Order of integration in triple integral












1












$begingroup$


Is there any hard and fast rule for what order you integrate for triple integrals. I know of Fubini's theorem but surely this doesn't cover all cases of triple integrals.



Say for example I have,



$$int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta dz dr $$



Why is it that I can integrate in this order as the first limit's are not a function of one of the variables the second are a function of $r$ and the last of no variable again, how would I ever know that this is the order I can integrate in apart from just inspecting.






calculus integration definite-integrals order-of-integration







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$endgroup$

















    1












    $begingroup$


    Is there any hard and fast rule for what order you integrate for triple integrals. I know of Fubini's theorem but surely this doesn't cover all cases of triple integrals.



    Say for example I have,



    $$int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta dz dr $$



    Why is it that I can integrate in this order as the first limit's are not a function of one of the variables the second are a function of $r$ and the last of no variable again, how would I ever know that this is the order I can integrate in apart from just inspecting.






    calculus integration definite-integrals order-of-integration







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Is there any hard and fast rule for what order you integrate for triple integrals. I know of Fubini's theorem but surely this doesn't cover all cases of triple integrals.



      Say for example I have,



      $$int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta dz dr $$



      Why is it that I can integrate in this order as the first limit's are not a function of one of the variables the second are a function of $r$ and the last of no variable again, how would I ever know that this is the order I can integrate in apart from just inspecting.






      calculus integration definite-integrals order-of-integration







      share|cite|improve this question









      $endgroup$




      Is there any hard and fast rule for what order you integrate for triple integrals. I know of Fubini's theorem but surely this doesn't cover all cases of triple integrals.



      Say for example I have,



      $$int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta dz dr $$



      Why is it that I can integrate in this order as the first limit's are not a function of one of the variables the second are a function of $r$ and the last of no variable again, how would I ever know that this is the order I can integrate in apart from just inspecting.






      calculus integration definite-integrals order-of-integration




      calculus integration definite-integrals order-of-integration






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      asked Jan 9 at 15:27







      user571032





























          2 Answers
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          0












          $begingroup$

          $theta$ is independent of $r$ and $z$ so the corresponding integral $int_0^{2pi}dots dtheta$ can be at any of positions (1st, 2nd or 3rd).



          A bound for $z$ depends on $r$ thus first has to be computed the integral $dz$, only then the integral $dr.$



          Convenient possibilities giving the same result without any further transformation are
          $$int_{0}^{2pi} int_{0}^{1} int_{0}^{1-r^{2}} r^{3} dz; dr; dtheta = int_{0}^{1} int_{0}^{2 pi} int_{0}^{1-r^{2}} r^{3} dz; dtheta ;dr=int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta; dz; dr.$$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            For the integral $int_0^{2pi}dtheta$, it is completely independent, as you said, from the other variables, so you can evaluate it at any time and multiply the resulting double integral by its results.



            For the integral $int_0^{1-r^2}dz$, although the integrand is just $1$, the limits on the integral depend on the other variables, so after you evaluate the integral, the result will be in terms of $r$. That means that no matter what, $int_0^1dr$ must be evaluated AFTER (on the outside) of $int_0^{1-r^2}dz$. And those are the only real restrictions on this specific example.



            In general, for most functions you will ever integrate in a multivariable calculus class, the $dtheta$ integral will be the outermost one, because it rarely (if ever) has bounds that depends on $z$ or $r$ in cylindrical coordinates or $rho$ or $phi$ in the case of spherical coordinates. As you said, you must inspect the integral before you start to determine if the order makes sense.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
              $endgroup$
              – user247327
              Jan 9 at 16:22











            Your Answer





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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            0












            $begingroup$

            $theta$ is independent of $r$ and $z$ so the corresponding integral $int_0^{2pi}dots dtheta$ can be at any of positions (1st, 2nd or 3rd).



            A bound for $z$ depends on $r$ thus first has to be computed the integral $dz$, only then the integral $dr.$



            Convenient possibilities giving the same result without any further transformation are
            $$int_{0}^{2pi} int_{0}^{1} int_{0}^{1-r^{2}} r^{3} dz; dr; dtheta = int_{0}^{1} int_{0}^{2 pi} int_{0}^{1-r^{2}} r^{3} dz; dtheta ;dr=int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta; dz; dr.$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              $theta$ is independent of $r$ and $z$ so the corresponding integral $int_0^{2pi}dots dtheta$ can be at any of positions (1st, 2nd or 3rd).



              A bound for $z$ depends on $r$ thus first has to be computed the integral $dz$, only then the integral $dr.$



              Convenient possibilities giving the same result without any further transformation are
              $$int_{0}^{2pi} int_{0}^{1} int_{0}^{1-r^{2}} r^{3} dz; dr; dtheta = int_{0}^{1} int_{0}^{2 pi} int_{0}^{1-r^{2}} r^{3} dz; dtheta ;dr=int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta; dz; dr.$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                $theta$ is independent of $r$ and $z$ so the corresponding integral $int_0^{2pi}dots dtheta$ can be at any of positions (1st, 2nd or 3rd).



                A bound for $z$ depends on $r$ thus first has to be computed the integral $dz$, only then the integral $dr.$



                Convenient possibilities giving the same result without any further transformation are
                $$int_{0}^{2pi} int_{0}^{1} int_{0}^{1-r^{2}} r^{3} dz; dr; dtheta = int_{0}^{1} int_{0}^{2 pi} int_{0}^{1-r^{2}} r^{3} dz; dtheta ;dr=int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta; dz; dr.$$






                share|cite|improve this answer











                $endgroup$



                $theta$ is independent of $r$ and $z$ so the corresponding integral $int_0^{2pi}dots dtheta$ can be at any of positions (1st, 2nd or 3rd).



                A bound for $z$ depends on $r$ thus first has to be computed the integral $dz$, only then the integral $dr.$



                Convenient possibilities giving the same result without any further transformation are
                $$int_{0}^{2pi} int_{0}^{1} int_{0}^{1-r^{2}} r^{3} dz; dr; dtheta = int_{0}^{1} int_{0}^{2 pi} int_{0}^{1-r^{2}} r^{3} dz; dtheta ;dr=int_{0}^{1} int_{0}^{1-r^{2}} int_{0}^{2 pi} r^{3} dtheta; dz; dr.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 9 at 16:15

























                answered Jan 9 at 15:41









                user376343user376343

                3,3582826




                3,3582826























                    0












                    $begingroup$

                    For the integral $int_0^{2pi}dtheta$, it is completely independent, as you said, from the other variables, so you can evaluate it at any time and multiply the resulting double integral by its results.



                    For the integral $int_0^{1-r^2}dz$, although the integrand is just $1$, the limits on the integral depend on the other variables, so after you evaluate the integral, the result will be in terms of $r$. That means that no matter what, $int_0^1dr$ must be evaluated AFTER (on the outside) of $int_0^{1-r^2}dz$. And those are the only real restrictions on this specific example.



                    In general, for most functions you will ever integrate in a multivariable calculus class, the $dtheta$ integral will be the outermost one, because it rarely (if ever) has bounds that depends on $z$ or $r$ in cylindrical coordinates or $rho$ or $phi$ in the case of spherical coordinates. As you said, you must inspect the integral before you start to determine if the order makes sense.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
                      $endgroup$
                      – user247327
                      Jan 9 at 16:22
















                    0












                    $begingroup$

                    For the integral $int_0^{2pi}dtheta$, it is completely independent, as you said, from the other variables, so you can evaluate it at any time and multiply the resulting double integral by its results.



                    For the integral $int_0^{1-r^2}dz$, although the integrand is just $1$, the limits on the integral depend on the other variables, so after you evaluate the integral, the result will be in terms of $r$. That means that no matter what, $int_0^1dr$ must be evaluated AFTER (on the outside) of $int_0^{1-r^2}dz$. And those are the only real restrictions on this specific example.



                    In general, for most functions you will ever integrate in a multivariable calculus class, the $dtheta$ integral will be the outermost one, because it rarely (if ever) has bounds that depends on $z$ or $r$ in cylindrical coordinates or $rho$ or $phi$ in the case of spherical coordinates. As you said, you must inspect the integral before you start to determine if the order makes sense.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
                      $endgroup$
                      – user247327
                      Jan 9 at 16:22














                    0












                    0








                    0





                    $begingroup$

                    For the integral $int_0^{2pi}dtheta$, it is completely independent, as you said, from the other variables, so you can evaluate it at any time and multiply the resulting double integral by its results.



                    For the integral $int_0^{1-r^2}dz$, although the integrand is just $1$, the limits on the integral depend on the other variables, so after you evaluate the integral, the result will be in terms of $r$. That means that no matter what, $int_0^1dr$ must be evaluated AFTER (on the outside) of $int_0^{1-r^2}dz$. And those are the only real restrictions on this specific example.



                    In general, for most functions you will ever integrate in a multivariable calculus class, the $dtheta$ integral will be the outermost one, because it rarely (if ever) has bounds that depends on $z$ or $r$ in cylindrical coordinates or $rho$ or $phi$ in the case of spherical coordinates. As you said, you must inspect the integral before you start to determine if the order makes sense.






                    share|cite|improve this answer









                    $endgroup$



                    For the integral $int_0^{2pi}dtheta$, it is completely independent, as you said, from the other variables, so you can evaluate it at any time and multiply the resulting double integral by its results.



                    For the integral $int_0^{1-r^2}dz$, although the integrand is just $1$, the limits on the integral depend on the other variables, so after you evaluate the integral, the result will be in terms of $r$. That means that no matter what, $int_0^1dr$ must be evaluated AFTER (on the outside) of $int_0^{1-r^2}dz$. And those are the only real restrictions on this specific example.



                    In general, for most functions you will ever integrate in a multivariable calculus class, the $dtheta$ integral will be the outermost one, because it rarely (if ever) has bounds that depends on $z$ or $r$ in cylindrical coordinates or $rho$ or $phi$ in the case of spherical coordinates. As you said, you must inspect the integral before you start to determine if the order makes sense.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 9 at 15:41









                    Calvin GodfreyCalvin Godfrey

                    633311




                    633311












                    • $begingroup$
                      In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
                      $endgroup$
                      – user247327
                      Jan 9 at 16:22


















                    • $begingroup$
                      In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
                      $endgroup$
                      – user247327
                      Jan 9 at 16:22
















                    $begingroup$
                    In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
                    $endgroup$
                    – user247327
                    Jan 9 at 16:22




                    $begingroup$
                    In fact, this integral can be done as a product: $left(int_0^{2pi} dthetaright)left(int_0^1r^3left(int_0^{1- r^2} dzright)drright)= 2pi int_0^1 r^3- r^5 dr$
                    $endgroup$
                    – user247327
                    Jan 9 at 16:22


















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