About a surface integral of type $1$
$begingroup$
Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
else $f=0$.
Here is my attempt:
I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.
calculus integration multivariable-calculus
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
asked Dec 26 '18 at 2:50
jacksonjackson
1149
$endgroup$
add a comment |
$begingroup$
Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
else $f=0$.
Here is my attempt:
I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.
calculus integration multivariable-calculus
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
asked Dec 26 '18 at 2:50
jacksonjackson
1149
$endgroup$
add a comment |
$begingroup$
Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
else $f=0$.
Here is my attempt:
I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.
calculus integration multivariable-calculus
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
asked Dec 26 '18 at 2:50
jacksonjackson
1149
$endgroup$
Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
else $f=0$.
Here is my attempt:
I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.
calculus integration multivariable-calculus
calculus integration multivariable-calculus
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
asked Dec 26 '18 at 2:50
jacksonjackson
1149
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
asked Dec 26 '18 at 2:50
jacksonjackson
1149
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
edited Dec 26 '18 at 3:48
Rócherz
2,7762721
2,7762721
asked Dec 26 '18 at 2:50
jacksonjackson
1149
asked Dec 26 '18 at 2:50
jacksonjackson
1149
asked Dec 26 '18 at 2:50
jacksonjackson
1149
1149
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.
Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then
$$
begin{matrix}
u=dfrac{1}{sqrt{6}}(2x-y-z) \
v=dfrac{1}{sqrt{2}}(y-z) \
w=dfrac{1}{sqrt{3}}(x+y+z) \
end{matrix}
$$
As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,
$$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$
Added
Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:
$$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$
$$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$
$$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$
$$=left{begin{matrix}
0&text{if}&t<-sqrt{3}\
2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
0&text{if}&sqrt{3}<t\
end{matrix}right.$$
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
$endgroup$
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
add a comment |
Your Answer
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1 Answer
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$begingroup$
The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.
Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then
$$
begin{matrix}
u=dfrac{1}{sqrt{6}}(2x-y-z) \
v=dfrac{1}{sqrt{2}}(y-z) \
w=dfrac{1}{sqrt{3}}(x+y+z) \
end{matrix}
$$
As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,
$$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$
Added
Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:
$$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$
$$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$
$$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$
$$=left{begin{matrix}
0&text{if}&t<-sqrt{3}\
2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
0&text{if}&sqrt{3}<t\
end{matrix}right.$$
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
$endgroup$
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
add a comment |
$begingroup$
The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.
Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then
$$
begin{matrix}
u=dfrac{1}{sqrt{6}}(2x-y-z) \
v=dfrac{1}{sqrt{2}}(y-z) \
w=dfrac{1}{sqrt{3}}(x+y+z) \
end{matrix}
$$
As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,
$$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$
Added
Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:
$$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$
$$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$
$$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$
$$=left{begin{matrix}
0&text{if}&t<-sqrt{3}\
2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
0&text{if}&sqrt{3}<t\
end{matrix}right.$$
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
$endgroup$
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
add a comment |
$begingroup$
The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.
Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then
$$
begin{matrix}
u=dfrac{1}{sqrt{6}}(2x-y-z) \
v=dfrac{1}{sqrt{2}}(y-z) \
w=dfrac{1}{sqrt{3}}(x+y+z) \
end{matrix}
$$
As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,
$$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$
Added
Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:
$$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$
$$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$
$$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$
$$=left{begin{matrix}
0&text{if}&t<-sqrt{3}\
2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
0&text{if}&sqrt{3}<t\
end{matrix}right.$$
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
$endgroup$
The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.
Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then
$$
begin{matrix}
u=dfrac{1}{sqrt{6}}(2x-y-z) \
v=dfrac{1}{sqrt{2}}(y-z) \
w=dfrac{1}{sqrt{3}}(x+y+z) \
end{matrix}
$$
As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,
$$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$
Added
Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:
$$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$
$$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$
$$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$
$$=left{begin{matrix}
0&text{if}&t<-sqrt{3}\
2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
0&text{if}&sqrt{3}<t\
end{matrix}right.$$
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
edited Dec 27 '18 at 21:51
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
answered Dec 26 '18 at 9:06
Rafa BudríaRafa Budría
5,7151825
5,7151825
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
add a comment |
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
$endgroup$
– jackson
Dec 26 '18 at 10:36
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
$endgroup$
– Rafa Budría
Dec 26 '18 at 10:43
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
$begingroup$
yes thanks a lot
$endgroup$
– jackson
Dec 27 '18 at 16:29
add a comment |
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