Mixing
How to calculate flux of vector field
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A vector field is given as $A = (yz, xz, xy)$ through surface $x+y+z=1$ where $x,y,z ge 0$, normal is chosen to be $hat{n} cdot e_z > 0$. Calculate the flux of the vector field.
I tried using Gauss theorem $ iint_S A cdot hat{n}dS = iiint_D nabla cdot A dV $, but $nabla cdot A $ gave the result of $0$, so I'm unsure how to tackle this problem. Any clues are welcome!
calculus integration multivariable-calculus surface-integrals divergence
asked Jan 15 at 19:38
user3221454user3221454
174
$endgroup$
add a comment |
$begingroup$
A vector field is given as $A = (yz, xz, xy)$ through surface $x+y+z=1$ where $x,y,z ge 0$, normal is chosen to be $hat{n} cdot e_z > 0$. Calculate the flux of the vector field.
I tried using Gauss theorem $ iint_S A cdot hat{n}dS = iiint_D nabla cdot A dV $, but $nabla cdot A $ gave the result of $0$, so I'm unsure how to tackle this problem. Any clues are welcome!
calculus integration multivariable-calculus surface-integrals divergence
asked Jan 15 at 19:38
user3221454user3221454
174
$endgroup$
$begingroup$
take the divergence of the function
$endgroup$
– user29418
Jan 15 at 19:42
$begingroup$
Gauss' theorem can only be used over closed surfaces. That doesn't mean that you can't use it, but if you do, you will need to find the flux across the surfaces that close up the volume.
$endgroup$
– Doug M
Jan 15 at 19:45
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You can calculate the flux passing through the surface.
$endgroup$
– copper.hat
Jan 15 at 19:49
$begingroup$
If you're going to use divergence, you'd best compute it correctly. It's a scalar function.
$endgroup$
– Ted Shifrin
Jan 15 at 19:53
add a comment |
$begingroup$
A vector field is given as $A = (yz, xz, xy)$ through surface $x+y+z=1$ where $x,y,z ge 0$, normal is chosen to be $hat{n} cdot e_z > 0$. Calculate the flux of the vector field.
I tried using Gauss theorem $ iint_S A cdot hat{n}dS = iiint_D nabla cdot A dV $, but $nabla cdot A $ gave the result of $0$, so I'm unsure how to tackle this problem. Any clues are welcome!
calculus integration multivariable-calculus surface-integrals divergence
asked Jan 15 at 19:38
user3221454user3221454
174
$endgroup$
A vector field is given as $A = (yz, xz, xy)$ through surface $x+y+z=1$ where $x,y,z ge 0$, normal is chosen to be $hat{n} cdot e_z > 0$. Calculate the flux of the vector field.
I tried using Gauss theorem $ iint_S A cdot hat{n}dS = iiint_D nabla cdot A dV $, but $nabla cdot A $ gave the result of $0$, so I'm unsure how to tackle this problem. Any clues are welcome!
calculus integration multivariable-calculus surface-integrals divergence
calculus integration multivariable-calculus surface-integrals divergence
asked Jan 15 at 19:38
user3221454user3221454
174
asked Jan 15 at 19:38
user3221454user3221454
174
edited Jan 15 at 19:56
user3221454
asked Jan 15 at 19:38
user3221454user3221454
174
asked Jan 15 at 19:38
user3221454user3221454
174
asked Jan 15 at 19:38
user3221454user3221454
174
174
$begingroup$
take the divergence of the function
$endgroup$
– user29418
Jan 15 at 19:42
$begingroup$
Gauss' theorem can only be used over closed surfaces. That doesn't mean that you can't use it, but if you do, you will need to find the flux across the surfaces that close up the volume.
$endgroup$
– Doug M
Jan 15 at 19:45
$begingroup$
You can calculate the flux passing through the surface.
$endgroup$
– copper.hat
Jan 15 at 19:49
$begingroup$
If you're going to use divergence, you'd best compute it correctly. It's a scalar function.
$endgroup$
– Ted Shifrin
Jan 15 at 19:53
add a comment |
$begingroup$
take the divergence of the function
$endgroup$
– user29418
Jan 15 at 19:42
$begingroup$
Gauss' theorem can only be used over closed surfaces. That doesn't mean that you can't use it, but if you do, you will need to find the flux across the surfaces that close up the volume.
$endgroup$
– Doug M
Jan 15 at 19:45
$begingroup$
You can calculate the flux passing through the surface.
$endgroup$
– copper.hat
Jan 15 at 19:49
$begingroup$
If you're going to use divergence, you'd best compute it correctly. It's a scalar function.
$endgroup$
– Ted Shifrin
Jan 15 at 19:53
$begingroup$
take the divergence of the function
$endgroup$
– user29418
Jan 15 at 19:42
$begingroup$
take the divergence of the function
$endgroup$
– user29418
Jan 15 at 19:42
$begingroup$
Gauss' theorem can only be used over closed surfaces. That doesn't mean that you can't use it, but if you do, you will need to find the flux across the surfaces that close up the volume.
$endgroup$
– Doug M
Jan 15 at 19:45
$begingroup$
Gauss' theorem can only be used over closed surfaces. That doesn't mean that you can't use it, but if you do, you will need to find the flux across the surfaces that close up the volume.
$endgroup$
– Doug M
Jan 15 at 19:45
$begingroup$
You can calculate the flux passing through the surface.
$endgroup$
– copper.hat
Jan 15 at 19:49
$begingroup$
You can calculate the flux passing through the surface.
$endgroup$
– copper.hat
Jan 15 at 19:49
$begingroup$
If you're going to use divergence, you'd best compute it correctly. It's a scalar function.
$endgroup$
– Ted Shifrin
Jan 15 at 19:53
$begingroup$
If you're going to use divergence, you'd best compute it correctly. It's a scalar function.
$endgroup$
– Ted Shifrin
Jan 15 at 19:53
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Calculating it directly
$z = 1 - x- y\
dS = (1,1,1)$
$int_0^1 int_0^{1-x} (y(1-x-y),x(1-x-y), xy)cdot(1,1,1) dx dy$
$int_0^1 int_0^{1-x} x+y - xy - x^2 - y^2 dx dy\
int_0^1 (x-x^2)(1-x)+ (1-x)(frac 12 (1-x)^2) - frac 13 (1-x)^3 dy\
int_0^1 x(1-x)^2+ frac 16 (1-x)^3 dy\
int_0^1 (1-x)^2- frac 56 (1-x)^3 dy\
frac 13 - frac 5{24} = frac 18$
As you suggested using the divergence theorem.
We can create a tetrahedron 3 triangles in the xy,yz, xz planes.
$iint f(x,y,z) dA_1 + iint f(x,y,z) dA_2 + iint f(x,y,z) dA_3 + iint f(x,y,z) dS = iint nabla cdot f dV$
$nabla cdot f = 0$
and by the symmetry of the figure.
$iint f(x,y,z) dA_1 = iint f(x,y,z) dA_2 = iint f(x,y,z) dA_3$
$iint f(x,y,z) dS = -3 iint f(x,y,z) dA_1$
We need our normals pointed outward. $(0,0,-1)$ would be the be the suitable normal
$iint f(x,y,z) dS = 3int_0^1 int_0^{1-x} xy dx dy$
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Calculating it directly
$z = 1 - x- y\
dS = (1,1,1)$
$int_0^1 int_0^{1-x} (y(1-x-y),x(1-x-y), xy)cdot(1,1,1) dx dy$
$int_0^1 int_0^{1-x} x+y - xy - x^2 - y^2 dx dy\
int_0^1 (x-x^2)(1-x)+ (1-x)(frac 12 (1-x)^2) - frac 13 (1-x)^3 dy\
int_0^1 x(1-x)^2+ frac 16 (1-x)^3 dy\
int_0^1 (1-x)^2- frac 56 (1-x)^3 dy\
frac 13 - frac 5{24} = frac 18$
As you suggested using the divergence theorem.
We can create a tetrahedron 3 triangles in the xy,yz, xz planes.
$iint f(x,y,z) dA_1 + iint f(x,y,z) dA_2 + iint f(x,y,z) dA_3 + iint f(x,y,z) dS = iint nabla cdot f dV$
$nabla cdot f = 0$
and by the symmetry of the figure.
$iint f(x,y,z) dA_1 = iint f(x,y,z) dA_2 = iint f(x,y,z) dA_3$
$iint f(x,y,z) dS = -3 iint f(x,y,z) dA_1$
We need our normals pointed outward. $(0,0,-1)$ would be the be the suitable normal
$iint f(x,y,z) dS = 3int_0^1 int_0^{1-x} xy dx dy$
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
add a comment |
$begingroup$
Calculating it directly
$z = 1 - x- y\
dS = (1,1,1)$
$int_0^1 int_0^{1-x} (y(1-x-y),x(1-x-y), xy)cdot(1,1,1) dx dy$
$int_0^1 int_0^{1-x} x+y - xy - x^2 - y^2 dx dy\
int_0^1 (x-x^2)(1-x)+ (1-x)(frac 12 (1-x)^2) - frac 13 (1-x)^3 dy\
int_0^1 x(1-x)^2+ frac 16 (1-x)^3 dy\
int_0^1 (1-x)^2- frac 56 (1-x)^3 dy\
frac 13 - frac 5{24} = frac 18$
As you suggested using the divergence theorem.
We can create a tetrahedron 3 triangles in the xy,yz, xz planes.
$iint f(x,y,z) dA_1 + iint f(x,y,z) dA_2 + iint f(x,y,z) dA_3 + iint f(x,y,z) dS = iint nabla cdot f dV$
$nabla cdot f = 0$
and by the symmetry of the figure.
$iint f(x,y,z) dA_1 = iint f(x,y,z) dA_2 = iint f(x,y,z) dA_3$
$iint f(x,y,z) dS = -3 iint f(x,y,z) dA_1$
We need our normals pointed outward. $(0,0,-1)$ would be the be the suitable normal
$iint f(x,y,z) dS = 3int_0^1 int_0^{1-x} xy dx dy$
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
add a comment |
$begingroup$
Calculating it directly
$z = 1 - x- y\
dS = (1,1,1)$
$int_0^1 int_0^{1-x} (y(1-x-y),x(1-x-y), xy)cdot(1,1,1) dx dy$
$int_0^1 int_0^{1-x} x+y - xy - x^2 - y^2 dx dy\
int_0^1 (x-x^2)(1-x)+ (1-x)(frac 12 (1-x)^2) - frac 13 (1-x)^3 dy\
int_0^1 x(1-x)^2+ frac 16 (1-x)^3 dy\
int_0^1 (1-x)^2- frac 56 (1-x)^3 dy\
frac 13 - frac 5{24} = frac 18$
As you suggested using the divergence theorem.
We can create a tetrahedron 3 triangles in the xy,yz, xz planes.
$iint f(x,y,z) dA_1 + iint f(x,y,z) dA_2 + iint f(x,y,z) dA_3 + iint f(x,y,z) dS = iint nabla cdot f dV$
$nabla cdot f = 0$
and by the symmetry of the figure.
$iint f(x,y,z) dA_1 = iint f(x,y,z) dA_2 = iint f(x,y,z) dA_3$
$iint f(x,y,z) dS = -3 iint f(x,y,z) dA_1$
We need our normals pointed outward. $(0,0,-1)$ would be the be the suitable normal
$iint f(x,y,z) dS = 3int_0^1 int_0^{1-x} xy dx dy$
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
$endgroup$
Calculating it directly
$z = 1 - x- y\
dS = (1,1,1)$
$int_0^1 int_0^{1-x} (y(1-x-y),x(1-x-y), xy)cdot(1,1,1) dx dy$
$int_0^1 int_0^{1-x} x+y - xy - x^2 - y^2 dx dy\
int_0^1 (x-x^2)(1-x)+ (1-x)(frac 12 (1-x)^2) - frac 13 (1-x)^3 dy\
int_0^1 x(1-x)^2+ frac 16 (1-x)^3 dy\
int_0^1 (1-x)^2- frac 56 (1-x)^3 dy\
frac 13 - frac 5{24} = frac 18$
As you suggested using the divergence theorem.
We can create a tetrahedron 3 triangles in the xy,yz, xz planes.
$iint f(x,y,z) dA_1 + iint f(x,y,z) dA_2 + iint f(x,y,z) dA_3 + iint f(x,y,z) dS = iint nabla cdot f dV$
$nabla cdot f = 0$
and by the symmetry of the figure.
$iint f(x,y,z) dA_1 = iint f(x,y,z) dA_2 = iint f(x,y,z) dA_3$
$iint f(x,y,z) dS = -3 iint f(x,y,z) dA_1$
We need our normals pointed outward. $(0,0,-1)$ would be the be the suitable normal
$iint f(x,y,z) dS = 3int_0^1 int_0^{1-x} xy dx dy$
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
answered Jan 15 at 20:01
Doug MDoug M
45.2k31854
45.2k31854
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
add a comment |
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Ah ok, I also tried to calculate directly. The mistake I had was for both integrals, I had both of them go from 0 to 1!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
$begingroup$
Thank you very much!
$endgroup$
– user3221454
Jan 15 at 20:06
add a comment |
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$begingroup$
take the divergence of the function
$endgroup$
– user29418
Jan 15 at 19:42
$begingroup$
Gauss' theorem can only be used over closed surfaces. That doesn't mean that you can't use it, but if you do, you will need to find the flux across the surfaces that close up the volume.
$endgroup$
– Doug M
Jan 15 at 19:45
$begingroup$
You can calculate the flux passing through the surface.
$endgroup$
– copper.hat
Jan 15 at 19:49
$begingroup$
If you're going to use divergence, you'd best compute it correctly. It's a scalar function.
$endgroup$
– Ted Shifrin
Jan 15 at 19:53