verifying way of determining a surface integral












2












$begingroup$


I want to determine



$ int_F f do $

with $f(x,y,z) = x^2z $



$F$ is cylinder surface (lateral surface) with
$ F= { (x,y,z) in mathbb{R}^3 : x^2+y^2= 4 , 0 leq z leq 1 } $



My Idea to solve this surface integral is so far:



Using cylindric coordinates
$phi(u_1, u_2) = begin{pmatrix} r cos (u_1) \ rsin(u_1) \u_2 end{pmatrix} $



$phi: (0,2 pi) times (o,z) $
$0 leq u_1 leq 2pi , 0leq u_2 leq z $



the derivates:



$ phi_{u_1}= begin{pmatrix} -r sin (u_1) \ rcos(u_1) \ 0 end{pmatrix}$



and
$ phi_{u_2}=begin{pmatrix} 0 \ 0 \1 end{pmatrix}$



the cross product comes to :



$ phi_{u_1} times phi_{u_2} = begin{pmatrix} r cos (u_1) \ rsin(u_1) \0 end{pmatrix}$



so $ ||phi_{u_1} times phi_{u_2} || = r $



so, it comes to calculate following integral:



$$ int_0^{2 pi} int_0^z (r cos^2 u_1 u_2) r du_2 du_1 $$



dont I miss both of the cover surfaces, or is that not needed?






real-analysis integration multiple-integral







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Can you give a context?
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 20:39










  • $begingroup$
    I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^
    $endgroup$
    – wondering1123
    Dec 2 '18 at 21:03










  • $begingroup$
    "$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you,
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 3:47










  • $begingroup$
    I added it to the text :)
    $endgroup$
    – wondering1123
    Dec 3 '18 at 17:39
















2












$begingroup$


I want to determine



$ int_F f do $

with $f(x,y,z) = x^2z $



$F$ is cylinder surface (lateral surface) with
$ F= { (x,y,z) in mathbb{R}^3 : x^2+y^2= 4 , 0 leq z leq 1 } $



My Idea to solve this surface integral is so far:



Using cylindric coordinates
$phi(u_1, u_2) = begin{pmatrix} r cos (u_1) \ rsin(u_1) \u_2 end{pmatrix} $



$phi: (0,2 pi) times (o,z) $
$0 leq u_1 leq 2pi , 0leq u_2 leq z $



the derivates:



$ phi_{u_1}= begin{pmatrix} -r sin (u_1) \ rcos(u_1) \ 0 end{pmatrix}$



and
$ phi_{u_2}=begin{pmatrix} 0 \ 0 \1 end{pmatrix}$



the cross product comes to :



$ phi_{u_1} times phi_{u_2} = begin{pmatrix} r cos (u_1) \ rsin(u_1) \0 end{pmatrix}$



so $ ||phi_{u_1} times phi_{u_2} || = r $



so, it comes to calculate following integral:



$$ int_0^{2 pi} int_0^z (r cos^2 u_1 u_2) r du_2 du_1 $$



dont I miss both of the cover surfaces, or is that not needed?






real-analysis integration multiple-integral







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Can you give a context?
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 20:39










  • $begingroup$
    I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^
    $endgroup$
    – wondering1123
    Dec 2 '18 at 21:03










  • $begingroup$
    "$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you,
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 3:47










  • $begingroup$
    I added it to the text :)
    $endgroup$
    – wondering1123
    Dec 3 '18 at 17:39














2












2








2


1



$begingroup$


I want to determine



$ int_F f do $

with $f(x,y,z) = x^2z $



$F$ is cylinder surface (lateral surface) with
$ F= { (x,y,z) in mathbb{R}^3 : x^2+y^2= 4 , 0 leq z leq 1 } $



My Idea to solve this surface integral is so far:



Using cylindric coordinates
$phi(u_1, u_2) = begin{pmatrix} r cos (u_1) \ rsin(u_1) \u_2 end{pmatrix} $



$phi: (0,2 pi) times (o,z) $
$0 leq u_1 leq 2pi , 0leq u_2 leq z $



the derivates:



$ phi_{u_1}= begin{pmatrix} -r sin (u_1) \ rcos(u_1) \ 0 end{pmatrix}$



and
$ phi_{u_2}=begin{pmatrix} 0 \ 0 \1 end{pmatrix}$



the cross product comes to :



$ phi_{u_1} times phi_{u_2} = begin{pmatrix} r cos (u_1) \ rsin(u_1) \0 end{pmatrix}$



so $ ||phi_{u_1} times phi_{u_2} || = r $



so, it comes to calculate following integral:



$$ int_0^{2 pi} int_0^z (r cos^2 u_1 u_2) r du_2 du_1 $$



dont I miss both of the cover surfaces, or is that not needed?






real-analysis integration multiple-integral







share|cite|improve this question











$endgroup$




I want to determine



$ int_F f do $

with $f(x,y,z) = x^2z $



$F$ is cylinder surface (lateral surface) with
$ F= { (x,y,z) in mathbb{R}^3 : x^2+y^2= 4 , 0 leq z leq 1 } $



My Idea to solve this surface integral is so far:



Using cylindric coordinates
$phi(u_1, u_2) = begin{pmatrix} r cos (u_1) \ rsin(u_1) \u_2 end{pmatrix} $



$phi: (0,2 pi) times (o,z) $
$0 leq u_1 leq 2pi , 0leq u_2 leq z $



the derivates:



$ phi_{u_1}= begin{pmatrix} -r sin (u_1) \ rcos(u_1) \ 0 end{pmatrix}$



and
$ phi_{u_2}=begin{pmatrix} 0 \ 0 \1 end{pmatrix}$



the cross product comes to :



$ phi_{u_1} times phi_{u_2} = begin{pmatrix} r cos (u_1) \ rsin(u_1) \0 end{pmatrix}$



so $ ||phi_{u_1} times phi_{u_2} || = r $



so, it comes to calculate following integral:



$$ int_0^{2 pi} int_0^z (r cos^2 u_1 u_2) r du_2 du_1 $$



dont I miss both of the cover surfaces, or is that not needed?






real-analysis integration multiple-integral




real-analysis integration multiple-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 12:01







wondering1123

















asked Dec 2 '18 at 20:37









wondering1123wondering1123

14911




14911








  • 1




    $begingroup$
    Can you give a context?
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 20:39










  • $begingroup$
    I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^
    $endgroup$
    – wondering1123
    Dec 2 '18 at 21:03










  • $begingroup$
    "$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you,
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 3:47










  • $begingroup$
    I added it to the text :)
    $endgroup$
    – wondering1123
    Dec 3 '18 at 17:39














  • 1




    $begingroup$
    Can you give a context?
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 20:39










  • $begingroup$
    I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^
    $endgroup$
    – wondering1123
    Dec 2 '18 at 21:03










  • $begingroup$
    "$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you,
    $endgroup$
    – Paul Sinclair
    Dec 3 '18 at 3:47










  • $begingroup$
    I added it to the text :)
    $endgroup$
    – wondering1123
    Dec 3 '18 at 17:39








1




1




$begingroup$
Can you give a context?
$endgroup$
– Tito Eliatron
Dec 2 '18 at 20:39




$begingroup$
Can you give a context?
$endgroup$
– Tito Eliatron
Dec 2 '18 at 20:39












$begingroup$
I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^
$endgroup$
– wondering1123
Dec 2 '18 at 21:03




$begingroup$
I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^
$endgroup$
– wondering1123
Dec 2 '18 at 21:03












$begingroup$
"$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you,
$endgroup$
– Paul Sinclair
Dec 3 '18 at 3:47




$begingroup$
"$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you,
$endgroup$
– Paul Sinclair
Dec 3 '18 at 3:47












$begingroup$
I added it to the text :)
$endgroup$
– wondering1123
Dec 3 '18 at 17:39




$begingroup$
I added it to the text :)
$endgroup$
– wondering1123
Dec 3 '18 at 17:39










1 Answer
1






active

oldest

votes


















1





+50







$begingroup$

Notation is strange, the reasoning is correct. Really,
$$x=rcos u_1,quad z=u_2,quad mathrm ds = r,mathrm du_1,mathrm du_2.$$
But the reasoning leads to the expression for the side surface
$$S_{side}=intlimits_0^{2pi}intlimits_0^{nothspace{1pt} zcolor{red}1}(r^{nothspace{1pt} 1color{red}2}cos^2u_1,u_2)r,du_1,du_2.$$



On the "floor" surface ($z=0$), the integrand is zero.



On the "ceiling" surface ($z=1$), there are the plain circle with the polar coordinates ($rho, u_1$)
$$x=rhocos u_1,quad x=rho sin u_1,quad z=1,quad mathrm ds = rho,mathrm drho,mathrm du_1,$$
$$S_{ceiling} = intlimits_0^{2pi}intlimits_0^r(rho^2cos^2u_1)rho,mathrm drho,mathrm du_1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the answere !! what is about both of the cover surfaces?
    $endgroup$
    – wondering1123
    Dec 5 '18 at 18:18






  • 1




    $begingroup$
    @wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
    $endgroup$
    – Yuri Negometyanov
    Dec 5 '18 at 18:57










  • $begingroup$
    yes, I do mean the ceiling and the floor!
    $endgroup$
    – wondering1123
    Dec 5 '18 at 19:50











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1





+50







$begingroup$

Notation is strange, the reasoning is correct. Really,
$$x=rcos u_1,quad z=u_2,quad mathrm ds = r,mathrm du_1,mathrm du_2.$$
But the reasoning leads to the expression for the side surface
$$S_{side}=intlimits_0^{2pi}intlimits_0^{nothspace{1pt} zcolor{red}1}(r^{nothspace{1pt} 1color{red}2}cos^2u_1,u_2)r,du_1,du_2.$$



On the "floor" surface ($z=0$), the integrand is zero.



On the "ceiling" surface ($z=1$), there are the plain circle with the polar coordinates ($rho, u_1$)
$$x=rhocos u_1,quad x=rho sin u_1,quad z=1,quad mathrm ds = rho,mathrm drho,mathrm du_1,$$
$$S_{ceiling} = intlimits_0^{2pi}intlimits_0^r(rho^2cos^2u_1)rho,mathrm drho,mathrm du_1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the answere !! what is about both of the cover surfaces?
    $endgroup$
    – wondering1123
    Dec 5 '18 at 18:18






  • 1




    $begingroup$
    @wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
    $endgroup$
    – Yuri Negometyanov
    Dec 5 '18 at 18:57










  • $begingroup$
    yes, I do mean the ceiling and the floor!
    $endgroup$
    – wondering1123
    Dec 5 '18 at 19:50
















1





+50







$begingroup$

Notation is strange, the reasoning is correct. Really,
$$x=rcos u_1,quad z=u_2,quad mathrm ds = r,mathrm du_1,mathrm du_2.$$
But the reasoning leads to the expression for the side surface
$$S_{side}=intlimits_0^{2pi}intlimits_0^{nothspace{1pt} zcolor{red}1}(r^{nothspace{1pt} 1color{red}2}cos^2u_1,u_2)r,du_1,du_2.$$



On the "floor" surface ($z=0$), the integrand is zero.



On the "ceiling" surface ($z=1$), there are the plain circle with the polar coordinates ($rho, u_1$)
$$x=rhocos u_1,quad x=rho sin u_1,quad z=1,quad mathrm ds = rho,mathrm drho,mathrm du_1,$$
$$S_{ceiling} = intlimits_0^{2pi}intlimits_0^r(rho^2cos^2u_1)rho,mathrm drho,mathrm du_1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the answere !! what is about both of the cover surfaces?
    $endgroup$
    – wondering1123
    Dec 5 '18 at 18:18






  • 1




    $begingroup$
    @wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
    $endgroup$
    – Yuri Negometyanov
    Dec 5 '18 at 18:57










  • $begingroup$
    yes, I do mean the ceiling and the floor!
    $endgroup$
    – wondering1123
    Dec 5 '18 at 19:50














1





+50







1





+50



1




+50



$begingroup$

Notation is strange, the reasoning is correct. Really,
$$x=rcos u_1,quad z=u_2,quad mathrm ds = r,mathrm du_1,mathrm du_2.$$
But the reasoning leads to the expression for the side surface
$$S_{side}=intlimits_0^{2pi}intlimits_0^{nothspace{1pt} zcolor{red}1}(r^{nothspace{1pt} 1color{red}2}cos^2u_1,u_2)r,du_1,du_2.$$



On the "floor" surface ($z=0$), the integrand is zero.



On the "ceiling" surface ($z=1$), there are the plain circle with the polar coordinates ($rho, u_1$)
$$x=rhocos u_1,quad x=rho sin u_1,quad z=1,quad mathrm ds = rho,mathrm drho,mathrm du_1,$$
$$S_{ceiling} = intlimits_0^{2pi}intlimits_0^r(rho^2cos^2u_1)rho,mathrm drho,mathrm du_1.$$






share|cite|improve this answer











$endgroup$



Notation is strange, the reasoning is correct. Really,
$$x=rcos u_1,quad z=u_2,quad mathrm ds = r,mathrm du_1,mathrm du_2.$$
But the reasoning leads to the expression for the side surface
$$S_{side}=intlimits_0^{2pi}intlimits_0^{nothspace{1pt} zcolor{red}1}(r^{nothspace{1pt} 1color{red}2}cos^2u_1,u_2)r,du_1,du_2.$$



On the "floor" surface ($z=0$), the integrand is zero.



On the "ceiling" surface ($z=1$), there are the plain circle with the polar coordinates ($rho, u_1$)
$$x=rhocos u_1,quad x=rho sin u_1,quad z=1,quad mathrm ds = rho,mathrm drho,mathrm du_1,$$
$$S_{ceiling} = intlimits_0^{2pi}intlimits_0^r(rho^2cos^2u_1)rho,mathrm drho,mathrm du_1.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 22:20

























answered Dec 5 '18 at 17:38









Yuri NegometyanovYuri Negometyanov

11.8k1729




11.8k1729












  • $begingroup$
    thanks for the answere !! what is about both of the cover surfaces?
    $endgroup$
    – wondering1123
    Dec 5 '18 at 18:18






  • 1




    $begingroup$
    @wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
    $endgroup$
    – Yuri Negometyanov
    Dec 5 '18 at 18:57










  • $begingroup$
    yes, I do mean the ceiling and the floor!
    $endgroup$
    – wondering1123
    Dec 5 '18 at 19:50


















  • $begingroup$
    thanks for the answere !! what is about both of the cover surfaces?
    $endgroup$
    – wondering1123
    Dec 5 '18 at 18:18






  • 1




    $begingroup$
    @wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
    $endgroup$
    – Yuri Negometyanov
    Dec 5 '18 at 18:57










  • $begingroup$
    yes, I do mean the ceiling and the floor!
    $endgroup$
    – wondering1123
    Dec 5 '18 at 19:50
















$begingroup$
thanks for the answere !! what is about both of the cover surfaces?
$endgroup$
– wondering1123
Dec 5 '18 at 18:18




$begingroup$
thanks for the answere !! what is about both of the cover surfaces?
$endgroup$
– wondering1123
Dec 5 '18 at 18:18




1




1




$begingroup$
@wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
$endgroup$
– Yuri Negometyanov
Dec 5 '18 at 18:57




$begingroup$
@wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too.
$endgroup$
– Yuri Negometyanov
Dec 5 '18 at 18:57












$begingroup$
yes, I do mean the ceiling and the floor!
$endgroup$
– wondering1123
Dec 5 '18 at 19:50




$begingroup$
yes, I do mean the ceiling and the floor!
$endgroup$
– wondering1123
Dec 5 '18 at 19:50


















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