 right]_{r=0}^{sqrt[3]{costheta}} sintheta dtheta dphi\
= frac{2pi}{3}int_0^{pi/2} costhetasintheta dtheta
= frac{pi}{3}
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get
    $$r^2=sqrt x - x^2$$
    There’s no need for a plus or minus because everything is squared and therefore positive.
    Using the classic volume of a solid of revolution technique, we get
    $$pi int_0^1 r^2 dx = int_0^1 sqrt x - x^2 dx$$
    (We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Try $int_0^1 pi r(x)^2,dx$ where $r(x)$ is obtained from @Aaron's hint.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        You can use polar coordinates to evaluate the integral. However, you should swap
        the roles of $x,y,z$ directions to simplify the expressions.



        In polar coordinates $(r,theta,phi) mapsto ( rcostheta, rsinthetacosphi, rsinthetasinphi)$ where $r ge 0$, $theta in [0,pi]$ and $phi in [0,2pi)$. The surface becomes
        $$(x^2+y^2+z^2)^2 = x iff r^4 = rcostheta$$
        Since $r ge 0$, the relevant range of $theta$ is $[0,pi/2]$ where $costheta ge 0$.
        The volume we seek becomes



        $$int_0^{2pi} int_0^{pi/2} left(int_0^{sqrt[3]{costheta}} r^2 drright) sintheta dtheta dphi
        =int_0^{2pi}int_0^{pi/2} left[ frac13 r^3 right]_{r=0}^{sqrt[3]{costheta}} sintheta dtheta dphi\
        = frac{2pi}{3}int_0^{pi/2} costhetasintheta dtheta
        = frac{pi}{3}
        $$






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          You can use polar coordinates to evaluate the integral. However, you should swap
          the roles of $x,y,z$ directions to simplify the expressions.



          In polar coordinates $(r,theta,phi) mapsto ( rcostheta, rsinthetacosphi, rsinthetasinphi)$ where $r ge 0$, $theta in [0,pi]$ and $phi in [0,2pi)$. The surface becomes
          $$(x^2+y^2+z^2)^2 = x iff r^4 = rcostheta$$
          Since $r ge 0$, the relevant range of $theta$ is $[0,pi/2]$ where $costheta ge 0$.
          The volume we seek becomes



          $$int_0^{2pi} int_0^{pi/2} left(int_0^{sqrt[3]{costheta}} r^2 drright) sintheta dtheta dphi
          =int_0^{2pi}int_0^{pi/2} left[ frac13 r^3 right]_{r=0}^{sqrt[3]{costheta}} sintheta dtheta dphi\
          = frac{2pi}{3}int_0^{pi/2} costhetasintheta dtheta
          = frac{pi}{3}
          $$






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            You can use polar coordinates to evaluate the integral. However, you should swap
            the roles of $x,y,z$ directions to simplify the expressions.



            In polar coordinates $(r,theta,phi) mapsto ( rcostheta, rsinthetacosphi, rsinthetasinphi)$ where $r ge 0$, $theta in [0,pi]$ and $phi in [0,2pi)$. The surface becomes
            $$(x^2+y^2+z^2)^2 = x iff r^4 = rcostheta$$
            Since $r ge 0$, the relevant range of $theta$ is $[0,pi/2]$ where $costheta ge 0$.
            The volume we seek becomes



            $$int_0^{2pi} int_0^{pi/2} left(int_0^{sqrt[3]{costheta}} r^2 drright) sintheta dtheta dphi
            =int_0^{2pi}int_0^{pi/2} left[ frac13 r^3 right]_{r=0}^{sqrt[3]{costheta}} sintheta dtheta dphi\
            = frac{2pi}{3}int_0^{pi/2} costhetasintheta dtheta
            = frac{pi}{3}
            $$






            share|cite|improve this answer











            $endgroup$



            You can use polar coordinates to evaluate the integral. However, you should swap
            the roles of $x,y,z$ directions to simplify the expressions.



            In polar coordinates $(r,theta,phi) mapsto ( rcostheta, rsinthetacosphi, rsinthetasinphi)$ where $r ge 0$, $theta in [0,pi]$ and $phi in [0,2pi)$. The surface becomes
            $$(x^2+y^2+z^2)^2 = x iff r^4 = rcostheta$$
            Since $r ge 0$, the relevant range of $theta$ is $[0,pi/2]$ where $costheta ge 0$.
            The volume we seek becomes



            $$int_0^{2pi} int_0^{pi/2} left(int_0^{sqrt[3]{costheta}} r^2 drright) sintheta dtheta dphi
            =int_0^{2pi}int_0^{pi/2} left[ frac13 r^3 right]_{r=0}^{sqrt[3]{costheta}} sintheta dtheta dphi\
            = frac{2pi}{3}int_0^{pi/2} costhetasintheta dtheta
            = frac{pi}{3}
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 22 '18 at 23:15









            Poujh

            616516




            616516










            answered Dec 11 '18 at 4:04









            achille huiachille hui

            96k5132258




            96k5132258























                1












                $begingroup$

                For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get
                $$r^2=sqrt x - x^2$$
                There’s no need for a plus or minus because everything is squared and therefore positive.
                Using the classic volume of a solid of revolution technique, we get
                $$pi int_0^1 r^2 dx = int_0^1 sqrt x - x^2 dx$$
                (We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get
                  $$r^2=sqrt x - x^2$$
                  There’s no need for a plus or minus because everything is squared and therefore positive.
                  Using the classic volume of a solid of revolution technique, we get
                  $$pi int_0^1 r^2 dx = int_0^1 sqrt x - x^2 dx$$
                  (We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get
                    $$r^2=sqrt x - x^2$$
                    There’s no need for a plus or minus because everything is squared and therefore positive.
                    Using the classic volume of a solid of revolution technique, we get
                    $$pi int_0^1 r^2 dx = int_0^1 sqrt x - x^2 dx$$
                    (We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)






                    share|cite|improve this answer









                    $endgroup$



                    For every x, you have a constant $y^2+z^2$. That means for every x we can make a cross section which looks like a circle on a plane parallel to the yz-plane, centered at that x value on the x axis with radius r such that $r^2=y^2+z^2$. If we take the square root in your expression, we get
                    $$r^2=sqrt x - x^2$$
                    There’s no need for a plus or minus because everything is squared and therefore positive.
                    Using the classic volume of a solid of revolution technique, we get
                    $$pi int_0^1 r^2 dx = int_0^1 sqrt x - x^2 dx$$
                    (We know bound is $x=1$ because for $x>1$, the left hand side is necessarily larger than the right hand side, because the left consists of $x^4+$non negative terms, while the right hand side is just $x$)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 11 '18 at 2:36









                    D.R.D.R.

                    1,465720




                    1,465720























                        0












                        $begingroup$

                        Try $int_0^1 pi r(x)^2,dx$ where $r(x)$ is obtained from @Aaron's hint.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Try $int_0^1 pi r(x)^2,dx$ where $r(x)$ is obtained from @Aaron's hint.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Try $int_0^1 pi r(x)^2,dx$ where $r(x)$ is obtained from @Aaron's hint.






                            share|cite|improve this answer









                            $endgroup$



                            Try $int_0^1 pi r(x)^2,dx$ where $r(x)$ is obtained from @Aaron's hint.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 11 '18 at 2:05









                            Bjørn Kjos-HanssenBjørn Kjos-Hanssen

                            2,096918




                            2,096918






























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