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How to calculate the flux through complicated surface

Calculate flux through a surfaceCalculate the flux through a closed surfaceFinding flux through a larger sphere given flux through a smaller, embedded sphereCompute the flux of the vector field $vecF$ through the surface SCalculation of flux through sphere when the vector field is not defined at the originDivergence theorem to calculate flux through an open cylindercalculate flux through surfaceHow to Calculate the flux of the Vector Field on the surface $z = 1-x^2-y^2$ ( getting normal vector $(0,0,0)$ at the point $(0,0,1)$ ?!!! )How to find the flux $int_S 2~dydz + dzdx + -3dxdy$ in the surface $x^2 + y^2 + z^2 +xyz = 1$ ( how to parametrize the surface ?)How to Find the integral $int_S−(xy^2)~dydz + (2x ^2 y)~dzdx − (zy^2)~dxdy$ where is the portion of the sphere $x^2 + y^2 + z^2 = 1$

1

$begingroup$

How can I find the integral $int_S 2~dydz + 1~dzdx - (3x)~dxdy$, where $S$ is the following 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's law, but I am not sure if the surface is closed. I thought if it isn't I shall close it with a sphere with radius $epsilon$.

Then I simply calculate the flux through the ball and say:

$$int_SFcdotvecn = int_Ball+SFcdotvecn - int_BallFcdotvecn = 0 :(textGauss's law) - int_BallFcdotvecn$$

If I can't use Gauss's law, what should I do?

multivariable-calculus vector-analysis divergence

share|cite|improve this question

edited Mar 10 at 20:18

Robert Howard

2,2112933

asked Jan 22 at 19:24

Mather Mather

4108

$endgroup$

add a comment | 

1

$begingroup$

How can I find the integral $int_S 2~dydz + 1~dzdx - (3x)~dxdy$, where $S$ is the following 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's law, but I am not sure if the surface is closed. I thought if it isn't I shall close it with a sphere with radius $epsilon$.

Then I simply calculate the flux through the ball and say:

$$int_SFcdotvecn = int_Ball+SFcdotvecn - int_BallFcdotvecn = 0 :(textGauss's law) - int_BallFcdotvecn$$

If I can't use Gauss's law, what should I do?

multivariable-calculus vector-analysis divergence

share|cite|improve this question

edited Mar 10 at 20:18

Robert Howard

2,2112933

asked Jan 22 at 19:24

Mather Mather

4108

$endgroup$

add a comment | 

$begingroup$

How can I find the integral $int_S 2~dydz + 1~dzdx - (3x)~dxdy$, where $S$ is the following 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's law, but I am not sure if the surface is closed. I thought if it isn't I shall close it with a sphere with radius $epsilon$.

Then I simply calculate the flux through the ball and say:

$$int_SFcdotvecn = int_Ball+SFcdotvecn - int_BallFcdotvecn = 0 :(textGauss's law) - int_BallFcdotvecn$$

If I can't use Gauss's law, what should I do?

multivariable-calculus vector-analysis divergence

share|cite|improve this question

edited Mar 10 at 20:18

Robert Howard

2,2112933

asked Jan 22 at 19:24

Mather Mather

4108

$endgroup$

share|cite|improve this question

edited Mar 10 at 20:18

Robert Howard

2,2112933

asked Jan 22 at 19:24

Mather Mather

4108

share|cite|improve this question

edited Mar 10 at 20:18

Robert Howard

2,2112933

asked Jan 22 at 19:24

Mather Mather

4108

edited Mar 10 at 20:18

Robert Howard

2,2112933

edited Mar 10 at 20:18

Robert Howard

2,2112933

asked Jan 22 at 19:24

Mather Mather

4108

asked Jan 22 at 19:24

Mather Mather

4108

1 Answer
1

active

oldest

votes

1

$begingroup$

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.

share|cite|improve this answer

answered Jan 22 at 21:03

GReyesGReyes

1,95515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

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1

$begingroup$

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.

share|cite|improve this answer

answered Jan 22 at 21:03

GReyesGReyes

1,95515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

1

$begingroup$

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.

share|cite|improve this answer

answered Jan 22 at 21:03

GReyesGReyes

1,95515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.

share|cite|improve this answer

answered Jan 22 at 21:03

GReyesGReyes

1,95515

$endgroup$

share|cite|improve this answer

answered Jan 22 at 21:03

GReyesGReyes

1,95515

answered Jan 22 at 21:03

GReyesGReyes

1,95515

answered Jan 22 at 21:03

GReyesGReyes

1,95515