Trouble with Spherical Coordinates and the Transformation Formula
$begingroup$
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?
real-analysis integration measure-theory diffeomorphism
asked Jan 17 at 15:58
SABOYSABOY
600311
$endgroup$
add a comment |
$begingroup$
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?
real-analysis integration measure-theory diffeomorphism
asked Jan 17 at 15:58
SABOYSABOY
600311
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
real-analysis integration measure-theory diffeomorphism
asked Jan 17 at 15:58
SABOYSABOY
600311
$endgroup$
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
real-analysis integration measure-theory diffeomorphism
asked Jan 17 at 15:58
SABOYSABOY
600311
asked Jan 17 at 15:58
SABOYSABOY
600311
asked Jan 17 at 15:58
SABOYSABOY
600311
asked Jan 17 at 15:58
SABOYSABOY
600311
asked Jan 17 at 15:58
SABOYSABOY
600311
600311
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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