Mixing
calculate flux through surface
$begingroup$
I need to calculate the flux of the vector field $vec{F}$ through the surface $D$, where
$$vec{F} = left<z, , y sqrt{x^2 + z^2}, , -x right> \
D = {x^2+6x+z^2le 0 ,| -1le y le 0}.$$
So there should be a cylinder (height on $y$ axis) shifted by 3 units (center $x=-3$), with cut on $ -1le y le 0$.
Cylindrical parametrization:
$$begin{cases}
x=rcostheta -3\
y=y\
z=rsintheta
end{cases}
$$
I think I can do this problem in two ways: The first one by calculating the flux for each of the 3 surfaces (1 cylinder, 2 disks), and the second one by using the divergence theorem.
$$operatorname{div}F = sqrt{x^2+y^2} overset{text{cylindrical}}{underset{text{coordinates}}{=}} sqrt{r^2-9-6rcostheta},$$
so with the divergence theorem, $$int_{0}^{2pi}int_{0}^{3sqrt{3}}int_{-1}^{0}{rsqrt{r^2-9-6rcostheta} , dy , dr , dtheta}.$$
Is this correct?
integration multivariable-calculus vector-analysis surface-integrals
edited 2 days ago
AEngineer
1,483215
asked Jan 12 at 18:18
NPLSNPLS
8812
$endgroup$
add a comment |
$begingroup$
I need to calculate the flux of the vector field $vec{F}$ through the surface $D$, where
$$vec{F} = left<z, , y sqrt{x^2 + z^2}, , -x right> \
D = {x^2+6x+z^2le 0 ,| -1le y le 0}.$$
So there should be a cylinder (height on $y$ axis) shifted by 3 units (center $x=-3$), with cut on $ -1le y le 0$.
Cylindrical parametrization:
$$begin{cases}
x=rcostheta -3\
y=y\
z=rsintheta
end{cases}
$$
I think I can do this problem in two ways: The first one by calculating the flux for each of the 3 surfaces (1 cylinder, 2 disks), and the second one by using the divergence theorem.
$$operatorname{div}F = sqrt{x^2+y^2} overset{text{cylindrical}}{underset{text{coordinates}}{=}} sqrt{r^2-9-6rcostheta},$$
so with the divergence theorem, $$int_{0}^{2pi}int_{0}^{3sqrt{3}}int_{-1}^{0}{rsqrt{r^2-9-6rcostheta} , dy , dr , dtheta}.$$
Is this correct?
integration multivariable-calculus vector-analysis surface-integrals
edited 2 days ago
AEngineer
1,483215
asked Jan 12 at 18:18
NPLSNPLS
8812
$endgroup$
$begingroup$
Why did you write down $;int_0^{-1};$ instead of the obvious (and correct) $;int_{-1}^0;$ ...? Then you get $;18pi;$ . But for this, the rest seems to be correct...
$endgroup$
– DonAntonio
Jan 12 at 18:37
$begingroup$
ohh yep :P , my bad
$endgroup$
– NPLS
Jan 12 at 18:47
add a comment |
$begingroup$
I need to calculate the flux of the vector field $vec{F}$ through the surface $D$, where
$$vec{F} = left<z, , y sqrt{x^2 + z^2}, , -x right> \
D = {x^2+6x+z^2le 0 ,| -1le y le 0}.$$
So there should be a cylinder (height on $y$ axis) shifted by 3 units (center $x=-3$), with cut on $ -1le y le 0$.
Cylindrical parametrization:
$$begin{cases}
x=rcostheta -3\
y=y\
z=rsintheta
end{cases}
$$
I think I can do this problem in two ways: The first one by calculating the flux for each of the 3 surfaces (1 cylinder, 2 disks), and the second one by using the divergence theorem.
$$operatorname{div}F = sqrt{x^2+y^2} overset{text{cylindrical}}{underset{text{coordinates}}{=}} sqrt{r^2-9-6rcostheta},$$
so with the divergence theorem, $$int_{0}^{2pi}int_{0}^{3sqrt{3}}int_{-1}^{0}{rsqrt{r^2-9-6rcostheta} , dy , dr , dtheta}.$$
Is this correct?
integration multivariable-calculus vector-analysis surface-integrals
edited 2 days ago
AEngineer
1,483215
asked Jan 12 at 18:18
NPLSNPLS
8812
$endgroup$
I need to calculate the flux of the vector field $vec{F}$ through the surface $D$, where
$$vec{F} = left<z, , y sqrt{x^2 + z^2}, , -x right> \
D = {x^2+6x+z^2le 0 ,| -1le y le 0}.$$
So there should be a cylinder (height on $y$ axis) shifted by 3 units (center $x=-3$), with cut on $ -1le y le 0$.
Cylindrical parametrization:
$$begin{cases}
x=rcostheta -3\
y=y\
z=rsintheta
end{cases}
$$
I think I can do this problem in two ways: The first one by calculating the flux for each of the 3 surfaces (1 cylinder, 2 disks), and the second one by using the divergence theorem.
$$operatorname{div}F = sqrt{x^2+y^2} overset{text{cylindrical}}{underset{text{coordinates}}{=}} sqrt{r^2-9-6rcostheta},$$
so with the divergence theorem, $$int_{0}^{2pi}int_{0}^{3sqrt{3}}int_{-1}^{0}{rsqrt{r^2-9-6rcostheta} , dy , dr , dtheta}.$$
Is this correct?
integration multivariable-calculus vector-analysis surface-integrals
integration multivariable-calculus vector-analysis surface-integrals
edited 2 days ago
AEngineer
1,483215
asked Jan 12 at 18:18
NPLSNPLS
8812
edited 2 days ago
AEngineer
1,483215
asked Jan 12 at 18:18
NPLSNPLS
8812
edited 2 days ago
AEngineer
1,483215
edited 2 days ago
AEngineer
1,483215
edited 2 days ago
AEngineer
1,483215
1,483215
asked Jan 12 at 18:18
NPLSNPLS
8812
asked Jan 12 at 18:18
NPLSNPLS
8812
asked Jan 12 at 18:18
NPLSNPLS
8812
8812
$begingroup$
Why did you write down $;int_0^{-1};$ instead of the obvious (and correct) $;int_{-1}^0;$ ...? Then you get $;18pi;$ . But for this, the rest seems to be correct...
$endgroup$
– DonAntonio
Jan 12 at 18:37
$begingroup$
ohh yep :P , my bad
$endgroup$
– NPLS
Jan 12 at 18:47
add a comment |
$begingroup$
Why did you write down $;int_0^{-1};$ instead of the obvious (and correct) $;int_{-1}^0;$ ...? Then you get $;18pi;$ . But for this, the rest seems to be correct...
$endgroup$
– DonAntonio
Jan 12 at 18:37
$begingroup$
ohh yep :P , my bad
$endgroup$
– NPLS
Jan 12 at 18:47
$begingroup$
Why did you write down $;int_0^{-1};$ instead of the obvious (and correct) $;int_{-1}^0;$ ...? Then you get $;18pi;$ . But for this, the rest seems to be correct...
$endgroup$
– DonAntonio
Jan 12 at 18:37
$begingroup$
Why did you write down $;int_0^{-1};$ instead of the obvious (and correct) $;int_{-1}^0;$ ...? Then you get $;18pi;$ . But for this, the rest seems to be correct...
$endgroup$
– DonAntonio
Jan 12 at 18:37
$begingroup$
ohh yep :P , my bad
$endgroup$
– NPLS
Jan 12 at 18:47
$begingroup$
ohh yep :P , my bad
$endgroup$
– NPLS
Jan 12 at 18:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not exactly sure where the $3sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flux of a vector field $vec{mathbf{V}}$ through a surface $Sigma$ be denoted $Phi$ and defined
$$
Phi := iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma.
$$
The vector $mathbf{hat{n}}$ is the unit outward normal to the surface $Sigma$. Suppose $Sigma$ is given by $z = f(x,y).$ Let $mathbf{vec{r}}(x,y)$ trace $Sigma$ such that
$$
mathbf{vec{r}}(x,y) = begin{pmatrix}
x \ y \ f(x,y)
end{pmatrix}.
$$
Then the unit normal $mathbf{vec{n}}$ is given by
$$
mathbf{vec{n}} = frac{mathbf{vec{r}}_x times mathbf{vec{r}}_y}{|| mathbf{vec{r}}_x times mathbf{vec{r}}_y ||} =
frac{1}{sqrt{f_x^{,2} + f_y^{,2} + 1}}
begin{pmatrix}
-f_x \
-f_y \
1
end{pmatrix}.
$$
So given that $mathbf{vec{V}} = u(x,y,z) mathbf{hat{i}} + v(x,y,z) mathbf{hat{j}} + w(x,y,z) mathbf{hat{k}}$, the corresponding flux of $mathbf{vec{V}}$ through $Sigma$ is
$$
Phi = iint_Sigma frac{-uf_x - vf_y + w}{sqrt{f_x^{,2} + f_y^{,2} + 1}} , dsigma.
$$
For the given field, we have
$$mathbf{vec{V}} = z mathbf{hat{i}} + y sqrt{x^2 + z^2} mathbf{hat{j}} - x mathbf{hat{k}},
$$
and the surface $Sigma$ is given such that $(x + 3)^2 + z^2 = 9 forall y in (-1,0).$
Thus we choose to trace the surface of the cylinder with
$$
mathbf{vec{r}}(x,z) =
begin{pmatrix}
x \
(x + 3)^2 + z^2 - 9 \
z
end{pmatrix},
$$
where the unit outward normal on the cylinder is
$$mathbf{hat{n}} = frac{1}{sqrt{4(x+3)^2 + 4z^2 + 1}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
=
frac{1}{sqrt{37}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
.
$$
The flux is then
begin{align}
Phi &= underbrace{iint_{Sigma_1} mathbf{vec{V}} cdot begin{pmatrix}
0 \ -1 \ 0
end{pmatrix} , dsigma}_{y = -1} + overbrace{iint_{Sigma_2} mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma}^{mathbf{vec{V}}_2text{ cannot contribute}} + underbrace{iint_{Sigma_3} mathbf{vec{V}} cdot begin{pmatrix}
0 \ 1 \ 0
end{pmatrix} , dsigma}_{text{nothing since }y = 0} \
&= iint_{Sigma_1} sqrt{x^2 + z^2} , dsigma + frac{1}{sqrt{37}}iint_{Sigma_2} -2z(x+3) + ysqrt{x^2 + z^2} + 2xz , dsigma \
&= int_0^{2pi}int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta - frac{6}{sqrt{37}} int_0^3 r^2 , dr , underbrace{int_0^{2pi} sintheta , dtheta}_{0} \
&= 96.
end{align}
(2) Divergence Theorem
Technical information about this method can also be found in MIT's open notes, and no visualization from the site this time, but the divergence theorem here is explained in casual language. Given everything is nice, the flux of the field through the surface is
$$
iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma = iiint_M nabla cdot mathbf{vec{V}} , dV,
$$
where $M$ is the bounded region contained within $Sigma$.
Applying it to this problem, the divergence theorem takes us straight to the end result of the direct approach.
begin{align}
Phi &= iint_Sigma nabla cdot mathbf{vec{V}} , dV \
&= iiint_M sqrt{x^2 + z^2} , dV \
&= int_{-1}^0 int_0^{2pi} int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta , dy \
&= 96.
end{align}
From the comments, $18pi$ falls out as the solution if $x$ is not properly shifted over from the origin. Also, Wikipedia is a fairly good source for this material as well.
answered yesterday
AEngineerAEngineer
1,483215
$endgroup$
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
add a comment |
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1 Answer
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votes
$begingroup$
I'm not exactly sure where the $3sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flux of a vector field $vec{mathbf{V}}$ through a surface $Sigma$ be denoted $Phi$ and defined
$$
Phi := iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma.
$$
The vector $mathbf{hat{n}}$ is the unit outward normal to the surface $Sigma$. Suppose $Sigma$ is given by $z = f(x,y).$ Let $mathbf{vec{r}}(x,y)$ trace $Sigma$ such that
$$
mathbf{vec{r}}(x,y) = begin{pmatrix}
x \ y \ f(x,y)
end{pmatrix}.
$$
Then the unit normal $mathbf{vec{n}}$ is given by
$$
mathbf{vec{n}} = frac{mathbf{vec{r}}_x times mathbf{vec{r}}_y}{|| mathbf{vec{r}}_x times mathbf{vec{r}}_y ||} =
frac{1}{sqrt{f_x^{,2} + f_y^{,2} + 1}}
begin{pmatrix}
-f_x \
-f_y \
1
end{pmatrix}.
$$
So given that $mathbf{vec{V}} = u(x,y,z) mathbf{hat{i}} + v(x,y,z) mathbf{hat{j}} + w(x,y,z) mathbf{hat{k}}$, the corresponding flux of $mathbf{vec{V}}$ through $Sigma$ is
$$
Phi = iint_Sigma frac{-uf_x - vf_y + w}{sqrt{f_x^{,2} + f_y^{,2} + 1}} , dsigma.
$$
For the given field, we have
$$mathbf{vec{V}} = z mathbf{hat{i}} + y sqrt{x^2 + z^2} mathbf{hat{j}} - x mathbf{hat{k}},
$$
and the surface $Sigma$ is given such that $(x + 3)^2 + z^2 = 9 forall y in (-1,0).$
Thus we choose to trace the surface of the cylinder with
$$
mathbf{vec{r}}(x,z) =
begin{pmatrix}
x \
(x + 3)^2 + z^2 - 9 \
z
end{pmatrix},
$$
where the unit outward normal on the cylinder is
$$mathbf{hat{n}} = frac{1}{sqrt{4(x+3)^2 + 4z^2 + 1}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
=
frac{1}{sqrt{37}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
.
$$
The flux is then
begin{align}
Phi &= underbrace{iint_{Sigma_1} mathbf{vec{V}} cdot begin{pmatrix}
0 \ -1 \ 0
end{pmatrix} , dsigma}_{y = -1} + overbrace{iint_{Sigma_2} mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma}^{mathbf{vec{V}}_2text{ cannot contribute}} + underbrace{iint_{Sigma_3} mathbf{vec{V}} cdot begin{pmatrix}
0 \ 1 \ 0
end{pmatrix} , dsigma}_{text{nothing since }y = 0} \
&= iint_{Sigma_1} sqrt{x^2 + z^2} , dsigma + frac{1}{sqrt{37}}iint_{Sigma_2} -2z(x+3) + ysqrt{x^2 + z^2} + 2xz , dsigma \
&= int_0^{2pi}int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta - frac{6}{sqrt{37}} int_0^3 r^2 , dr , underbrace{int_0^{2pi} sintheta , dtheta}_{0} \
&= 96.
end{align}
(2) Divergence Theorem
Technical information about this method can also be found in MIT's open notes, and no visualization from the site this time, but the divergence theorem here is explained in casual language. Given everything is nice, the flux of the field through the surface is
$$
iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma = iiint_M nabla cdot mathbf{vec{V}} , dV,
$$
where $M$ is the bounded region contained within $Sigma$.
Applying it to this problem, the divergence theorem takes us straight to the end result of the direct approach.
begin{align}
Phi &= iint_Sigma nabla cdot mathbf{vec{V}} , dV \
&= iiint_M sqrt{x^2 + z^2} , dV \
&= int_{-1}^0 int_0^{2pi} int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta , dy \
&= 96.
end{align}
From the comments, $18pi$ falls out as the solution if $x$ is not properly shifted over from the origin. Also, Wikipedia is a fairly good source for this material as well.
answered yesterday
AEngineerAEngineer
1,483215
$endgroup$
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
add a comment |
$begingroup$
I'm not exactly sure where the $3sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flux of a vector field $vec{mathbf{V}}$ through a surface $Sigma$ be denoted $Phi$ and defined
$$
Phi := iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma.
$$
The vector $mathbf{hat{n}}$ is the unit outward normal to the surface $Sigma$. Suppose $Sigma$ is given by $z = f(x,y).$ Let $mathbf{vec{r}}(x,y)$ trace $Sigma$ such that
$$
mathbf{vec{r}}(x,y) = begin{pmatrix}
x \ y \ f(x,y)
end{pmatrix}.
$$
Then the unit normal $mathbf{vec{n}}$ is given by
$$
mathbf{vec{n}} = frac{mathbf{vec{r}}_x times mathbf{vec{r}}_y}{|| mathbf{vec{r}}_x times mathbf{vec{r}}_y ||} =
frac{1}{sqrt{f_x^{,2} + f_y^{,2} + 1}}
begin{pmatrix}
-f_x \
-f_y \
1
end{pmatrix}.
$$
So given that $mathbf{vec{V}} = u(x,y,z) mathbf{hat{i}} + v(x,y,z) mathbf{hat{j}} + w(x,y,z) mathbf{hat{k}}$, the corresponding flux of $mathbf{vec{V}}$ through $Sigma$ is
$$
Phi = iint_Sigma frac{-uf_x - vf_y + w}{sqrt{f_x^{,2} + f_y^{,2} + 1}} , dsigma.
$$
For the given field, we have
$$mathbf{vec{V}} = z mathbf{hat{i}} + y sqrt{x^2 + z^2} mathbf{hat{j}} - x mathbf{hat{k}},
$$
and the surface $Sigma$ is given such that $(x + 3)^2 + z^2 = 9 forall y in (-1,0).$
Thus we choose to trace the surface of the cylinder with
$$
mathbf{vec{r}}(x,z) =
begin{pmatrix}
x \
(x + 3)^2 + z^2 - 9 \
z
end{pmatrix},
$$
where the unit outward normal on the cylinder is
$$mathbf{hat{n}} = frac{1}{sqrt{4(x+3)^2 + 4z^2 + 1}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
=
frac{1}{sqrt{37}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
.
$$
The flux is then
begin{align}
Phi &= underbrace{iint_{Sigma_1} mathbf{vec{V}} cdot begin{pmatrix}
0 \ -1 \ 0
end{pmatrix} , dsigma}_{y = -1} + overbrace{iint_{Sigma_2} mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma}^{mathbf{vec{V}}_2text{ cannot contribute}} + underbrace{iint_{Sigma_3} mathbf{vec{V}} cdot begin{pmatrix}
0 \ 1 \ 0
end{pmatrix} , dsigma}_{text{nothing since }y = 0} \
&= iint_{Sigma_1} sqrt{x^2 + z^2} , dsigma + frac{1}{sqrt{37}}iint_{Sigma_2} -2z(x+3) + ysqrt{x^2 + z^2} + 2xz , dsigma \
&= int_0^{2pi}int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta - frac{6}{sqrt{37}} int_0^3 r^2 , dr , underbrace{int_0^{2pi} sintheta , dtheta}_{0} \
&= 96.
end{align}
(2) Divergence Theorem
Technical information about this method can also be found in MIT's open notes, and no visualization from the site this time, but the divergence theorem here is explained in casual language. Given everything is nice, the flux of the field through the surface is
$$
iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma = iiint_M nabla cdot mathbf{vec{V}} , dV,
$$
where $M$ is the bounded region contained within $Sigma$.
Applying it to this problem, the divergence theorem takes us straight to the end result of the direct approach.
begin{align}
Phi &= iint_Sigma nabla cdot mathbf{vec{V}} , dV \
&= iiint_M sqrt{x^2 + z^2} , dV \
&= int_{-1}^0 int_0^{2pi} int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta , dy \
&= 96.
end{align}
From the comments, $18pi$ falls out as the solution if $x$ is not properly shifted over from the origin. Also, Wikipedia is a fairly good source for this material as well.
answered yesterday
AEngineerAEngineer
1,483215
$endgroup$
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
add a comment |
$begingroup$
I'm not exactly sure where the $3sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flux of a vector field $vec{mathbf{V}}$ through a surface $Sigma$ be denoted $Phi$ and defined
$$
Phi := iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma.
$$
The vector $mathbf{hat{n}}$ is the unit outward normal to the surface $Sigma$. Suppose $Sigma$ is given by $z = f(x,y).$ Let $mathbf{vec{r}}(x,y)$ trace $Sigma$ such that
$$
mathbf{vec{r}}(x,y) = begin{pmatrix}
x \ y \ f(x,y)
end{pmatrix}.
$$
Then the unit normal $mathbf{vec{n}}$ is given by
$$
mathbf{vec{n}} = frac{mathbf{vec{r}}_x times mathbf{vec{r}}_y}{|| mathbf{vec{r}}_x times mathbf{vec{r}}_y ||} =
frac{1}{sqrt{f_x^{,2} + f_y^{,2} + 1}}
begin{pmatrix}
-f_x \
-f_y \
1
end{pmatrix}.
$$
So given that $mathbf{vec{V}} = u(x,y,z) mathbf{hat{i}} + v(x,y,z) mathbf{hat{j}} + w(x,y,z) mathbf{hat{k}}$, the corresponding flux of $mathbf{vec{V}}$ through $Sigma$ is
$$
Phi = iint_Sigma frac{-uf_x - vf_y + w}{sqrt{f_x^{,2} + f_y^{,2} + 1}} , dsigma.
$$
For the given field, we have
$$mathbf{vec{V}} = z mathbf{hat{i}} + y sqrt{x^2 + z^2} mathbf{hat{j}} - x mathbf{hat{k}},
$$
and the surface $Sigma$ is given such that $(x + 3)^2 + z^2 = 9 forall y in (-1,0).$
Thus we choose to trace the surface of the cylinder with
$$
mathbf{vec{r}}(x,z) =
begin{pmatrix}
x \
(x + 3)^2 + z^2 - 9 \
z
end{pmatrix},
$$
where the unit outward normal on the cylinder is
$$mathbf{hat{n}} = frac{1}{sqrt{4(x+3)^2 + 4z^2 + 1}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
=
frac{1}{sqrt{37}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
.
$$
The flux is then
begin{align}
Phi &= underbrace{iint_{Sigma_1} mathbf{vec{V}} cdot begin{pmatrix}
0 \ -1 \ 0
end{pmatrix} , dsigma}_{y = -1} + overbrace{iint_{Sigma_2} mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma}^{mathbf{vec{V}}_2text{ cannot contribute}} + underbrace{iint_{Sigma_3} mathbf{vec{V}} cdot begin{pmatrix}
0 \ 1 \ 0
end{pmatrix} , dsigma}_{text{nothing since }y = 0} \
&= iint_{Sigma_1} sqrt{x^2 + z^2} , dsigma + frac{1}{sqrt{37}}iint_{Sigma_2} -2z(x+3) + ysqrt{x^2 + z^2} + 2xz , dsigma \
&= int_0^{2pi}int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta - frac{6}{sqrt{37}} int_0^3 r^2 , dr , underbrace{int_0^{2pi} sintheta , dtheta}_{0} \
&= 96.
end{align}
(2) Divergence Theorem
Technical information about this method can also be found in MIT's open notes, and no visualization from the site this time, but the divergence theorem here is explained in casual language. Given everything is nice, the flux of the field through the surface is
$$
iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma = iiint_M nabla cdot mathbf{vec{V}} , dV,
$$
where $M$ is the bounded region contained within $Sigma$.
Applying it to this problem, the divergence theorem takes us straight to the end result of the direct approach.
begin{align}
Phi &= iint_Sigma nabla cdot mathbf{vec{V}} , dV \
&= iiint_M sqrt{x^2 + z^2} , dV \
&= int_{-1}^0 int_0^{2pi} int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta , dy \
&= 96.
end{align}
From the comments, $18pi$ falls out as the solution if $x$ is not properly shifted over from the origin. Also, Wikipedia is a fairly good source for this material as well.
answered yesterday
AEngineerAEngineer
1,483215
$endgroup$
I'm not exactly sure where the $3sqrt{3}$ comes from in your result, but there is indeed more than one way to evaluate this problem.
(1) Direct method
Here is some technical information about this method from MIT's open notes, and some visualization for what the flux of a vector field through a surface is. Let the flux of a vector field $vec{mathbf{V}}$ through a surface $Sigma$ be denoted $Phi$ and defined
$$
Phi := iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma.
$$
The vector $mathbf{hat{n}}$ is the unit outward normal to the surface $Sigma$. Suppose $Sigma$ is given by $z = f(x,y).$ Let $mathbf{vec{r}}(x,y)$ trace $Sigma$ such that
$$
mathbf{vec{r}}(x,y) = begin{pmatrix}
x \ y \ f(x,y)
end{pmatrix}.
$$
Then the unit normal $mathbf{vec{n}}$ is given by
$$
mathbf{vec{n}} = frac{mathbf{vec{r}}_x times mathbf{vec{r}}_y}{|| mathbf{vec{r}}_x times mathbf{vec{r}}_y ||} =
frac{1}{sqrt{f_x^{,2} + f_y^{,2} + 1}}
begin{pmatrix}
-f_x \
-f_y \
1
end{pmatrix}.
$$
So given that $mathbf{vec{V}} = u(x,y,z) mathbf{hat{i}} + v(x,y,z) mathbf{hat{j}} + w(x,y,z) mathbf{hat{k}}$, the corresponding flux of $mathbf{vec{V}}$ through $Sigma$ is
$$
Phi = iint_Sigma frac{-uf_x - vf_y + w}{sqrt{f_x^{,2} + f_y^{,2} + 1}} , dsigma.
$$
For the given field, we have
$$mathbf{vec{V}} = z mathbf{hat{i}} + y sqrt{x^2 + z^2} mathbf{hat{j}} - x mathbf{hat{k}},
$$
and the surface $Sigma$ is given such that $(x + 3)^2 + z^2 = 9 forall y in (-1,0).$
Thus we choose to trace the surface of the cylinder with
$$
mathbf{vec{r}}(x,z) =
begin{pmatrix}
x \
(x + 3)^2 + z^2 - 9 \
z
end{pmatrix},
$$
where the unit outward normal on the cylinder is
$$mathbf{hat{n}} = frac{1}{sqrt{4(x+3)^2 + 4z^2 + 1}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
=
frac{1}{sqrt{37}}
begin{pmatrix}
-2(x+3) \
1 \
-2z
end{pmatrix}
.
$$
The flux is then
begin{align}
Phi &= underbrace{iint_{Sigma_1} mathbf{vec{V}} cdot begin{pmatrix}
0 \ -1 \ 0
end{pmatrix} , dsigma}_{y = -1} + overbrace{iint_{Sigma_2} mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma}^{mathbf{vec{V}}_2text{ cannot contribute}} + underbrace{iint_{Sigma_3} mathbf{vec{V}} cdot begin{pmatrix}
0 \ 1 \ 0
end{pmatrix} , dsigma}_{text{nothing since }y = 0} \
&= iint_{Sigma_1} sqrt{x^2 + z^2} , dsigma + frac{1}{sqrt{37}}iint_{Sigma_2} -2z(x+3) + ysqrt{x^2 + z^2} + 2xz , dsigma \
&= int_0^{2pi}int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta - frac{6}{sqrt{37}} int_0^3 r^2 , dr , underbrace{int_0^{2pi} sintheta , dtheta}_{0} \
&= 96.
end{align}
(2) Divergence Theorem
Technical information about this method can also be found in MIT's open notes, and no visualization from the site this time, but the divergence theorem here is explained in casual language. Given everything is nice, the flux of the field through the surface is
$$
iint_Sigma mathbf{vec{V}} cdot mathbf{hat{n}} , dsigma = iiint_M nabla cdot mathbf{vec{V}} , dV,
$$
where $M$ is the bounded region contained within $Sigma$.
Applying it to this problem, the divergence theorem takes us straight to the end result of the direct approach.
begin{align}
Phi &= iint_Sigma nabla cdot mathbf{vec{V}} , dV \
&= iiint_M sqrt{x^2 + z^2} , dV \
&= int_{-1}^0 int_0^{2pi} int_0^3 rsqrt{(r costheta - 3)^2 + (r sintheta)^2} , dr , dtheta , dy \
&= 96.
end{align}
From the comments, $18pi$ falls out as the solution if $x$ is not properly shifted over from the origin. Also, Wikipedia is a fairly good source for this material as well.
answered yesterday
AEngineerAEngineer
1,483215
answered yesterday
AEngineerAEngineer
1,483215
answered yesterday
AEngineerAEngineer
1,483215
answered yesterday
AEngineerAEngineer
1,483215
1,483215
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
add a comment |
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
I don't understand why is it from 0 to $2pi$. The cylinder is shifted (and tangent to the y-z plane) so It's not centred in the origin (if so, that would be from $0$ to $2pi$). So with this thought, the angle should be from $0$ to $pi$. Could you explain how do you set the angle range ?
$endgroup$
– NPLS
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
We're essentially shifting the perspective of where the origin is in the $xz$ plane by saying $x = r costheta - 3.$ The Cartesian limits on the integrals would properly be, with no shifts and in order from outside to inside, $y in (-1,0)$, $x in (-6,0)$, $z in (-sqrt{9 - (x + 3)^2},sqrt{9 - (x + 3)^2}).$ So if we go to shifted polar coordinates $x = r costheta - 3$ and $z = r sintheta,$ these limits on $x$ and $z$ are only achieved if $r in (0, 3)$ and $theta in (0, 2pi).$
$endgroup$
– AEngineer
yesterday
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
ohh, Eureka! you made me understand! Now everything is at the right place. I know how to simplify the integral, without using the shifted polar coordinates It's much more simple. The Flux is also equal to $int_{pi/2}^{3/2pi}int_{0}^{-6cos(theta)}int_{-1}^{0}r^2dydrdtheta=96$!
$endgroup$
– NPLS
22 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
$begingroup$
That's it, you got it! Unless you mean 96 factorial - then we might have a little problem on our hands, ha. Anyways, glad I could help.
$endgroup$
– AEngineer
20 hours ago
add a comment |
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$begingroup$
Why did you write down $;int_0^{-1};$ instead of the obvious (and correct) $;int_{-1}^0;$ ...? Then you get $;18pi;$ . But for this, the rest seems to be correct...
$endgroup$
– DonAntonio
Jan 12 at 18:37
$begingroup$
ohh yep :P , my bad
$endgroup$
– NPLS
Jan 12 at 18:47