Trouble with Spherical Coordinates and the Transformation Formula












0












$begingroup$


I know for a function $f: mathbb R^{3} to mathbb R$ that is integrable



Then we have a coordinate map:



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}, Phi(r,theta, phi):=(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})$



and



$int_{mathbb R^{3}}f(x,y,z)dlambda^{3}(x,y,z)=int_{mathbb R_{geq 0}}int_{[0,2pi]}int_{[0,pi]}f(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})r^2sin{theta}dtheta dphi dr$



In our proof of the above, we had to select a $V$ to exclude in order to find a diffeomorphism $Phi: mathbb R_{> 0}times]0,pi[times]0,2pi[tomathbb R^{3}-V$



In our proof we defined $V$ as $V:=mathbb R_{geq 0}times {0}times mathbb R$



Problem: How exactly did we determine this $V$? I understand that the map $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$ is surjective, but why did we have to change



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}$ to open intervals $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$



What purpose does it serve?










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$endgroup$












  • $begingroup$
    Could you clarify your definition of $V$? Do you mean the coefficient $r^2sintheta$? And I think sometimes your $]0,pi[$ really means $[0, pi]$. Am I right?
    $endgroup$
    – MoonKnight
    Jan 17 at 17:50
















0












$begingroup$


I know for a function $f: mathbb R^{3} to mathbb R$ that is integrable



Then we have a coordinate map:



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}, Phi(r,theta, phi):=(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})$



and



$int_{mathbb R^{3}}f(x,y,z)dlambda^{3}(x,y,z)=int_{mathbb R_{geq 0}}int_{[0,2pi]}int_{[0,pi]}f(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})r^2sin{theta}dtheta dphi dr$



In our proof of the above, we had to select a $V$ to exclude in order to find a diffeomorphism $Phi: mathbb R_{> 0}times]0,pi[times]0,2pi[tomathbb R^{3}-V$



In our proof we defined $V$ as $V:=mathbb R_{geq 0}times {0}times mathbb R$



Problem: How exactly did we determine this $V$? I understand that the map $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$ is surjective, but why did we have to change



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}$ to open intervals $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$



What purpose does it serve?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Could you clarify your definition of $V$? Do you mean the coefficient $r^2sintheta$? And I think sometimes your $]0,pi[$ really means $[0, pi]$. Am I right?
    $endgroup$
    – MoonKnight
    Jan 17 at 17:50














0












0








0





$begingroup$


I know for a function $f: mathbb R^{3} to mathbb R$ that is integrable



Then we have a coordinate map:



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}, Phi(r,theta, phi):=(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})$



and



$int_{mathbb R^{3}}f(x,y,z)dlambda^{3}(x,y,z)=int_{mathbb R_{geq 0}}int_{[0,2pi]}int_{[0,pi]}f(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})r^2sin{theta}dtheta dphi dr$



In our proof of the above, we had to select a $V$ to exclude in order to find a diffeomorphism $Phi: mathbb R_{> 0}times]0,pi[times]0,2pi[tomathbb R^{3}-V$



In our proof we defined $V$ as $V:=mathbb R_{geq 0}times {0}times mathbb R$



Problem: How exactly did we determine this $V$? I understand that the map $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$ is surjective, but why did we have to change



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}$ to open intervals $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$



What purpose does it serve?










share|cite|improve this question









$endgroup$




I know for a function $f: mathbb R^{3} to mathbb R$ that is integrable



Then we have a coordinate map:



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}, Phi(r,theta, phi):=(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})$



and



$int_{mathbb R^{3}}f(x,y,z)dlambda^{3}(x,y,z)=int_{mathbb R_{geq 0}}int_{[0,2pi]}int_{[0,pi]}f(rsin{theta}cos{phi},rsin{theta}sin{phi},rcos{theta})r^2sin{theta}dtheta dphi dr$



In our proof of the above, we had to select a $V$ to exclude in order to find a diffeomorphism $Phi: mathbb R_{> 0}times]0,pi[times]0,2pi[tomathbb R^{3}-V$



In our proof we defined $V$ as $V:=mathbb R_{geq 0}times {0}times mathbb R$



Problem: How exactly did we determine this $V$? I understand that the map $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$ is surjective, but why did we have to change



$Phi: mathbb R_{geq 0}times[0,pi]times[0,2pi] to mathbb R^{3}$ to open intervals $Phi: mathbb R_{>0 }times]0,pi[times]0,2pi[tomathbb R^{3}-V$



What purpose does it serve?







real-analysis integration measure-theory diffeomorphism






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asked Jan 17 at 15:58









SABOYSABOY

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  • $begingroup$
    Could you clarify your definition of $V$? Do you mean the coefficient $r^2sintheta$? And I think sometimes your $]0,pi[$ really means $[0, pi]$. Am I right?
    $endgroup$
    – MoonKnight
    Jan 17 at 17:50


















  • $begingroup$
    Could you clarify your definition of $V$? Do you mean the coefficient $r^2sintheta$? And I think sometimes your $]0,pi[$ really means $[0, pi]$. Am I right?
    $endgroup$
    – MoonKnight
    Jan 17 at 17:50
















$begingroup$
Could you clarify your definition of $V$? Do you mean the coefficient $r^2sintheta$? And I think sometimes your $]0,pi[$ really means $[0, pi]$. Am I right?
$endgroup$
– MoonKnight
Jan 17 at 17:50




$begingroup$
Could you clarify your definition of $V$? Do you mean the coefficient $r^2sintheta$? And I think sometimes your $]0,pi[$ really means $[0, pi]$. Am I right?
$endgroup$
– MoonKnight
Jan 17 at 17:50










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