How to calculate the flux through complicated Surface
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How to Find the integral $int_{S} 2~dydz + 1~dzdx - (3x)~dxdy$
Where $S$ is the surface :
$x^2 + 2y^2 + 3z^2 + xyze^{(x+y+z)sin(x^2 -y+z)} = 1$
$0leq x,y,z $
i think that i must use Gauss therom. but i am not sure if the surface is closed, I though if it isn't I Shall close it with sphere with radius $~ huge epsilon$
then i simply calculate the flux through the ball and say :
$ int_{S}F_dot~vec{n} = int_{Ball+S}F_dot~vec{n} - int_{Ball}F_dot~vec{n} = 0 ~( Gauss ~law) - int_{Ball}F_dot~vec{n} $
if i cant use Gauss law what should i do ?
multivariable-calculus vector-analysis divergence
asked Jan 22 at 19:24
Mather Mather
3418
$endgroup$
add a comment |
$begingroup$
How to Find the integral $int_{S} 2~dydz + 1~dzdx - (3x)~dxdy$
Where $S$ is the surface :
$x^2 + 2y^2 + 3z^2 + xyze^{(x+y+z)sin(x^2 -y+z)} = 1$
$0leq x,y,z $
i think that i must use Gauss therom. but i am not sure if the surface is closed, I though if it isn't I Shall close it with sphere with radius $~ huge epsilon$
then i simply calculate the flux through the ball and say :
$ int_{S}F_dot~vec{n} = int_{Ball+S}F_dot~vec{n} - int_{Ball}F_dot~vec{n} = 0 ~( Gauss ~law) - int_{Ball}F_dot~vec{n} $
if i cant use Gauss law what should i do ?
multivariable-calculus vector-analysis divergence
asked Jan 22 at 19:24
Mather Mather
3418
$endgroup$
add a comment |
$begingroup$
How to Find the integral $int_{S} 2~dydz + 1~dzdx - (3x)~dxdy$
Where $S$ is the surface :
$x^2 + 2y^2 + 3z^2 + xyze^{(x+y+z)sin(x^2 -y+z)} = 1$
$0leq x,y,z $
i think that i must use Gauss therom. but i am not sure if the surface is closed, I though if it isn't I Shall close it with sphere with radius $~ huge epsilon$
then i simply calculate the flux through the ball and say :
$ int_{S}F_dot~vec{n} = int_{Ball+S}F_dot~vec{n} - int_{Ball}F_dot~vec{n} = 0 ~( Gauss ~law) - int_{Ball}F_dot~vec{n} $
if i cant use Gauss law what should i do ?
multivariable-calculus vector-analysis divergence
asked Jan 22 at 19:24
Mather Mather
3418
$endgroup$
How to Find the integral $int_{S} 2~dydz + 1~dzdx - (3x)~dxdy$
Where $S$ is the surface :
$x^2 + 2y^2 + 3z^2 + xyze^{(x+y+z)sin(x^2 -y+z)} = 1$
$0leq x,y,z $
i think that i must use Gauss therom. but i am not sure if the surface is closed, I though if it isn't I Shall close it with sphere with radius $~ huge epsilon$
then i simply calculate the flux through the ball and say :
$ int_{S}F_dot~vec{n} = int_{Ball+S}F_dot~vec{n} - int_{Ball}F_dot~vec{n} = 0 ~( Gauss ~law) - int_{Ball}F_dot~vec{n} $
if i cant use Gauss law what should i do ?
multivariable-calculus vector-analysis divergence
multivariable-calculus vector-analysis divergence
asked Jan 22 at 19:24
Mather Mather
3418
asked Jan 22 at 19:24
Mather Mather
3418
edited Jan 22 at 19:34
Mather
asked Jan 22 at 19:24
Mather Mather
3418
asked Jan 22 at 19:24
Mather Mather
3418
asked Jan 22 at 19:24
Mather Mather
3418
3418
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Your surface is not closed. Its traces on the $OXY$, $OXZ$ and $OYZ$ planes are quarters of ellipses. You can make it closed by adding the coordinate planes. Your total flux through the closed surface is zero by Gauss' Theorem, so the flux through your surface is the opposite of the flux through the elliptical pieces of coordinate planes that you added. The latter is easily found as double integrals.
answered Jan 22 at 21:03
GReyesGReyes
1,59515
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$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Your surface is not closed. Its traces on the $OXY$, $OXZ$ and $OYZ$ planes are quarters of ellipses. You can make it closed by adding the coordinate planes. Your total flux through the closed surface is zero by Gauss' Theorem, so the flux through your surface is the opposite of the flux through the elliptical pieces of coordinate planes that you added. The latter is easily found as double integrals.
answered Jan 22 at 21:03
GReyesGReyes
1,59515
$endgroup$
$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
add a comment |
$begingroup$
Your surface is not closed. Its traces on the $OXY$, $OXZ$ and $OYZ$ planes are quarters of ellipses. You can make it closed by adding the coordinate planes. Your total flux through the closed surface is zero by Gauss' Theorem, so the flux through your surface is the opposite of the flux through the elliptical pieces of coordinate planes that you added. The latter is easily found as double integrals.
answered Jan 22 at 21:03
GReyesGReyes
1,59515
$endgroup$
$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
add a comment |
$begingroup$
Your surface is not closed. Its traces on the $OXY$, $OXZ$ and $OYZ$ planes are quarters of ellipses. You can make it closed by adding the coordinate planes. Your total flux through the closed surface is zero by Gauss' Theorem, so the flux through your surface is the opposite of the flux through the elliptical pieces of coordinate planes that you added. The latter is easily found as double integrals.
answered Jan 22 at 21:03
GReyesGReyes
1,59515
$endgroup$
Your surface is not closed. Its traces on the $OXY$, $OXZ$ and $OYZ$ planes are quarters of ellipses. You can make it closed by adding the coordinate planes. Your total flux through the closed surface is zero by Gauss' Theorem, so the flux through your surface is the opposite of the flux through the elliptical pieces of coordinate planes that you added. The latter is easily found as double integrals.
answered Jan 22 at 21:03
GReyesGReyes
1,59515
answered Jan 22 at 21:03
GReyesGReyes
1,59515
answered Jan 22 at 21:03
GReyesGReyes
1,59515
answered Jan 22 at 21:03
GReyesGReyes
1,59515
1,59515
$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
add a comment |
$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
i dont understand your analysis , how did you know that it isn't closed ?
$endgroup$
– Mather
Jan 22 at 21:19
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
you said that $ x=0 $ gives you an ellipse , $ y =0$ gives you an ellipse and $ z = 0$ gives you an ellipse on $[YZ]$ , $[XZ]$ , $[XY]$ but how did you know that its not closed
$endgroup$
– Mather
Jan 22 at 21:22
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Those elliptical intersections with the coordinate planes constitute the boundary of your surface, therefore it is not closed. A way to visualize it is the following: imagine the eight of an ellipsoid centered at the origin of coordinates (in canonical position) contained in the first octant. Your surface is a complicated surface that has the same elliptical intersections with the planes as the aforementioned ellipsoid.
$endgroup$
– GReyes
Jan 22 at 21:52
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
Your surface is contained in the first octant, so it has edges on the coordinate planes.
$endgroup$
– GReyes
Jan 22 at 21:54
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
$begingroup$
oh nice i got it thank you
$endgroup$
– Mather
Jan 22 at 21:56
add a comment |
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