Error in calculation of intersection of three cylinders












0












$begingroup$


Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



I want to find the volume. I have to use symmetry and without polar coordinates.



My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



$lambda^{d}(A)=int_{A}dxdydz$



And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






real-analysis integration measure-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



    I want to find the volume. I have to use symmetry and without polar coordinates.



    My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



    $lambda^{d}(A)=int_{A}dxdydz$



    And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



    So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



    And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



    So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






    real-analysis integration measure-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



      I want to find the volume. I have to use symmetry and without polar coordinates.



      My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



      $lambda^{d}(A)=int_{A}dxdydz$



      And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



      So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



      And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



      So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






      real-analysis integration measure-theory







      share|cite|improve this question









      $endgroup$




      Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



      I want to find the volume. I have to use symmetry and without polar coordinates.



      My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



      $lambda^{d}(A)=int_{A}dxdydz$



      And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



      So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



      And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



      So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






      real-analysis integration measure-theory




      real-analysis integration measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 19 at 20:14









      SABOYSABOY

      656311




      656311






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
          $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
          hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



          $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
          i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






          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%2f3079769%2ferror-in-calculation-of-intersection-of-three-cylinders%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












            $begingroup$

            $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
            $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
            hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



            $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
            i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
              $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
              hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



              $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
              i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
                $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
                hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



                $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
                i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






                share|cite|improve this answer









                $endgroup$



                $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
                $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
                hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



                $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
                i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 at 22:20









                Jack D'AurizioJack D'Aurizio

                290k33283664




                290k33283664






























                    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%2f3079769%2ferror-in-calculation-of-intersection-of-three-cylinders%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

                    'app-layout' is not a known element: how to share Component with different Modules

                    Image
                    0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                    Read more

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                    Read more

                    WPF add header to Image with URL pettitions [duplicate]

                    Image
                    up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                    Read more