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












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$begingroup$


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







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    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






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      asked 1 hour ago









      Hugh N.Hugh N.

      404




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          2 Answers
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          $begingroup$


          “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









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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

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              active

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              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






























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