Volume inside sphere bounded by plane in double integrals











up vote
0
down vote

favorite












I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.






integration multivariable-calculus polar-coordinates







share|cite|improve this question
























  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 at 8:36















up vote
0
down vote

favorite












I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.






integration multivariable-calculus polar-coordinates







share|cite|improve this question
























  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 at 8:36













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.






integration multivariable-calculus polar-coordinates







share|cite|improve this question















I am trying to solve for volume below a plane bounded by a sphere given by



$$x^2+y^2+z^2 = 9$$ below a plane z $in$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be



$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!



EDIT:
I ended up taking the volume $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.






integration multivariable-calculus polar-coordinates




integration multivariable-calculus polar-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 8:41

























asked Nov 20 at 15:19









mstro

11




11












  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 at 8:36


















  • Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
    – Andrei
    Nov 20 at 16:27










  • I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
    – mstro
    Nov 22 at 8:36
















Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27




Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$.
– Andrei
Nov 20 at 16:27












I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36




I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere!
– mstro
Nov 22 at 8:36










1 Answer
1






active

oldest

votes

















up vote
0
down vote













The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
$$frac{1}{3}pi(9-c^2)c$$
The second is the volume between the cone and z = 0 which can be obtained by
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






share|cite|improve this answer





















    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',
    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%2f3006445%2fvolume-inside-sphere-bounded-by-plane-in-double-integrals%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
    Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
    It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
    $$frac{1}{3}pi(9-c^2)c$$
    The second is the volume between the cone and z = 0 which can be obtained by
    $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
    Then the total volume is
    $$pi(9c - frac{c^3}{3})$$
    In fact it is easier to use xyz co-ordinates.
    $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






    share|cite|improve this answer

























      up vote
      0
      down vote













      The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
      Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
      It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
      $$frac{1}{3}pi(9-c^2)c$$
      The second is the volume between the cone and z = 0 which can be obtained by
      $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
      Then the total volume is
      $$pi(9c - frac{c^3}{3})$$
      In fact it is easier to use xyz co-ordinates.
      $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
        Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
        It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
        $$frac{1}{3}pi(9-c^2)c$$
        The second is the volume between the cone and z = 0 which can be obtained by
        $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
        Then the total volume is
        $$pi(9c - frac{c^3}{3})$$
        In fact it is easier to use xyz co-ordinates.
        $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$






        share|cite|improve this answer












        The following is to calculate the volume of a sphere $x^2+y^2+z^2=9$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $rsin{phi} = c$.
        Then the intersection of the plane and the sphere is $sin{phi} = frac{c}{r}$
        It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $sqrt{9-c^2}$. The volume is
        $$frac{1}{3}pi(9-c^2)c$$
        The second is the volume between the cone and z = 0 which can be obtained by
        $$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
        Then the total volume is
        $$pi(9c - frac{c^3}{3})$$
        In fact it is easier to use xyz co-ordinates.
        $$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 22:47









        KY Tang

        113




        113






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3006445%2fvolume-inside-sphere-bounded-by-plane-in-double-integrals%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