Coordination change for integrals and determinant of the jacobian matrix
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so I am currently learning about flow integrals and my book had an example which confused me. (Example 1) As you see, in Example 1, they don't care about the determinant of the jacobian of the coordination change. They just switch coordinats without transforming the integral properly. In Example 2 they actually use the determinant of the jacobian of the coordination change.
Let me make two examples:
Example 1:
Calculate the flow of $v=(0,0,1-z)$ from bottom up through
$$H={(x,y,z)in mathbb R^3 | x^2+y^2+z^2=1, z>0}$$
Parametrization:
$$Phi:[0,2pi]times[0,pi/2]tomathbb R^3, quad (u,v)mapsto begin{pmatrix}sin vcos u \ sin vsin u\ cos vend{pmatrix}$$
Normal vector:
$$Phi_vtimes Phi_u = begin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$$
Flow integraL:
$int_H vcdot ndo = int_0^{2pi}duint_0^{pi/2}dv begin{pmatrix}0\0\1-cos vend{pmatrix}cdotbegin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$
$=2piint_0^{pi/2}dv sin v cos v(1-cos v)$
$=2piint_0^{pi/2}dv sin v cos v - 2pi int_0^{pi/2}dv sin v cos^2 v=...=pi/3$
Example 2: We want to calculate the flow of $v=(xz,z,y)$ through the unit ball, centered around $O=(0,0,0)$ using Gauss's Divergene Theorem.
We have to calculate $int_S vcdot n do = int_{B_1} div(v) dmu$
$div(v)=z$
We use spherical coordinates. So: $dxdydz=r^2drsinphi dphi dvarphi$
$int_s v cdot n do = int_{B_1} zdmu = int_0^1 r^2 dr int_0^pi sinphi dphi int_0^{2pi}dvarphi cdot rcos phi=...=0$
calculus integration
asked Dec 31 '18 at 15:23
xotixxotix
291411
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add a comment |
$begingroup$
so I am currently learning about flow integrals and my book had an example which confused me. (Example 1) As you see, in Example 1, they don't care about the determinant of the jacobian of the coordination change. They just switch coordinats without transforming the integral properly. In Example 2 they actually use the determinant of the jacobian of the coordination change.
Let me make two examples:
Example 1:
Calculate the flow of $v=(0,0,1-z)$ from bottom up through
$$H={(x,y,z)in mathbb R^3 | x^2+y^2+z^2=1, z>0}$$
Parametrization:
$$Phi:[0,2pi]times[0,pi/2]tomathbb R^3, quad (u,v)mapsto begin{pmatrix}sin vcos u \ sin vsin u\ cos vend{pmatrix}$$
Normal vector:
$$Phi_vtimes Phi_u = begin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$$
Flow integraL:
$int_H vcdot ndo = int_0^{2pi}duint_0^{pi/2}dv begin{pmatrix}0\0\1-cos vend{pmatrix}cdotbegin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$
$=2piint_0^{pi/2}dv sin v cos v(1-cos v)$
$=2piint_0^{pi/2}dv sin v cos v - 2pi int_0^{pi/2}dv sin v cos^2 v=...=pi/3$
Example 2: We want to calculate the flow of $v=(xz,z,y)$ through the unit ball, centered around $O=(0,0,0)$ using Gauss's Divergene Theorem.
We have to calculate $int_S vcdot n do = int_{B_1} div(v) dmu$
$div(v)=z$
We use spherical coordinates. So: $dxdydz=r^2drsinphi dphi dvarphi$
$int_s v cdot n do = int_{B_1} zdmu = int_0^1 r^2 dr int_0^pi sinphi dphi int_0^{2pi}dvarphi cdot rcos phi=...=0$
calculus integration
asked Dec 31 '18 at 15:23
xotixxotix
291411
$endgroup$
1
$begingroup$
The jacobian is absorbed into $Phi_vtimes Phi_u$. If you go through the derivation of the formula, you will see where it arises
$endgroup$
– Matematleta
Dec 31 '18 at 15:45
$begingroup$
Yeah I just went through everything again and the normal vector$n$ absorbs it. So if we calculate a given flow using Gauss's Divergence Theorem, we actually have to take it into account but if we use the normal definition, we don't - right?
$endgroup$
– xotix
Dec 31 '18 at 15:59
add a comment |
$begingroup$
so I am currently learning about flow integrals and my book had an example which confused me. (Example 1) As you see, in Example 1, they don't care about the determinant of the jacobian of the coordination change. They just switch coordinats without transforming the integral properly. In Example 2 they actually use the determinant of the jacobian of the coordination change.
Let me make two examples:
Example 1:
Calculate the flow of $v=(0,0,1-z)$ from bottom up through
$$H={(x,y,z)in mathbb R^3 | x^2+y^2+z^2=1, z>0}$$
Parametrization:
$$Phi:[0,2pi]times[0,pi/2]tomathbb R^3, quad (u,v)mapsto begin{pmatrix}sin vcos u \ sin vsin u\ cos vend{pmatrix}$$
Normal vector:
$$Phi_vtimes Phi_u = begin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$$
Flow integraL:
$int_H vcdot ndo = int_0^{2pi}duint_0^{pi/2}dv begin{pmatrix}0\0\1-cos vend{pmatrix}cdotbegin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$
$=2piint_0^{pi/2}dv sin v cos v(1-cos v)$
$=2piint_0^{pi/2}dv sin v cos v - 2pi int_0^{pi/2}dv sin v cos^2 v=...=pi/3$
Example 2: We want to calculate the flow of $v=(xz,z,y)$ through the unit ball, centered around $O=(0,0,0)$ using Gauss's Divergene Theorem.
We have to calculate $int_S vcdot n do = int_{B_1} div(v) dmu$
$div(v)=z$
We use spherical coordinates. So: $dxdydz=r^2drsinphi dphi dvarphi$
$int_s v cdot n do = int_{B_1} zdmu = int_0^1 r^2 dr int_0^pi sinphi dphi int_0^{2pi}dvarphi cdot rcos phi=...=0$
calculus integration
asked Dec 31 '18 at 15:23
xotixxotix
291411
$endgroup$
so I am currently learning about flow integrals and my book had an example which confused me. (Example 1) As you see, in Example 1, they don't care about the determinant of the jacobian of the coordination change. They just switch coordinats without transforming the integral properly. In Example 2 they actually use the determinant of the jacobian of the coordination change.
Let me make two examples:
Example 1:
Calculate the flow of $v=(0,0,1-z)$ from bottom up through
$$H={(x,y,z)in mathbb R^3 | x^2+y^2+z^2=1, z>0}$$
Parametrization:
$$Phi:[0,2pi]times[0,pi/2]tomathbb R^3, quad (u,v)mapsto begin{pmatrix}sin vcos u \ sin vsin u\ cos vend{pmatrix}$$
Normal vector:
$$Phi_vtimes Phi_u = begin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$$
Flow integraL:
$int_H vcdot ndo = int_0^{2pi}duint_0^{pi/2}dv begin{pmatrix}0\0\1-cos vend{pmatrix}cdotbegin{pmatrix}sin^2 vcos u \ sin vsin^2 u\ sin v cos vend{pmatrix}$
$=2piint_0^{pi/2}dv sin v cos v(1-cos v)$
$=2piint_0^{pi/2}dv sin v cos v - 2pi int_0^{pi/2}dv sin v cos^2 v=...=pi/3$
Example 2: We want to calculate the flow of $v=(xz,z,y)$ through the unit ball, centered around $O=(0,0,0)$ using Gauss's Divergene Theorem.
We have to calculate $int_S vcdot n do = int_{B_1} div(v) dmu$
$div(v)=z$
We use spherical coordinates. So: $dxdydz=r^2drsinphi dphi dvarphi$
$int_s v cdot n do = int_{B_1} zdmu = int_0^1 r^2 dr int_0^pi sinphi dphi int_0^{2pi}dvarphi cdot rcos phi=...=0$
calculus integration
calculus integration
asked Dec 31 '18 at 15:23
xotixxotix
291411
asked Dec 31 '18 at 15:23
xotixxotix
291411
asked Dec 31 '18 at 15:23
xotixxotix
291411
asked Dec 31 '18 at 15:23
xotixxotix
291411
asked Dec 31 '18 at 15:23
xotixxotix
291411
291411
1
$begingroup$
The jacobian is absorbed into $Phi_vtimes Phi_u$. If you go through the derivation of the formula, you will see where it arises
$endgroup$
– Matematleta
Dec 31 '18 at 15:45
$begingroup$
Yeah I just went through everything again and the normal vector$n$ absorbs it. So if we calculate a given flow using Gauss's Divergence Theorem, we actually have to take it into account but if we use the normal definition, we don't - right?
$endgroup$
– xotix
Dec 31 '18 at 15:59
add a comment |
1
$begingroup$
The jacobian is absorbed into $Phi_vtimes Phi_u$. If you go through the derivation of the formula, you will see where it arises
$endgroup$
– Matematleta
Dec 31 '18 at 15:45
$begingroup$
Yeah I just went through everything again and the normal vector$n$ absorbs it. So if we calculate a given flow using Gauss's Divergence Theorem, we actually have to take it into account but if we use the normal definition, we don't - right?
$endgroup$
– xotix
Dec 31 '18 at 15:59
1
1
$begingroup$
The jacobian is absorbed into $Phi_vtimes Phi_u$. If you go through the derivation of the formula, you will see where it arises
$endgroup$
– Matematleta
Dec 31 '18 at 15:45
$begingroup$
The jacobian is absorbed into $Phi_vtimes Phi_u$. If you go through the derivation of the formula, you will see where it arises
$endgroup$
– Matematleta
Dec 31 '18 at 15:45
$begingroup$
Yeah I just went through everything again and the normal vector$n$ absorbs it. So if we calculate a given flow using Gauss's Divergence Theorem, we actually have to take it into account but if we use the normal definition, we don't - right?
$endgroup$
– xotix
Dec 31 '18 at 15:59
$begingroup$
Yeah I just went through everything again and the normal vector$n$ absorbs it. So if we calculate a given flow using Gauss's Divergence Theorem, we actually have to take it into account but if we use the normal definition, we don't - right?
$endgroup$
– xotix
Dec 31 '18 at 15:59
add a comment |
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1
$begingroup$
The jacobian is absorbed into $Phi_vtimes Phi_u$. If you go through the derivation of the formula, you will see where it arises
$endgroup$
– Matematleta
Dec 31 '18 at 15:45
$begingroup$
Yeah I just went through everything again and the normal vector$n$ absorbs it. So if we calculate a given flow using Gauss's Divergence Theorem, we actually have to take it into account but if we use the normal definition, we don't - right?
$endgroup$
– xotix
Dec 31 '18 at 15:59