Difficult Region of Integration Involving Gauss's Theorem
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I'm told to use Gauss's Theorem to compute the flux of a field $vec F = <x,y^2,y+z>$ along the boundary of the cylindrical solid $x^2+y^2 le 4$ below $z=8$ and above $z=x$.
I know by Gauss's Theorem that:
Net Flux = $iint_{partial D} vec F cdot vec ndS = iiint_D nabla cdot vec FdV$
This computation is pretty straight forward. $nabla cdot vec F = 2+2y$. But the region of integration is particularly difficult to map out.
I thought to use cylindrical coordinates and setting the bounds to $0 le theta le 2 pi$, $0 le z le 8$, and $0 le r le 4$, but this seems like it would just give me the area of the cylinder of height 8--and wouldn't include the part where z=x slices through the cylinder.
What would be the right way to go in terms of the bounds of integration?
calculus integration multivariable-calculus physics divergence
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
$endgroup$
add a comment |
$begingroup$
I'm told to use Gauss's Theorem to compute the flux of a field $vec F = <x,y^2,y+z>$ along the boundary of the cylindrical solid $x^2+y^2 le 4$ below $z=8$ and above $z=x$.
I know by Gauss's Theorem that:
Net Flux = $iint_{partial D} vec F cdot vec ndS = iiint_D nabla cdot vec FdV$
This computation is pretty straight forward. $nabla cdot vec F = 2+2y$. But the region of integration is particularly difficult to map out.
I thought to use cylindrical coordinates and setting the bounds to $0 le theta le 2 pi$, $0 le z le 8$, and $0 le r le 4$, but this seems like it would just give me the area of the cylinder of height 8--and wouldn't include the part where z=x slices through the cylinder.
What would be the right way to go in terms of the bounds of integration?
calculus integration multivariable-calculus physics divergence
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
$endgroup$
add a comment |
$begingroup$
I'm told to use Gauss's Theorem to compute the flux of a field $vec F = <x,y^2,y+z>$ along the boundary of the cylindrical solid $x^2+y^2 le 4$ below $z=8$ and above $z=x$.
I know by Gauss's Theorem that:
Net Flux = $iint_{partial D} vec F cdot vec ndS = iiint_D nabla cdot vec FdV$
This computation is pretty straight forward. $nabla cdot vec F = 2+2y$. But the region of integration is particularly difficult to map out.
I thought to use cylindrical coordinates and setting the bounds to $0 le theta le 2 pi$, $0 le z le 8$, and $0 le r le 4$, but this seems like it would just give me the area of the cylinder of height 8--and wouldn't include the part where z=x slices through the cylinder.
What would be the right way to go in terms of the bounds of integration?
calculus integration multivariable-calculus physics divergence
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
$endgroup$
I'm told to use Gauss's Theorem to compute the flux of a field $vec F = <x,y^2,y+z>$ along the boundary of the cylindrical solid $x^2+y^2 le 4$ below $z=8$ and above $z=x$.
I know by Gauss's Theorem that:
Net Flux = $iint_{partial D} vec F cdot vec ndS = iiint_D nabla cdot vec FdV$
This computation is pretty straight forward. $nabla cdot vec F = 2+2y$. But the region of integration is particularly difficult to map out.
I thought to use cylindrical coordinates and setting the bounds to $0 le theta le 2 pi$, $0 le z le 8$, and $0 le r le 4$, but this seems like it would just give me the area of the cylinder of height 8--and wouldn't include the part where z=x slices through the cylinder.
What would be the right way to go in terms of the bounds of integration?
calculus integration multivariable-calculus physics divergence
calculus integration multivariable-calculus physics divergence
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
edited Nov 30 '18 at 5:38
Robert Howard
1,9261822
1,9261822
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
asked Nov 30 '18 at 4:40
Jackson JoffeJackson Joffe
575
575
add a comment |
add a comment |
1 Answer
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oldest
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You have the $theta$ and $r$ bounds right, as well as the upper bound for $z$. The lower bound can be thought of as the lower "boundary" of your region, i.e.,
$$
z = x
$$
or
$$
z = r cos theta
$$
in cylindrical coordinates.
So your bounds would be
$$rcos theta leq z leq 8$$
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have the $theta$ and $r$ bounds right, as well as the upper bound for $z$. The lower bound can be thought of as the lower "boundary" of your region, i.e.,
$$
z = x
$$
or
$$
z = r cos theta
$$
in cylindrical coordinates.
So your bounds would be
$$rcos theta leq z leq 8$$
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
You have the $theta$ and $r$ bounds right, as well as the upper bound for $z$. The lower bound can be thought of as the lower "boundary" of your region, i.e.,
$$
z = x
$$
or
$$
z = r cos theta
$$
in cylindrical coordinates.
So your bounds would be
$$rcos theta leq z leq 8$$
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
You have the $theta$ and $r$ bounds right, as well as the upper bound for $z$. The lower bound can be thought of as the lower "boundary" of your region, i.e.,
$$
z = x
$$
or
$$
z = r cos theta
$$
in cylindrical coordinates.
So your bounds would be
$$rcos theta leq z leq 8$$
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
$endgroup$
You have the $theta$ and $r$ bounds right, as well as the upper bound for $z$. The lower bound can be thought of as the lower "boundary" of your region, i.e.,
$$
z = x
$$
or
$$
z = r cos theta
$$
in cylindrical coordinates.
So your bounds would be
$$rcos theta leq z leq 8$$
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
answered Nov 30 '18 at 4:45
user25959user25959
1,573816
1,573816
add a comment |
add a comment |
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