How to evaluate $int_S(x^4+y^4+z^4) , dS$ over surface of the unit sphere.












3












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Question. Let $S$ denote the unit sphere in $mathbb{R}^3$. Evaluate: $$int_S (x^4+y^4+z^4) , dS$$




My Solution. First I parametrize $S$ by $$r(u,v)=(cos v cos u, cos v sin u, sin v)$$ $0le u le 2 pi;~-frac{pi}{2}le v le frac{pi}{2}$



Let $~f(x,y,z)=x^4+y^4+z^4$. Here $~|frac{partial r}{partial u} times frac{partial r}{partial v}|=|cos v|$



Then $displaystyle int_S(x^4+y^4+z^4),dS = int_{-pi/2}^{pi/2} int_0^{2pi} f[r(u,v)] left|frac{partial r}{partial u} times frac{partial r}{partial v}right|~du~dv$



Thus I try to calculate this integral directly using the definition of surface integral.



But I had so much calculations in this way. Does this particular problem can be solved using any theorems e.g-Gauss' Divergence (By writing $f$ as $Fcdot n$ for some vector field $F$?
Thank you.






multivariable-calculus surface-integrals







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  • $begingroup$
    The outer normal vector $n$ is $(x,y,z)$. This may help.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:13










  • $begingroup$
    but we should take the unit normal to use divergence theorem...@xbh
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:17










  • $begingroup$
    Normally you need to make it a unit one, but on the unit sphere, this is truly the unit vector at $(x,y,z)$.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:18










  • $begingroup$
    ohh...Right ......
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:19










  • $begingroup$
    What was the problem in direct calculation? All you need to do is calculate integrals of powers of sin or cos .of u and v.
    $endgroup$
    – herb steinberg
    Aug 12 '18 at 15:26
















3












$begingroup$



Question. Let $S$ denote the unit sphere in $mathbb{R}^3$. Evaluate: $$int_S (x^4+y^4+z^4) , dS$$




My Solution. First I parametrize $S$ by $$r(u,v)=(cos v cos u, cos v sin u, sin v)$$ $0le u le 2 pi;~-frac{pi}{2}le v le frac{pi}{2}$



Let $~f(x,y,z)=x^4+y^4+z^4$. Here $~|frac{partial r}{partial u} times frac{partial r}{partial v}|=|cos v|$



Then $displaystyle int_S(x^4+y^4+z^4),dS = int_{-pi/2}^{pi/2} int_0^{2pi} f[r(u,v)] left|frac{partial r}{partial u} times frac{partial r}{partial v}right|~du~dv$



Thus I try to calculate this integral directly using the definition of surface integral.



But I had so much calculations in this way. Does this particular problem can be solved using any theorems e.g-Gauss' Divergence (By writing $f$ as $Fcdot n$ for some vector field $F$?
Thank you.






multivariable-calculus surface-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    The outer normal vector $n$ is $(x,y,z)$. This may help.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:13










  • $begingroup$
    but we should take the unit normal to use divergence theorem...@xbh
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:17










  • $begingroup$
    Normally you need to make it a unit one, but on the unit sphere, this is truly the unit vector at $(x,y,z)$.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:18










  • $begingroup$
    ohh...Right ......
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:19










  • $begingroup$
    What was the problem in direct calculation? All you need to do is calculate integrals of powers of sin or cos .of u and v.
    $endgroup$
    – herb steinberg
    Aug 12 '18 at 15:26














3












3








3


3



$begingroup$



Question. Let $S$ denote the unit sphere in $mathbb{R}^3$. Evaluate: $$int_S (x^4+y^4+z^4) , dS$$




My Solution. First I parametrize $S$ by $$r(u,v)=(cos v cos u, cos v sin u, sin v)$$ $0le u le 2 pi;~-frac{pi}{2}le v le frac{pi}{2}$



Let $~f(x,y,z)=x^4+y^4+z^4$. Here $~|frac{partial r}{partial u} times frac{partial r}{partial v}|=|cos v|$



Then $displaystyle int_S(x^4+y^4+z^4),dS = int_{-pi/2}^{pi/2} int_0^{2pi} f[r(u,v)] left|frac{partial r}{partial u} times frac{partial r}{partial v}right|~du~dv$



Thus I try to calculate this integral directly using the definition of surface integral.



But I had so much calculations in this way. Does this particular problem can be solved using any theorems e.g-Gauss' Divergence (By writing $f$ as $Fcdot n$ for some vector field $F$?
Thank you.






multivariable-calculus surface-integrals







share|cite|improve this question











$endgroup$





Question. Let $S$ denote the unit sphere in $mathbb{R}^3$. Evaluate: $$int_S (x^4+y^4+z^4) , dS$$




My Solution. First I parametrize $S$ by $$r(u,v)=(cos v cos u, cos v sin u, sin v)$$ $0le u le 2 pi;~-frac{pi}{2}le v le frac{pi}{2}$



Let $~f(x,y,z)=x^4+y^4+z^4$. Here $~|frac{partial r}{partial u} times frac{partial r}{partial v}|=|cos v|$



Then $displaystyle int_S(x^4+y^4+z^4),dS = int_{-pi/2}^{pi/2} int_0^{2pi} f[r(u,v)] left|frac{partial r}{partial u} times frac{partial r}{partial v}right|~du~dv$



Thus I try to calculate this integral directly using the definition of surface integral.



But I had so much calculations in this way. Does this particular problem can be solved using any theorems e.g-Gauss' Divergence (By writing $f$ as $Fcdot n$ for some vector field $F$?
Thank you.






multivariable-calculus surface-integrals




multivariable-calculus surface-integrals






share|cite|improve this question















share|cite|improve this question













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edited Aug 12 '18 at 16:22









Michael Hardy

1




1










asked Aug 12 '18 at 15:07









Indrajit GhoshIndrajit Ghosh

1,0921718




1,0921718












  • $begingroup$
    The outer normal vector $n$ is $(x,y,z)$. This may help.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:13










  • $begingroup$
    but we should take the unit normal to use divergence theorem...@xbh
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:17










  • $begingroup$
    Normally you need to make it a unit one, but on the unit sphere, this is truly the unit vector at $(x,y,z)$.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:18










  • $begingroup$
    ohh...Right ......
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:19










  • $begingroup$
    What was the problem in direct calculation? All you need to do is calculate integrals of powers of sin or cos .of u and v.
    $endgroup$
    – herb steinberg
    Aug 12 '18 at 15:26


















  • $begingroup$
    The outer normal vector $n$ is $(x,y,z)$. This may help.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:13










  • $begingroup$
    but we should take the unit normal to use divergence theorem...@xbh
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:17










  • $begingroup$
    Normally you need to make it a unit one, but on the unit sphere, this is truly the unit vector at $(x,y,z)$.
    $endgroup$
    – xbh
    Aug 12 '18 at 15:18










  • $begingroup$
    ohh...Right ......
    $endgroup$
    – Indrajit Ghosh
    Aug 12 '18 at 15:19










  • $begingroup$
    What was the problem in direct calculation? All you need to do is calculate integrals of powers of sin or cos .of u and v.
    $endgroup$
    – herb steinberg
    Aug 12 '18 at 15:26
















$begingroup$
The outer normal vector $n$ is $(x,y,z)$. This may help.
$endgroup$
– xbh
Aug 12 '18 at 15:13




$begingroup$
The outer normal vector $n$ is $(x,y,z)$. This may help.
$endgroup$
– xbh
Aug 12 '18 at 15:13












$begingroup$
but we should take the unit normal to use divergence theorem...@xbh
$endgroup$
– Indrajit Ghosh
Aug 12 '18 at 15:17




$begingroup$
but we should take the unit normal to use divergence theorem...@xbh
$endgroup$
– Indrajit Ghosh
Aug 12 '18 at 15:17












$begingroup$
Normally you need to make it a unit one, but on the unit sphere, this is truly the unit vector at $(x,y,z)$.
$endgroup$
– xbh
Aug 12 '18 at 15:18




$begingroup$
Normally you need to make it a unit one, but on the unit sphere, this is truly the unit vector at $(x,y,z)$.
$endgroup$
– xbh
Aug 12 '18 at 15:18












$begingroup$
ohh...Right ......
$endgroup$
– Indrajit Ghosh
Aug 12 '18 at 15:19




$begingroup$
ohh...Right ......
$endgroup$
– Indrajit Ghosh
Aug 12 '18 at 15:19












$begingroup$
What was the problem in direct calculation? All you need to do is calculate integrals of powers of sin or cos .of u and v.
$endgroup$
– herb steinberg
Aug 12 '18 at 15:26




$begingroup$
What was the problem in direct calculation? All you need to do is calculate integrals of powers of sin or cos .of u and v.
$endgroup$
– herb steinberg
Aug 12 '18 at 15:26










3 Answers
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I shall denote by $text{d}Sigma$ the surface area element. In the polar form $$(x,y,z)=big(rcos(phi)sin(theta),rsin(phi)cos(theta),rcos(theta)big),,$$ I would only compute the integral $$int_{partial B_1(boldsymbol{0})},z^4,text{d}Sigma=int_0^{2pi},int_0^pi,cos^4(theta),sin(theta),text{d}theta,text{d}phi=2pi,int_{-1}^{+1},t^4,text{d}t=frac{4pi}{5},,$$ where $t:=cos(theta)$, and then multiply that by $3$ to get the final answer. Due to symmetry, this is justified.






share|cite|improve this answer











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    1












    $begingroup$

    Let $~f(x,y,z)=x^4+y^4+z^4$



    The unit outward drawn normal to $S$ is $n=(x,y,z)$. Taking $F=(x^3,y^3,z^3)$ we have $f=F.n$. Hence the given integral:



    $int_{S}f(x,y,z)dS=int_{S}F.n~dS=int_{V}div F~dV =3int_{0}^{1}int_{-pi/2}^{-pi/2}int_{0}^{2pi}r^2|cos v|~dr~du~dv=frac{12pi}{5}.$






    share|cite|improve this answer









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      0












      $begingroup$

      Direct calculation (unit sphere). $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=(1-2h)$



      where $h=cos^4vcos^2usin^2u+cos^2vsin^2v=h_1+h_2$. The surface integral is $S=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(1-2h)cosvdvdu$. These can be calculated directly.
      $int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cosvdvdu=4pi$. Let
      $I_1=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cos^5vdvcos^2usin^2udu$. Let $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}sin^2vcos^3vdv$. At this point note that $I_1=frac{I_2}{2}$. $I_2$ is easier to calculate, so we don't need to calculate $I_1$. $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(sin^2v-sin^4v)cosvdv=frac{8pi}{15}$. Therefore $S=frac{12pi}{5}$.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        I shall denote by $text{d}Sigma$ the surface area element. In the polar form $$(x,y,z)=big(rcos(phi)sin(theta),rsin(phi)cos(theta),rcos(theta)big),,$$ I would only compute the integral $$int_{partial B_1(boldsymbol{0})},z^4,text{d}Sigma=int_0^{2pi},int_0^pi,cos^4(theta),sin(theta),text{d}theta,text{d}phi=2pi,int_{-1}^{+1},t^4,text{d}t=frac{4pi}{5},,$$ where $t:=cos(theta)$, and then multiply that by $3$ to get the final answer. Due to symmetry, this is justified.






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          I shall denote by $text{d}Sigma$ the surface area element. In the polar form $$(x,y,z)=big(rcos(phi)sin(theta),rsin(phi)cos(theta),rcos(theta)big),,$$ I would only compute the integral $$int_{partial B_1(boldsymbol{0})},z^4,text{d}Sigma=int_0^{2pi},int_0^pi,cos^4(theta),sin(theta),text{d}theta,text{d}phi=2pi,int_{-1}^{+1},t^4,text{d}t=frac{4pi}{5},,$$ where $t:=cos(theta)$, and then multiply that by $3$ to get the final answer. Due to symmetry, this is justified.






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            I shall denote by $text{d}Sigma$ the surface area element. In the polar form $$(x,y,z)=big(rcos(phi)sin(theta),rsin(phi)cos(theta),rcos(theta)big),,$$ I would only compute the integral $$int_{partial B_1(boldsymbol{0})},z^4,text{d}Sigma=int_0^{2pi},int_0^pi,cos^4(theta),sin(theta),text{d}theta,text{d}phi=2pi,int_{-1}^{+1},t^4,text{d}t=frac{4pi}{5},,$$ where $t:=cos(theta)$, and then multiply that by $3$ to get the final answer. Due to symmetry, this is justified.






            share|cite|improve this answer











            $endgroup$



            I shall denote by $text{d}Sigma$ the surface area element. In the polar form $$(x,y,z)=big(rcos(phi)sin(theta),rsin(phi)cos(theta),rcos(theta)big),,$$ I would only compute the integral $$int_{partial B_1(boldsymbol{0})},z^4,text{d}Sigma=int_0^{2pi},int_0^pi,cos^4(theta),sin(theta),text{d}theta,text{d}phi=2pi,int_{-1}^{+1},t^4,text{d}t=frac{4pi}{5},,$$ where $t:=cos(theta)$, and then multiply that by $3$ to get the final answer. Due to symmetry, this is justified.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 1:34

























            answered Aug 12 '18 at 16:47









            BatominovskiBatominovski

            33.1k33293




            33.1k33293























                1












                $begingroup$

                Let $~f(x,y,z)=x^4+y^4+z^4$



                The unit outward drawn normal to $S$ is $n=(x,y,z)$. Taking $F=(x^3,y^3,z^3)$ we have $f=F.n$. Hence the given integral:



                $int_{S}f(x,y,z)dS=int_{S}F.n~dS=int_{V}div F~dV =3int_{0}^{1}int_{-pi/2}^{-pi/2}int_{0}^{2pi}r^2|cos v|~dr~du~dv=frac{12pi}{5}.$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Let $~f(x,y,z)=x^4+y^4+z^4$



                  The unit outward drawn normal to $S$ is $n=(x,y,z)$. Taking $F=(x^3,y^3,z^3)$ we have $f=F.n$. Hence the given integral:



                  $int_{S}f(x,y,z)dS=int_{S}F.n~dS=int_{V}div F~dV =3int_{0}^{1}int_{-pi/2}^{-pi/2}int_{0}^{2pi}r^2|cos v|~dr~du~dv=frac{12pi}{5}.$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Let $~f(x,y,z)=x^4+y^4+z^4$



                    The unit outward drawn normal to $S$ is $n=(x,y,z)$. Taking $F=(x^3,y^3,z^3)$ we have $f=F.n$. Hence the given integral:



                    $int_{S}f(x,y,z)dS=int_{S}F.n~dS=int_{V}div F~dV =3int_{0}^{1}int_{-pi/2}^{-pi/2}int_{0}^{2pi}r^2|cos v|~dr~du~dv=frac{12pi}{5}.$






                    share|cite|improve this answer









                    $endgroup$



                    Let $~f(x,y,z)=x^4+y^4+z^4$



                    The unit outward drawn normal to $S$ is $n=(x,y,z)$. Taking $F=(x^3,y^3,z^3)$ we have $f=F.n$. Hence the given integral:



                    $int_{S}f(x,y,z)dS=int_{S}F.n~dS=int_{V}div F~dV =3int_{0}^{1}int_{-pi/2}^{-pi/2}int_{0}^{2pi}r^2|cos v|~dr~du~dv=frac{12pi}{5}.$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 12 '18 at 16:57









                    Indrajit GhoshIndrajit Ghosh

                    1,0921718




                    1,0921718























                        0












                        $begingroup$

                        Direct calculation (unit sphere). $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=(1-2h)$



                        where $h=cos^4vcos^2usin^2u+cos^2vsin^2v=h_1+h_2$. The surface integral is $S=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(1-2h)cosvdvdu$. These can be calculated directly.
                        $int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cosvdvdu=4pi$. Let
                        $I_1=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cos^5vdvcos^2usin^2udu$. Let $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}sin^2vcos^3vdv$. At this point note that $I_1=frac{I_2}{2}$. $I_2$ is easier to calculate, so we don't need to calculate $I_1$. $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(sin^2v-sin^4v)cosvdv=frac{8pi}{15}$. Therefore $S=frac{12pi}{5}$.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          Direct calculation (unit sphere). $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=(1-2h)$



                          where $h=cos^4vcos^2usin^2u+cos^2vsin^2v=h_1+h_2$. The surface integral is $S=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(1-2h)cosvdvdu$. These can be calculated directly.
                          $int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cosvdvdu=4pi$. Let
                          $I_1=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cos^5vdvcos^2usin^2udu$. Let $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}sin^2vcos^3vdv$. At this point note that $I_1=frac{I_2}{2}$. $I_2$ is easier to calculate, so we don't need to calculate $I_1$. $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(sin^2v-sin^4v)cosvdv=frac{8pi}{15}$. Therefore $S=frac{12pi}{5}$.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Direct calculation (unit sphere). $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=(1-2h)$



                            where $h=cos^4vcos^2usin^2u+cos^2vsin^2v=h_1+h_2$. The surface integral is $S=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(1-2h)cosvdvdu$. These can be calculated directly.
                            $int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cosvdvdu=4pi$. Let
                            $I_1=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cos^5vdvcos^2usin^2udu$. Let $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}sin^2vcos^3vdv$. At this point note that $I_1=frac{I_2}{2}$. $I_2$ is easier to calculate, so we don't need to calculate $I_1$. $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(sin^2v-sin^4v)cosvdv=frac{8pi}{15}$. Therefore $S=frac{12pi}{5}$.






                            share|cite|improve this answer











                            $endgroup$



                            Direct calculation (unit sphere). $x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=(1-2h)$



                            where $h=cos^4vcos^2usin^2u+cos^2vsin^2v=h_1+h_2$. The surface integral is $S=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(1-2h)cosvdvdu$. These can be calculated directly.
                            $int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cosvdvdu=4pi$. Let
                            $I_1=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}cos^5vdvcos^2usin^2udu$. Let $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}sin^2vcos^3vdv$. At this point note that $I_1=frac{I_2}{2}$. $I_2$ is easier to calculate, so we don't need to calculate $I_1$. $I_2=int_0^{2pi}int_{frac{-pi}{2}}^{frac{pi}{2}}(sin^2v-sin^4v)cosvdv=frac{8pi}{15}$. Therefore $S=frac{12pi}{5}$.







                            share|cite|improve this answer














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                            edited Aug 12 '18 at 21:26

























                            answered Aug 12 '18 at 21:20









                            herb steinbergherb steinberg

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