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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034761%2fvolume-of-surface-given-by-x2y2z22-x%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034761%2fvolume-of-surface-given-by-x2y2z22-x%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Plaza Victoria

                                Image
                                Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                                Read more

                                Puebla de Zaragoza

                                Image
                                Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
                                Read more

                                Musa

                                Image
                                Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
                                Read more