Having trouble setting up the triple integral inside the region(sphere) $;x^2+y^2+z^2 = 4$












0












$begingroup$


I'm having trouble setting up the bounds and the integral. The question asked to find the mass of the 3D region $ x^2 +y^2 + z^2 ≤ 4,x ≥ 0, y≥ 0, z ≥ 0 $ if the density is equal to xyz. I don't want to evaluate it, I just want to clear some few things so I may never have trouble with such problems again.




  1. Can this question be done with Polar coordinates? If yes, then why and what will be the limits?


  2. In case of rectangular coordinates, how would the limits be set up and what would be the integrand?



My attempt:



To find x and y intercepts, taking y and z = 0, so we have $x = 2$, $-2$. Are these the limits of x? I don't think that -2 will be the lower limit because the region specifies that x ≥ 0 so the limits will be [0,2]. Kindly confirm.



Using the same logic for y and making y depend on x, the limits will be $ y = [0,sqrt{4-x^2}] $ disregarding the out of bound limit $-sqrt{4-x^2} $ . Is this a correct assumption?



Finally the limits for z will be $[0, sqrt{4-x^2-y^2}]$



Now I'm not sure how to set up the integral but I would assume that it would be in the order of dzdydx with the integrand being 1 but coming back to the question, I have integrate xyz over the region to calculate the mass. Thank you in advance.






integration multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is (by far) most easily done in spherical coordinates.
    $endgroup$
    – T. Bongers
    Nov 27 '18 at 18:47










  • $begingroup$
    You should edit your title so that it makes sense. The region is not $x^2+y^2+z^2$. It's the inside of the sphere $x^2+y^2+z^2=4$. :)
    $endgroup$
    – Ted Shifrin
    Nov 27 '18 at 18:49










  • $begingroup$
    Computing such integrals in Cartesian coordinates is usually a bad idea. Spherical coordinates work wonders in cases like these because they greatly simplify expressions such as $x^2+y^2+z^2$. What you're essentially doing when using spherical coordinates is just a change of coordinates, which is valid since every point of $(x,y,z)inmathbb{R}^3$ in Cartesian coordinates is uniquely determined by the triplet $(r, theta, phi)$ in spherical coordinates.
    $endgroup$
    – MisterRiemann
    Nov 27 '18 at 18:50












  • $begingroup$
    @TedShifrin edited, now kindly clear my doubts that I have mentioned in my attempt. Thank you
    $endgroup$
    – tNotr
    Nov 27 '18 at 18:58










  • $begingroup$
    @MisterRiemann so regions having equations involving the square of 3 variables can be always solved by spherical coordinates, ok but what if the question has the region lying under the sphere, above the z plane and inside the cylinder $x^2 + y^2 = 5 $ ? How would you set and evaluate the integral then> I have intentionally left out numbers as I just want to get the idea. Thanks
    $endgroup$
    – tNotr
    Nov 27 '18 at 19:04
















0












$begingroup$


I'm having trouble setting up the bounds and the integral. The question asked to find the mass of the 3D region $ x^2 +y^2 + z^2 ≤ 4,x ≥ 0, y≥ 0, z ≥ 0 $ if the density is equal to xyz. I don't want to evaluate it, I just want to clear some few things so I may never have trouble with such problems again.




  1. Can this question be done with Polar coordinates? If yes, then why and what will be the limits?


  2. In case of rectangular coordinates, how would the limits be set up and what would be the integrand?



My attempt:



To find x and y intercepts, taking y and z = 0, so we have $x = 2$, $-2$. Are these the limits of x? I don't think that -2 will be the lower limit because the region specifies that x ≥ 0 so the limits will be [0,2]. Kindly confirm.



Using the same logic for y and making y depend on x, the limits will be $ y = [0,sqrt{4-x^2}] $ disregarding the out of bound limit $-sqrt{4-x^2} $ . Is this a correct assumption?



Finally the limits for z will be $[0, sqrt{4-x^2-y^2}]$



Now I'm not sure how to set up the integral but I would assume that it would be in the order of dzdydx with the integrand being 1 but coming back to the question, I have integrate xyz over the region to calculate the mass. Thank you in advance.






integration multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is (by far) most easily done in spherical coordinates.
    $endgroup$
    – T. Bongers
    Nov 27 '18 at 18:47










  • $begingroup$
    You should edit your title so that it makes sense. The region is not $x^2+y^2+z^2$. It's the inside of the sphere $x^2+y^2+z^2=4$. :)
    $endgroup$
    – Ted Shifrin
    Nov 27 '18 at 18:49










  • $begingroup$
    Computing such integrals in Cartesian coordinates is usually a bad idea. Spherical coordinates work wonders in cases like these because they greatly simplify expressions such as $x^2+y^2+z^2$. What you're essentially doing when using spherical coordinates is just a change of coordinates, which is valid since every point of $(x,y,z)inmathbb{R}^3$ in Cartesian coordinates is uniquely determined by the triplet $(r, theta, phi)$ in spherical coordinates.
    $endgroup$
    – MisterRiemann
    Nov 27 '18 at 18:50












  • $begingroup$
    @TedShifrin edited, now kindly clear my doubts that I have mentioned in my attempt. Thank you
    $endgroup$
    – tNotr
    Nov 27 '18 at 18:58










  • $begingroup$
    @MisterRiemann so regions having equations involving the square of 3 variables can be always solved by spherical coordinates, ok but what if the question has the region lying under the sphere, above the z plane and inside the cylinder $x^2 + y^2 = 5 $ ? How would you set and evaluate the integral then> I have intentionally left out numbers as I just want to get the idea. Thanks
    $endgroup$
    – tNotr
    Nov 27 '18 at 19:04














0












0








0





$begingroup$


I'm having trouble setting up the bounds and the integral. The question asked to find the mass of the 3D region $ x^2 +y^2 + z^2 ≤ 4,x ≥ 0, y≥ 0, z ≥ 0 $ if the density is equal to xyz. I don't want to evaluate it, I just want to clear some few things so I may never have trouble with such problems again.




  1. Can this question be done with Polar coordinates? If yes, then why and what will be the limits?


  2. In case of rectangular coordinates, how would the limits be set up and what would be the integrand?



My attempt:



To find x and y intercepts, taking y and z = 0, so we have $x = 2$, $-2$. Are these the limits of x? I don't think that -2 will be the lower limit because the region specifies that x ≥ 0 so the limits will be [0,2]. Kindly confirm.



Using the same logic for y and making y depend on x, the limits will be $ y = [0,sqrt{4-x^2}] $ disregarding the out of bound limit $-sqrt{4-x^2} $ . Is this a correct assumption?



Finally the limits for z will be $[0, sqrt{4-x^2-y^2}]$



Now I'm not sure how to set up the integral but I would assume that it would be in the order of dzdydx with the integrand being 1 but coming back to the question, I have integrate xyz over the region to calculate the mass. Thank you in advance.






integration multivariable-calculus







share|cite|improve this question











$endgroup$




I'm having trouble setting up the bounds and the integral. The question asked to find the mass of the 3D region $ x^2 +y^2 + z^2 ≤ 4,x ≥ 0, y≥ 0, z ≥ 0 $ if the density is equal to xyz. I don't want to evaluate it, I just want to clear some few things so I may never have trouble with such problems again.




  1. Can this question be done with Polar coordinates? If yes, then why and what will be the limits?


  2. In case of rectangular coordinates, how would the limits be set up and what would be the integrand?



My attempt:



To find x and y intercepts, taking y and z = 0, so we have $x = 2$, $-2$. Are these the limits of x? I don't think that -2 will be the lower limit because the region specifies that x ≥ 0 so the limits will be [0,2]. Kindly confirm.



Using the same logic for y and making y depend on x, the limits will be $ y = [0,sqrt{4-x^2}] $ disregarding the out of bound limit $-sqrt{4-x^2} $ . Is this a correct assumption?



Finally the limits for z will be $[0, sqrt{4-x^2-y^2}]$



Now I'm not sure how to set up the integral but I would assume that it would be in the order of dzdydx with the integrand being 1 but coming back to the question, I have integrate xyz over the region to calculate the mass. Thank you in advance.






integration multivariable-calculus




integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 19:17









user376343

3,4033826




3,4033826










asked Nov 27 '18 at 18:46









tNotrtNotr

142




142








  • 1




    $begingroup$
    This is (by far) most easily done in spherical coordinates.
    $endgroup$
    – T. Bongers
    Nov 27 '18 at 18:47










  • $begingroup$
    You should edit your title so that it makes sense. The region is not $x^2+y^2+z^2$. It's the inside of the sphere $x^2+y^2+z^2=4$. :)
    $endgroup$
    – Ted Shifrin
    Nov 27 '18 at 18:49










  • $begingroup$
    Computing such integrals in Cartesian coordinates is usually a bad idea. Spherical coordinates work wonders in cases like these because they greatly simplify expressions such as $x^2+y^2+z^2$. What you're essentially doing when using spherical coordinates is just a change of coordinates, which is valid since every point of $(x,y,z)inmathbb{R}^3$ in Cartesian coordinates is uniquely determined by the triplet $(r, theta, phi)$ in spherical coordinates.
    $endgroup$
    – MisterRiemann
    Nov 27 '18 at 18:50












  • $begingroup$
    @TedShifrin edited, now kindly clear my doubts that I have mentioned in my attempt. Thank you
    $endgroup$
    – tNotr
    Nov 27 '18 at 18:58










  • $begingroup$
    @MisterRiemann so regions having equations involving the square of 3 variables can be always solved by spherical coordinates, ok but what if the question has the region lying under the sphere, above the z plane and inside the cylinder $x^2 + y^2 = 5 $ ? How would you set and evaluate the integral then> I have intentionally left out numbers as I just want to get the idea. Thanks
    $endgroup$
    – tNotr
    Nov 27 '18 at 19:04














  • 1




    $begingroup$
    This is (by far) most easily done in spherical coordinates.
    $endgroup$
    – T. Bongers
    Nov 27 '18 at 18:47










  • $begingroup$
    You should edit your title so that it makes sense. The region is not $x^2+y^2+z^2$. It's the inside of the sphere $x^2+y^2+z^2=4$. :)
    $endgroup$
    – Ted Shifrin
    Nov 27 '18 at 18:49










  • $begingroup$
    Computing such integrals in Cartesian coordinates is usually a bad idea. Spherical coordinates work wonders in cases like these because they greatly simplify expressions such as $x^2+y^2+z^2$. What you're essentially doing when using spherical coordinates is just a change of coordinates, which is valid since every point of $(x,y,z)inmathbb{R}^3$ in Cartesian coordinates is uniquely determined by the triplet $(r, theta, phi)$ in spherical coordinates.
    $endgroup$
    – MisterRiemann
    Nov 27 '18 at 18:50












  • $begingroup$
    @TedShifrin edited, now kindly clear my doubts that I have mentioned in my attempt. Thank you
    $endgroup$
    – tNotr
    Nov 27 '18 at 18:58










  • $begingroup$
    @MisterRiemann so regions having equations involving the square of 3 variables can be always solved by spherical coordinates, ok but what if the question has the region lying under the sphere, above the z plane and inside the cylinder $x^2 + y^2 = 5 $ ? How would you set and evaluate the integral then> I have intentionally left out numbers as I just want to get the idea. Thanks
    $endgroup$
    – tNotr
    Nov 27 '18 at 19:04








1




1




$begingroup$
This is (by far) most easily done in spherical coordinates.
$endgroup$
– T. Bongers
Nov 27 '18 at 18:47




$begingroup$
This is (by far) most easily done in spherical coordinates.
$endgroup$
– T. Bongers
Nov 27 '18 at 18:47












$begingroup$
You should edit your title so that it makes sense. The region is not $x^2+y^2+z^2$. It's the inside of the sphere $x^2+y^2+z^2=4$. :)
$endgroup$
– Ted Shifrin
Nov 27 '18 at 18:49




$begingroup$
You should edit your title so that it makes sense. The region is not $x^2+y^2+z^2$. It's the inside of the sphere $x^2+y^2+z^2=4$. :)
$endgroup$
– Ted Shifrin
Nov 27 '18 at 18:49












$begingroup$
Computing such integrals in Cartesian coordinates is usually a bad idea. Spherical coordinates work wonders in cases like these because they greatly simplify expressions such as $x^2+y^2+z^2$. What you're essentially doing when using spherical coordinates is just a change of coordinates, which is valid since every point of $(x,y,z)inmathbb{R}^3$ in Cartesian coordinates is uniquely determined by the triplet $(r, theta, phi)$ in spherical coordinates.
$endgroup$
– MisterRiemann
Nov 27 '18 at 18:50






$begingroup$
Computing such integrals in Cartesian coordinates is usually a bad idea. Spherical coordinates work wonders in cases like these because they greatly simplify expressions such as $x^2+y^2+z^2$. What you're essentially doing when using spherical coordinates is just a change of coordinates, which is valid since every point of $(x,y,z)inmathbb{R}^3$ in Cartesian coordinates is uniquely determined by the triplet $(r, theta, phi)$ in spherical coordinates.
$endgroup$
– MisterRiemann
Nov 27 '18 at 18:50














$begingroup$
@TedShifrin edited, now kindly clear my doubts that I have mentioned in my attempt. Thank you
$endgroup$
– tNotr
Nov 27 '18 at 18:58




$begingroup$
@TedShifrin edited, now kindly clear my doubts that I have mentioned in my attempt. Thank you
$endgroup$
– tNotr
Nov 27 '18 at 18:58












$begingroup$
@MisterRiemann so regions having equations involving the square of 3 variables can be always solved by spherical coordinates, ok but what if the question has the region lying under the sphere, above the z plane and inside the cylinder $x^2 + y^2 = 5 $ ? How would you set and evaluate the integral then> I have intentionally left out numbers as I just want to get the idea. Thanks
$endgroup$
– tNotr
Nov 27 '18 at 19:04




$begingroup$
@MisterRiemann so regions having equations involving the square of 3 variables can be always solved by spherical coordinates, ok but what if the question has the region lying under the sphere, above the z plane and inside the cylinder $x^2 + y^2 = 5 $ ? How would you set and evaluate the integral then> I have intentionally left out numbers as I just want to get the idea. Thanks
$endgroup$
– tNotr
Nov 27 '18 at 19:04










1 Answer
1






active

oldest

votes


















0












$begingroup$

Since $x=rsinthetacosvarphi,,y=rsinthetasinvarphi,,z=rcostheta$ has Jacobian $r^2sin theta$, an infinitesimal region has mass $xyz dxdydz=r^5 drsin^3thetacostheta dthetasinvarphicosvarphi dvarphi$. We integrate $r$ from $0$ to $2$, $theta$ from $0$ to $pi/2$ and $varphi$ from $0$ to $pi/2$, obtaining (if my arithmetic's right) $$int_0^2 r^5 drint_0^{pi/2}sin^3thetacostheta dthetaint_0^{pi/2}frac{1}{2}sin 2varphi dvarphi=frac{2^6}{6}frac{1}{4}frac{1}{4}=frac{2}{3}.$$






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%2f3016141%2fhaving-trouble-setting-up-the-triple-integral-inside-the-regionsphere-x2y%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









    0












    $begingroup$

    Since $x=rsinthetacosvarphi,,y=rsinthetasinvarphi,,z=rcostheta$ has Jacobian $r^2sin theta$, an infinitesimal region has mass $xyz dxdydz=r^5 drsin^3thetacostheta dthetasinvarphicosvarphi dvarphi$. We integrate $r$ from $0$ to $2$, $theta$ from $0$ to $pi/2$ and $varphi$ from $0$ to $pi/2$, obtaining (if my arithmetic's right) $$int_0^2 r^5 drint_0^{pi/2}sin^3thetacostheta dthetaint_0^{pi/2}frac{1}{2}sin 2varphi dvarphi=frac{2^6}{6}frac{1}{4}frac{1}{4}=frac{2}{3}.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since $x=rsinthetacosvarphi,,y=rsinthetasinvarphi,,z=rcostheta$ has Jacobian $r^2sin theta$, an infinitesimal region has mass $xyz dxdydz=r^5 drsin^3thetacostheta dthetasinvarphicosvarphi dvarphi$. We integrate $r$ from $0$ to $2$, $theta$ from $0$ to $pi/2$ and $varphi$ from $0$ to $pi/2$, obtaining (if my arithmetic's right) $$int_0^2 r^5 drint_0^{pi/2}sin^3thetacostheta dthetaint_0^{pi/2}frac{1}{2}sin 2varphi dvarphi=frac{2^6}{6}frac{1}{4}frac{1}{4}=frac{2}{3}.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since $x=rsinthetacosvarphi,,y=rsinthetasinvarphi,,z=rcostheta$ has Jacobian $r^2sin theta$, an infinitesimal region has mass $xyz dxdydz=r^5 drsin^3thetacostheta dthetasinvarphicosvarphi dvarphi$. We integrate $r$ from $0$ to $2$, $theta$ from $0$ to $pi/2$ and $varphi$ from $0$ to $pi/2$, obtaining (if my arithmetic's right) $$int_0^2 r^5 drint_0^{pi/2}sin^3thetacostheta dthetaint_0^{pi/2}frac{1}{2}sin 2varphi dvarphi=frac{2^6}{6}frac{1}{4}frac{1}{4}=frac{2}{3}.$$






        share|cite|improve this answer









        $endgroup$



        Since $x=rsinthetacosvarphi,,y=rsinthetasinvarphi,,z=rcostheta$ has Jacobian $r^2sin theta$, an infinitesimal region has mass $xyz dxdydz=r^5 drsin^3thetacostheta dthetasinvarphicosvarphi dvarphi$. We integrate $r$ from $0$ to $2$, $theta$ from $0$ to $pi/2$ and $varphi$ from $0$ to $pi/2$, obtaining (if my arithmetic's right) $$int_0^2 r^5 drint_0^{pi/2}sin^3thetacostheta dthetaint_0^{pi/2}frac{1}{2}sin 2varphi dvarphi=frac{2^6}{6}frac{1}{4}frac{1}{4}=frac{2}{3}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 19:06









        J.G.J.G.

        25.2k22539




        25.2k22539






























            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%2f3016141%2fhaving-trouble-setting-up-the-triple-integral-inside-the-regionsphere-x2y%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more