how to find the limits for this type of integrals using spherical coordinates












0












$begingroup$



move the integral



$int_0^1 int_0^{sqrt{1 - x^2}} int_0^{sqrt{1 - x^2 - y^2}} sqrt{x^2 + y^2 + z^2},mathrm dz,mathrm dy,mathrm dx = frac{pi}{8}$



to spherical coordinates integral.




i dont know how to find the limits of the integral when moving to $theta, r , phi $



can you please guide me with finding the limits because i always get a mistakes there




my trial :




reading the $mathrm dz$ we see that $ 0 leq z^2 leq 1-x^2-y^2 $
or $ 0leq x^2 + y^2 + z^2 leq 1 $



reading $mathrm dy$ we see that $ 0 leq y^2 leq sqrt{1-x^2} $ or $ 0leq x^2 + y^2 leq 1$



reading $dx$ we see that $ 0 leq x leq 1$



so the spherical limits :



$ 0leq r^2 leq 1$



$ 0 leq r^2sin(theta)^2 leq 1$



$ 0 leq rcos(theta)sin(phi) leq 1$




how do i continue from here ? i think i made a mistake with $phi$ because i didn't get the proper value for the integral after calculating in the spherical co-ordinates.







integration multivariable-calculus spherical-coordinates jacobian







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You lost the point that $x,y,z$ are all positive
    $endgroup$
    – Damien
    Dec 25 '18 at 11:20
















0












$begingroup$



move the integral



$int_0^1 int_0^{sqrt{1 - x^2}} int_0^{sqrt{1 - x^2 - y^2}} sqrt{x^2 + y^2 + z^2},mathrm dz,mathrm dy,mathrm dx = frac{pi}{8}$



to spherical coordinates integral.




i dont know how to find the limits of the integral when moving to $theta, r , phi $



can you please guide me with finding the limits because i always get a mistakes there




my trial :




reading the $mathrm dz$ we see that $ 0 leq z^2 leq 1-x^2-y^2 $
or $ 0leq x^2 + y^2 + z^2 leq 1 $



reading $mathrm dy$ we see that $ 0 leq y^2 leq sqrt{1-x^2} $ or $ 0leq x^2 + y^2 leq 1$



reading $dx$ we see that $ 0 leq x leq 1$



so the spherical limits :



$ 0leq r^2 leq 1$



$ 0 leq r^2sin(theta)^2 leq 1$



$ 0 leq rcos(theta)sin(phi) leq 1$




how do i continue from here ? i think i made a mistake with $phi$ because i didn't get the proper value for the integral after calculating in the spherical co-ordinates.







integration multivariable-calculus spherical-coordinates jacobian







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You lost the point that $x,y,z$ are all positive
    $endgroup$
    – Damien
    Dec 25 '18 at 11:20














0












0








0





$begingroup$



move the integral



$int_0^1 int_0^{sqrt{1 - x^2}} int_0^{sqrt{1 - x^2 - y^2}} sqrt{x^2 + y^2 + z^2},mathrm dz,mathrm dy,mathrm dx = frac{pi}{8}$



to spherical coordinates integral.




i dont know how to find the limits of the integral when moving to $theta, r , phi $



can you please guide me with finding the limits because i always get a mistakes there




my trial :




reading the $mathrm dz$ we see that $ 0 leq z^2 leq 1-x^2-y^2 $
or $ 0leq x^2 + y^2 + z^2 leq 1 $



reading $mathrm dy$ we see that $ 0 leq y^2 leq sqrt{1-x^2} $ or $ 0leq x^2 + y^2 leq 1$



reading $dx$ we see that $ 0 leq x leq 1$



so the spherical limits :



$ 0leq r^2 leq 1$



$ 0 leq r^2sin(theta)^2 leq 1$



$ 0 leq rcos(theta)sin(phi) leq 1$




how do i continue from here ? i think i made a mistake with $phi$ because i didn't get the proper value for the integral after calculating in the spherical co-ordinates.







integration multivariable-calculus spherical-coordinates jacobian







share|cite|improve this question











$endgroup$





move the integral



$int_0^1 int_0^{sqrt{1 - x^2}} int_0^{sqrt{1 - x^2 - y^2}} sqrt{x^2 + y^2 + z^2},mathrm dz,mathrm dy,mathrm dx = frac{pi}{8}$



to spherical coordinates integral.




i dont know how to find the limits of the integral when moving to $theta, r , phi $



can you please guide me with finding the limits because i always get a mistakes there




my trial :




reading the $mathrm dz$ we see that $ 0 leq z^2 leq 1-x^2-y^2 $
or $ 0leq x^2 + y^2 + z^2 leq 1 $



reading $mathrm dy$ we see that $ 0 leq y^2 leq sqrt{1-x^2} $ or $ 0leq x^2 + y^2 leq 1$



reading $dx$ we see that $ 0 leq x leq 1$



so the spherical limits :



$ 0leq r^2 leq 1$



$ 0 leq r^2sin(theta)^2 leq 1$



$ 0 leq rcos(theta)sin(phi) leq 1$




how do i continue from here ? i think i made a mistake with $phi$ because i didn't get the proper value for the integral after calculating in the spherical co-ordinates.







integration multivariable-calculus spherical-coordinates jacobian




integration multivariable-calculus spherical-coordinates jacobian






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 11:44









José Carlos Santos

173k23133241




173k23133241










asked Dec 25 '18 at 10:51









Mather Mather

4028




4028








  • 1




    $begingroup$
    You lost the point that $x,y,z$ are all positive
    $endgroup$
    – Damien
    Dec 25 '18 at 11:20














  • 1




    $begingroup$
    You lost the point that $x,y,z$ are all positive
    $endgroup$
    – Damien
    Dec 25 '18 at 11:20








1




1




$begingroup$
You lost the point that $x,y,z$ are all positive
$endgroup$
– Damien
Dec 25 '18 at 11:20




$begingroup$
You lost the point that $x,y,z$ are all positive
$endgroup$
– Damien
Dec 25 '18 at 11:20










1 Answer
1






active

oldest

votes


















0












$begingroup$

It follows from your computations that your integral is equal to$$int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr.$$Butbegin{align}int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr&=left(int_0^1r^3,mathrm drright)left(int_0^{fracpi2}sin(phi),mathrm dphiright)left(int_0^{fracpi2},mathrm dthetaright)\&=frac14times1timesfracpi2\&=fracpi8.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    how did you know that $ 0 leq theta leq frac{pi}{2} $
    $endgroup$
    – Mather
    Dec 25 '18 at 11:36












  • $begingroup$
    Because $x,y,zleqslant0$.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:38










  • $begingroup$
    also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
    $endgroup$
    – Mather
    Dec 25 '18 at 11:38












  • $begingroup$
    Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:42












  • $begingroup$
    yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
    $endgroup$
    – Mather
    Dec 25 '18 at 11:46














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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

It follows from your computations that your integral is equal to$$int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr.$$Butbegin{align}int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr&=left(int_0^1r^3,mathrm drright)left(int_0^{fracpi2}sin(phi),mathrm dphiright)left(int_0^{fracpi2},mathrm dthetaright)\&=frac14times1timesfracpi2\&=fracpi8.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    how did you know that $ 0 leq theta leq frac{pi}{2} $
    $endgroup$
    – Mather
    Dec 25 '18 at 11:36












  • $begingroup$
    Because $x,y,zleqslant0$.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:38










  • $begingroup$
    also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
    $endgroup$
    – Mather
    Dec 25 '18 at 11:38












  • $begingroup$
    Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:42












  • $begingroup$
    yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
    $endgroup$
    – Mather
    Dec 25 '18 at 11:46


















0












$begingroup$

It follows from your computations that your integral is equal to$$int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr.$$Butbegin{align}int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr&=left(int_0^1r^3,mathrm drright)left(int_0^{fracpi2}sin(phi),mathrm dphiright)left(int_0^{fracpi2},mathrm dthetaright)\&=frac14times1timesfracpi2\&=fracpi8.end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    how did you know that $ 0 leq theta leq frac{pi}{2} $
    $endgroup$
    – Mather
    Dec 25 '18 at 11:36












  • $begingroup$
    Because $x,y,zleqslant0$.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:38










  • $begingroup$
    also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
    $endgroup$
    – Mather
    Dec 25 '18 at 11:38












  • $begingroup$
    Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:42












  • $begingroup$
    yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
    $endgroup$
    – Mather
    Dec 25 '18 at 11:46
















0












0








0





$begingroup$

It follows from your computations that your integral is equal to$$int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr.$$Butbegin{align}int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr&=left(int_0^1r^3,mathrm drright)left(int_0^{fracpi2}sin(phi),mathrm dphiright)left(int_0^{fracpi2},mathrm dthetaright)\&=frac14times1timesfracpi2\&=fracpi8.end{align}






share|cite|improve this answer









$endgroup$



It follows from your computations that your integral is equal to$$int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr.$$Butbegin{align}int_0^1int_0^{fracpi2}int_0^{fracpi2}r^3sin(phi),mathrm dtheta,mathrm dphi,mathrm dr&=left(int_0^1r^3,mathrm drright)left(int_0^{fracpi2}sin(phi),mathrm dphiright)left(int_0^{fracpi2},mathrm dthetaright)\&=frac14times1timesfracpi2\&=fracpi8.end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 11:34









José Carlos SantosJosé Carlos Santos

173k23133241




173k23133241












  • $begingroup$
    how did you know that $ 0 leq theta leq frac{pi}{2} $
    $endgroup$
    – Mather
    Dec 25 '18 at 11:36












  • $begingroup$
    Because $x,y,zleqslant0$.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:38










  • $begingroup$
    also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
    $endgroup$
    – Mather
    Dec 25 '18 at 11:38












  • $begingroup$
    Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:42












  • $begingroup$
    yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
    $endgroup$
    – Mather
    Dec 25 '18 at 11:46




















  • $begingroup$
    how did you know that $ 0 leq theta leq frac{pi}{2} $
    $endgroup$
    – Mather
    Dec 25 '18 at 11:36












  • $begingroup$
    Because $x,y,zleqslant0$.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:38










  • $begingroup$
    also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
    $endgroup$
    – Mather
    Dec 25 '18 at 11:38












  • $begingroup$
    Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
    $endgroup$
    – José Carlos Santos
    Dec 25 '18 at 11:42












  • $begingroup$
    yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
    $endgroup$
    – Mather
    Dec 25 '18 at 11:46


















$begingroup$
how did you know that $ 0 leq theta leq frac{pi}{2} $
$endgroup$
– Mather
Dec 25 '18 at 11:36






$begingroup$
how did you know that $ 0 leq theta leq frac{pi}{2} $
$endgroup$
– Mather
Dec 25 '18 at 11:36














$begingroup$
Because $x,y,zleqslant0$.
$endgroup$
– José Carlos Santos
Dec 25 '18 at 11:38




$begingroup$
Because $x,y,zleqslant0$.
$endgroup$
– José Carlos Santos
Dec 25 '18 at 11:38












$begingroup$
also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
$endgroup$
– Mather
Dec 25 '18 at 11:38






$begingroup$
also why is $ 0leq r leq 1 $ and not $ 0leq r leq frac{1}{sin{theta} }$
$endgroup$
– Mather
Dec 25 '18 at 11:38














$begingroup$
Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
$endgroup$
– José Carlos Santos
Dec 25 '18 at 11:42






$begingroup$
Why should we have that? The region that you are interested in is the part of the unit sphere in which all Cartesian coordinates are non-negative. So, $r$ can take any value from $0$ to $1$, whatever $theta$ may be.
$endgroup$
– José Carlos Santos
Dec 25 '18 at 11:42














$begingroup$
yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
$endgroup$
– Mather
Dec 25 '18 at 11:46






$begingroup$
yes but the inequality says (by moving from xyz , to spherical ) that $ 0 leq r^2sin(theta)^2 leq 1 $ so dividing makes the ineuality $(a)$ $ 0leq r leq frac{1}{sin(theta)} $ doesn't it ?
$endgroup$
– Mather
Dec 25 '18 at 11:46




















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