Looking for where I went wrong: Finding the volume of a solid that lies within both a cylinder and sphere












4












$begingroup$


I'm currently working on this question:



Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.



I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.



Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:



$int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$



However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?



enter image description here






calculus integration multivariable-calculus volume multiple-integral







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    I'm currently working on this question:



    Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.



    I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.



    Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:



    $int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$



    However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?



    enter image description here






    calculus integration multivariable-calculus volume multiple-integral







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      I'm currently working on this question:



      Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.



      I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.



      Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:



      $int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$



      However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?



      enter image description here






      calculus integration multivariable-calculus volume multiple-integral







      share|cite|improve this question









      $endgroup$




      I'm currently working on this question:



      Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.



      I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.



      Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:



      $int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$



      However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?



      enter image description here






      calculus integration multivariable-calculus volume multiple-integral




      calculus integration multivariable-calculus volume multiple-integral






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      Hugh N.Hugh N.

      404




      404






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$


          “Isn’t $r^2$ always 1?”




          Yes, on the surface of the cylinder.



          But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with



          $$rin[0,1]$$
          $$thetain[0,2pi)$$
          $$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.



            The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.






            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%2f3112329%2flooking-for-where-i-went-wrong-finding-the-volume-of-a-solid-that-lies-within-b%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$


              “Isn’t $r^2$ always 1?”




              Yes, on the surface of the cylinder.



              But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with



              $$rin[0,1]$$
              $$thetain[0,2pi)$$
              $$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$


                “Isn’t $r^2$ always 1?”




                Yes, on the surface of the cylinder.



                But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with



                $$rin[0,1]$$
                $$thetain[0,2pi)$$
                $$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$


                  “Isn’t $r^2$ always 1?”




                  Yes, on the surface of the cylinder.



                  But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with



                  $$rin[0,1]$$
                  $$thetain[0,2pi)$$
                  $$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$






                  share|cite|improve this answer









                  $endgroup$




                  “Isn’t $r^2$ always 1?”




                  Yes, on the surface of the cylinder.



                  But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with



                  $$rin[0,1]$$
                  $$thetain[0,2pi)$$
                  $$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  symplectomorphicsymplectomorphic

                  12.4k22039




                  12.4k22039























                      3












                      $begingroup$

                      No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.



                      The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.



                        The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.



                          The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.






                          share|cite|improve this answer









                          $endgroup$



                          No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.



                          The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          jmerryjmerry

                          8,7581123




                          8,7581123






























                              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%2f3112329%2flooking-for-where-i-went-wrong-finding-the-volume-of-a-solid-that-lies-within-b%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

                              Knooppunt Holsloot

                              Image
                              Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                              Read more

                              Altaar (religie)

                              Image
                              Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                              Read more

                              Gregoriusmis

                              Image
                              Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                              Read more