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How to find the flux $int_{S} 2~dydz + dzdx + -3dxdy$ in the surface $x^2 + y^2 + z^2 +xyz = 1$ ( how to...
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Find the integral $int_{S} 2~dydz + dzdx + -3dxdy$ where $S$ is the surface $x^2 + y^2 + z^2 +xyz = 1$ , $0 leq x,y,z$.
choose the direction of the normal as you like.
i am having hard time parametrizing the equation . how can i parametrize it to calculate the flux .
multivariable-calculus vector-analysis surface-integrals
asked Jan 16 at 10:58
Mather Mather
3437
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add a comment |
$begingroup$
Find the integral $int_{S} 2~dydz + dzdx + -3dxdy$ where $S$ is the surface $x^2 + y^2 + z^2 +xyz = 1$ , $0 leq x,y,z$.
choose the direction of the normal as you like.
i am having hard time parametrizing the equation . how can i parametrize it to calculate the flux .
multivariable-calculus vector-analysis surface-integrals
asked Jan 16 at 10:58
Mather Mather
3437
$endgroup$
add a comment |
$begingroup$
Find the integral $int_{S} 2~dydz + dzdx + -3dxdy$ where $S$ is the surface $x^2 + y^2 + z^2 +xyz = 1$ , $0 leq x,y,z$.
choose the direction of the normal as you like.
i am having hard time parametrizing the equation . how can i parametrize it to calculate the flux .
multivariable-calculus vector-analysis surface-integrals
asked Jan 16 at 10:58
Mather Mather
3437
$endgroup$
Find the integral $int_{S} 2~dydz + dzdx + -3dxdy$ where $S$ is the surface $x^2 + y^2 + z^2 +xyz = 1$ , $0 leq x,y,z$.
choose the direction of the normal as you like.
i am having hard time parametrizing the equation . how can i parametrize it to calculate the flux .
multivariable-calculus vector-analysis surface-integrals
multivariable-calculus vector-analysis surface-integrals
asked Jan 16 at 10:58
Mather Mather
3437
asked Jan 16 at 10:58
Mather Mather
3437
edited Jan 16 at 11:23
Mather
asked Jan 16 at 10:58
Mather Mather
3437
asked Jan 16 at 10:58
Mather Mather
3437
asked Jan 16 at 10:58
Mather Mather
3437
3437
add a comment |
add a comment |
1 Answer
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Although you can parametrize the surface (for example, $z=frac{1}{2} left(sqrt{x^2 y^2-4 x^2-4 y^2+4}-x yright)$), I believe, it's not the way the problem is intended to be solved.
First, you want to show that the surface projects onto one of the basis planes uniquely (without folds). For example, if you consider the surface equation as an equation on $z$, there is only one positive root.
Now we can think of our integral. We divide the surface into small patches. Expression $dx,dy$ is an area of a projection of a small patch on plane $xy$. Since unique projection of each patch, sum of projection areas of all patches is the area of the projection $A_{xy}$:
$$
int_Sleft(2dy,dz+dz,dx-3dx,dyright)=2int_S dy,dz+int_S dz,dx-3int_Sdx,dy =
2A_{yz}+A_{xz}-3A_{xy}.
$$
But due to symmetry, areas of all three projections are the same, and the asnwer is $0$.
P.S. If there were other numbers, we needed to calculate the area of the projections, which is easy, since it's a quarter-circle.
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Although you can parametrize the surface (for example, $z=frac{1}{2} left(sqrt{x^2 y^2-4 x^2-4 y^2+4}-x yright)$), I believe, it's not the way the problem is intended to be solved.
First, you want to show that the surface projects onto one of the basis planes uniquely (without folds). For example, if you consider the surface equation as an equation on $z$, there is only one positive root.
Now we can think of our integral. We divide the surface into small patches. Expression $dx,dy$ is an area of a projection of a small patch on plane $xy$. Since unique projection of each patch, sum of projection areas of all patches is the area of the projection $A_{xy}$:
$$
int_Sleft(2dy,dz+dz,dx-3dx,dyright)=2int_S dy,dz+int_S dz,dx-3int_Sdx,dy =
2A_{yz}+A_{xz}-3A_{xy}.
$$
But due to symmetry, areas of all three projections are the same, and the asnwer is $0$.
P.S. If there were other numbers, we needed to calculate the area of the projections, which is easy, since it's a quarter-circle.
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
add a comment |
$begingroup$
Although you can parametrize the surface (for example, $z=frac{1}{2} left(sqrt{x^2 y^2-4 x^2-4 y^2+4}-x yright)$), I believe, it's not the way the problem is intended to be solved.
First, you want to show that the surface projects onto one of the basis planes uniquely (without folds). For example, if you consider the surface equation as an equation on $z$, there is only one positive root.
Now we can think of our integral. We divide the surface into small patches. Expression $dx,dy$ is an area of a projection of a small patch on plane $xy$. Since unique projection of each patch, sum of projection areas of all patches is the area of the projection $A_{xy}$:
$$
int_Sleft(2dy,dz+dz,dx-3dx,dyright)=2int_S dy,dz+int_S dz,dx-3int_Sdx,dy =
2A_{yz}+A_{xz}-3A_{xy}.
$$
But due to symmetry, areas of all three projections are the same, and the asnwer is $0$.
P.S. If there were other numbers, we needed to calculate the area of the projections, which is easy, since it's a quarter-circle.
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
add a comment |
$begingroup$
Although you can parametrize the surface (for example, $z=frac{1}{2} left(sqrt{x^2 y^2-4 x^2-4 y^2+4}-x yright)$), I believe, it's not the way the problem is intended to be solved.
First, you want to show that the surface projects onto one of the basis planes uniquely (without folds). For example, if you consider the surface equation as an equation on $z$, there is only one positive root.
Now we can think of our integral. We divide the surface into small patches. Expression $dx,dy$ is an area of a projection of a small patch on plane $xy$. Since unique projection of each patch, sum of projection areas of all patches is the area of the projection $A_{xy}$:
$$
int_Sleft(2dy,dz+dz,dx-3dx,dyright)=2int_S dy,dz+int_S dz,dx-3int_Sdx,dy =
2A_{yz}+A_{xz}-3A_{xy}.
$$
But due to symmetry, areas of all three projections are the same, and the asnwer is $0$.
P.S. If there were other numbers, we needed to calculate the area of the projections, which is easy, since it's a quarter-circle.
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
$endgroup$
Although you can parametrize the surface (for example, $z=frac{1}{2} left(sqrt{x^2 y^2-4 x^2-4 y^2+4}-x yright)$), I believe, it's not the way the problem is intended to be solved.
First, you want to show that the surface projects onto one of the basis planes uniquely (without folds). For example, if you consider the surface equation as an equation on $z$, there is only one positive root.
Now we can think of our integral. We divide the surface into small patches. Expression $dx,dy$ is an area of a projection of a small patch on plane $xy$. Since unique projection of each patch, sum of projection areas of all patches is the area of the projection $A_{xy}$:
$$
int_Sleft(2dy,dz+dz,dx-3dx,dyright)=2int_S dy,dz+int_S dz,dx-3int_Sdx,dy =
2A_{yz}+A_{xz}-3A_{xy}.
$$
But due to symmetry, areas of all three projections are the same, and the asnwer is $0$.
P.S. If there were other numbers, we needed to calculate the area of the projections, which is easy, since it's a quarter-circle.
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
answered Jan 16 at 11:36
Vasily MitchVasily Mitch
2,3141311
2,3141311
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
add a comment |
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
i dont understand all the subject of projection is and how it is related to the surface integrals / flux , etc . at all . do you have some kind of source that explains that deeply it will save my life !!
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
for instance orthogonal projections what it means and how it is related to integrals and calcluas
$endgroup$
– Mather
Jan 16 at 11:44
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
$begingroup$
Maybe the picture from this thread helps.
$endgroup$
– Vasily Mitch
Jan 16 at 11:56
add a comment |
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