Why doesn't integrating over a sphere with $phi$ between $0$ and $2pi$ and $theta$ between $0$ and $pi$ work?












2














One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$



Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/



If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.



I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.

Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.

However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?






integration multivariable-calculus spherical-coordinates multiple-integral







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    2














    One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$



    Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/



    If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.



    I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.

    Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.

    However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?






    integration multivariable-calculus spherical-coordinates multiple-integral







    share|cite|improve this question



























      2












      2








      2







      One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$



      Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/



      If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.



      I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.

      Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.

      However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?






      integration multivariable-calculus spherical-coordinates multiple-integral







      share|cite|improve this question















      One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$



      Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/



      If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.



      I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.

      Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.

      However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?






      integration multivariable-calculus spherical-coordinates multiple-integral




      integration multivariable-calculus spherical-coordinates multiple-integral






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      edited Nov 23 at 9:01









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          When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.






          share|cite|improve this answer





























            1














            The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by




            • $x=Rcos theta sin phi$

            • $y=Rsin theta sin phi$

            • $z=Rcos phi$


            with




            • $theta in [0,2pi)$

            • $phi in[0,pi]$


            and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.



            Of course we can also use different range for the parametrization as for example




            • $theta in [-pi,pi)$

            • $phi in[0,pi]$


            enter image description here



            (credits http://mathworld.wolfram.com/SphericalCoordinates.html)






            share|cite|improve this answer























            • I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
              – Hans Lundmark
              Nov 23 at 8:38










            • @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
              – gimusi
              Nov 23 at 8:45











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            2 Answers
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            active

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            2 Answers
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            active

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            active

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            active

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            1














            When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.






            share|cite|improve this answer


























              1














              When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.






              share|cite|improve this answer
























                1












                1








                1






                When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.






                share|cite|improve this answer












                When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 8:36









                Hans Lundmark

                35k564112




                35k564112























                    1














                    The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by




                    • $x=Rcos theta sin phi$

                    • $y=Rsin theta sin phi$

                    • $z=Rcos phi$


                    with




                    • $theta in [0,2pi)$

                    • $phi in[0,pi]$


                    and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.



                    Of course we can also use different range for the parametrization as for example




                    • $theta in [-pi,pi)$

                    • $phi in[0,pi]$


                    enter image description here



                    (credits http://mathworld.wolfram.com/SphericalCoordinates.html)






                    share|cite|improve this answer























                    • I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
                      – Hans Lundmark
                      Nov 23 at 8:38










                    • @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
                      – gimusi
                      Nov 23 at 8:45
















                    1














                    The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by




                    • $x=Rcos theta sin phi$

                    • $y=Rsin theta sin phi$

                    • $z=Rcos phi$


                    with




                    • $theta in [0,2pi)$

                    • $phi in[0,pi]$


                    and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.



                    Of course we can also use different range for the parametrization as for example




                    • $theta in [-pi,pi)$

                    • $phi in[0,pi]$


                    enter image description here



                    (credits http://mathworld.wolfram.com/SphericalCoordinates.html)






                    share|cite|improve this answer























                    • I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
                      – Hans Lundmark
                      Nov 23 at 8:38










                    • @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
                      – gimusi
                      Nov 23 at 8:45














                    1












                    1








                    1






                    The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by




                    • $x=Rcos theta sin phi$

                    • $y=Rsin theta sin phi$

                    • $z=Rcos phi$


                    with




                    • $theta in [0,2pi)$

                    • $phi in[0,pi]$


                    and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.



                    Of course we can also use different range for the parametrization as for example




                    • $theta in [-pi,pi)$

                    • $phi in[0,pi]$


                    enter image description here



                    (credits http://mathworld.wolfram.com/SphericalCoordinates.html)






                    share|cite|improve this answer














                    The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by




                    • $x=Rcos theta sin phi$

                    • $y=Rsin theta sin phi$

                    • $z=Rcos phi$


                    with




                    • $theta in [0,2pi)$

                    • $phi in[0,pi]$


                    and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.



                    Of course we can also use different range for the parametrization as for example




                    • $theta in [-pi,pi)$

                    • $phi in[0,pi]$


                    enter image description here



                    (credits http://mathworld.wolfram.com/SphericalCoordinates.html)







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 23 at 8:47

























                    answered Nov 23 at 7:34









                    gimusi

                    1




                    1












                    • I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
                      – Hans Lundmark
                      Nov 23 at 8:38










                    • @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
                      – gimusi
                      Nov 23 at 8:45


















                    • I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
                      – Hans Lundmark
                      Nov 23 at 8:38










                    • @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
                      – gimusi
                      Nov 23 at 8:45
















                    I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
                    – Hans Lundmark
                    Nov 23 at 8:38




                    I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
                    – Hans Lundmark
                    Nov 23 at 8:38












                    @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
                    – gimusi
                    Nov 23 at 8:45




                    @HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
                    – gimusi
                    Nov 23 at 8:45


















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