About surface of integral of type one












0












$begingroup$


Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$



$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,



$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$



My attempt.



If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.






real-analysis calculus integration limits analysis







share|cite|improve this question











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  • $begingroup$
    ooyes thanks a lot
    $endgroup$
    – jackson
    Jan 8 at 16:34
















0












$begingroup$


Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$



$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,



$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$



My attempt.



If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.






real-analysis calculus integration limits analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    ooyes thanks a lot
    $endgroup$
    – jackson
    Jan 8 at 16:34














0












0








0


0



$begingroup$


Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$



$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,



$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$



My attempt.



If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.






real-analysis calculus integration limits analysis







share|cite|improve this question











$endgroup$




Compute F(x,y,z,t)=$iint _Sf(xi ,eta ,zeta) dS$,S is surface
$$(xi -x)^2+(eta-y)^2+(zeta-z)^2=t^2$$,suppose $ r^2={x^2+y^2+z^2}>a^2>0$



$f(xi ,eta ,zeta)=1$
,when $xi^ 2+eta ^2+zeta ^2<a^2$,



$f=0$ , when $\ xi^ 2+eta ^2+zeta ^2ge a^2$



My attempt.



If r$ge t+a$,then the integral is zero,but when r$<t+a$ ,the integral is difficult .i think maybe I should rotate the Rectangular Coordinates.






real-analysis calculus integration limits analysis




real-analysis calculus integration limits analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 16:34







jackson

















asked Jan 8 at 10:14









jacksonjackson

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1499












  • $begingroup$
    ooyes thanks a lot
    $endgroup$
    – jackson
    Jan 8 at 16:34


















  • $begingroup$
    ooyes thanks a lot
    $endgroup$
    – jackson
    Jan 8 at 16:34
















$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34




$begingroup$
ooyes thanks a lot
$endgroup$
– jackson
Jan 8 at 16:34










1 Answer
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The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.



enter image description here



We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
and then
$$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$






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    $begingroup$

    The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.



    enter image description here



    We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
    and then
    $$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.



      enter image description here



      We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
      and then
      $$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.



        enter image description here



        We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
        and then
        $$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$






        share|cite|improve this answer











        $endgroup$



        The situation is rotation symmetric with respect to the axis through $O=(0,0,0)$ and $P=(x,y,z)$. It follows that the value $F(x,y,z,t)$ depends only on $a$, $r>a$, and $t$. You may assume $P=(0,0,r)$, and then have to compute the surface area of the part of the large sphere $S_t: xi^2+eta^2+(zeta-r)^2=t^2$ lying within the small sphere $xi^2+eta^2+zeta^2=a^2$. To this end it is sufficient to draw two circles in the $(rho,zeta)$-half plane with $rho:=sqrt{xi^2+eta^2}geq0$, and to do some elementary geometry. Note that the surface area of a sphere cap is equal to the height of the cap times the length of the equator.



        enter image description here



        We have to compute $$h:=zeta_*-(r-t)={1over2r}bigl(a^2-(r-t)^2bigr) ,$$
        and then
        $$F(r,t)=int_{S_t}f(xi,eta,zeta)>{rm d}S=2pi t> h .$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 9 at 19:35

























        answered Jan 8 at 16:53









        Christian BlatterChristian Blatter

        176k9115328




        176k9115328






























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