Integral of multivariable function












1












$begingroup$


I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.



I am suppose to create Integral definition from it.



Now i am not sure if u understood correctly how to read the boundaries.



The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).



However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.



That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



However how to decide a ,b and c? I am having trouble deciding this trivial problem..



The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:



$$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



However this is quite simple case so i am not sure how to proceed or think about it in more complex cases



Thanks for help!






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.



    I am suppose to create Integral definition from it.



    Now i am not sure if u understood correctly how to read the boundaries.



    The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).



    However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.



    That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



    However how to decide a ,b and c? I am having trouble deciding this trivial problem..



    The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:



    $$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



    However this is quite simple case so i am not sure how to proceed or think about it in more complex cases



    Thanks for help!






    calculus integration multivariable-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.



      I am suppose to create Integral definition from it.



      Now i am not sure if u understood correctly how to read the boundaries.



      The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).



      However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.



      That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



      However how to decide a ,b and c? I am having trouble deciding this trivial problem..



      The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:



      $$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



      However this is quite simple case so i am not sure how to proceed or think about it in more complex cases



      Thanks for help!






      calculus integration multivariable-calculus







      share|cite|improve this question











      $endgroup$




      I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.



      I am suppose to create Integral definition from it.



      Now i am not sure if u understood correctly how to read the boundaries.



      The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).



      However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.



      That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



      However how to decide a ,b and c? I am having trouble deciding this trivial problem..



      The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:



      $$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$



      However this is quite simple case so i am not sure how to proceed or think about it in more complex cases



      Thanks for help!






      calculus integration multivariable-calculus




      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 Dec 12 '18 at 16:22









      gt6989b

      35.3k22557




      35.3k22557










      asked Dec 12 '18 at 16:17









      trolkuratrolkura

      24519




      24519






















          3 Answers
          3






          active

          oldest

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          2












          $begingroup$

          Note that it can be useful to switch to radial coordinates:
          $$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$



          Edit
          correction after comment below






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
            $endgroup$
            – Hans Lundmark
            Dec 12 '18 at 16:34



















          1












          $begingroup$

          In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.



          So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Here is how I think about it.



            First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.



            enter image description here



            Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.



            In this case, it doesn't make a difference. So, lets integrate by $y$ first.



            We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$



            If we were integrating by $x$ first it would be a horizontal line.



            What will be the formula for those points?



            $-sqrt {4-x^2}, sqrt {4-x^2}$



            Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.



            How far can $x$ rage before you bust the limits of your region?



            $int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$



            When you have more experience, you can trust your intuition. Until then, don't.



            Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Note that it can be useful to switch to radial coordinates:
              $$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$



              Edit
              correction after comment below






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
                $endgroup$
                – Hans Lundmark
                Dec 12 '18 at 16:34
















              2












              $begingroup$

              Note that it can be useful to switch to radial coordinates:
              $$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$



              Edit
              correction after comment below






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
                $endgroup$
                – Hans Lundmark
                Dec 12 '18 at 16:34














              2












              2








              2





              $begingroup$

              Note that it can be useful to switch to radial coordinates:
              $$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$



              Edit
              correction after comment below






              share|cite|improve this answer











              $endgroup$



              Note that it can be useful to switch to radial coordinates:
              $$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$



              Edit
              correction after comment below







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 12 '18 at 16:40

























              answered Dec 12 '18 at 16:31









              SebapiSebapi

              515




              515








              • 1




                $begingroup$
                You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
                $endgroup$
                – Hans Lundmark
                Dec 12 '18 at 16:34














              • 1




                $begingroup$
                You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
                $endgroup$
                – Hans Lundmark
                Dec 12 '18 at 16:34








              1




              1




              $begingroup$
              You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
              $endgroup$
              – Hans Lundmark
              Dec 12 '18 at 16:34




              $begingroup$
              You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
              $endgroup$
              – Hans Lundmark
              Dec 12 '18 at 16:34











              1












              $begingroup$

              In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.



              So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.



                So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.



                  So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$






                  share|cite|improve this answer











                  $endgroup$



                  In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.



                  So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 18:13

























                  answered Dec 12 '18 at 16:21









                  José Carlos SantosJosé Carlos Santos

                  172k22132239




                  172k22132239























                      1












                      $begingroup$

                      Here is how I think about it.



                      First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.



                      enter image description here



                      Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.



                      In this case, it doesn't make a difference. So, lets integrate by $y$ first.



                      We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$



                      If we were integrating by $x$ first it would be a horizontal line.



                      What will be the formula for those points?



                      $-sqrt {4-x^2}, sqrt {4-x^2}$



                      Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.



                      How far can $x$ rage before you bust the limits of your region?



                      $int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$



                      When you have more experience, you can trust your intuition. Until then, don't.



                      Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Here is how I think about it.



                        First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.



                        enter image description here



                        Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.



                        In this case, it doesn't make a difference. So, lets integrate by $y$ first.



                        We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$



                        If we were integrating by $x$ first it would be a horizontal line.



                        What will be the formula for those points?



                        $-sqrt {4-x^2}, sqrt {4-x^2}$



                        Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.



                        How far can $x$ rage before you bust the limits of your region?



                        $int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$



                        When you have more experience, you can trust your intuition. Until then, don't.



                        Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Here is how I think about it.



                          First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.



                          enter image description here



                          Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.



                          In this case, it doesn't make a difference. So, lets integrate by $y$ first.



                          We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$



                          If we were integrating by $x$ first it would be a horizontal line.



                          What will be the formula for those points?



                          $-sqrt {4-x^2}, sqrt {4-x^2}$



                          Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.



                          How far can $x$ rage before you bust the limits of your region?



                          $int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$



                          When you have more experience, you can trust your intuition. Until then, don't.



                          Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.






                          share|cite|improve this answer









                          $endgroup$



                          Here is how I think about it.



                          First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.



                          enter image description here



                          Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.



                          In this case, it doesn't make a difference. So, lets integrate by $y$ first.



                          We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$



                          If we were integrating by $x$ first it would be a horizontal line.



                          What will be the formula for those points?



                          $-sqrt {4-x^2}, sqrt {4-x^2}$



                          Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.



                          How far can $x$ rage before you bust the limits of your region?



                          $int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$



                          When you have more experience, you can trust your intuition. Until then, don't.



                          Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 12 '18 at 18:46









                          Doug MDoug M

                          45.3k31954




                          45.3k31954






























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