Mixing
What is the difference between two questions?
$begingroup$
Find the flux of $F=(x-y)i+xj $ across the circle $x^{2}+y^{2}=1$ in the $xy$-plane.
Find the circulation of the field $F=(x-y)i+xj$ around the circle $r(t)=(cos (t))i +(sin(t)) j,0leq t leq 2 pi.$
I think both are same.
Both can be solved by Green's theorem.
calculus multivariable-calculus vectors vector-analysis
edited Jan 10 at 15:48
saulspatz
14.8k21329
asked Jan 10 at 15:36
sejysejy
1589
$endgroup$
add a comment |
$begingroup$
Find the flux of $F=(x-y)i+xj $ across the circle $x^{2}+y^{2}=1$ in the $xy$-plane.
Find the circulation of the field $F=(x-y)i+xj$ around the circle $r(t)=(cos (t))i +(sin(t)) j,0leq t leq 2 pi.$
I think both are same.
Both can be solved by Green's theorem.
calculus multivariable-calculus vectors vector-analysis
edited Jan 10 at 15:48
saulspatz
14.8k21329
asked Jan 10 at 15:36
sejysejy
1589
$endgroup$
add a comment |
$begingroup$
Find the flux of $F=(x-y)i+xj $ across the circle $x^{2}+y^{2}=1$ in the $xy$-plane.
Find the circulation of the field $F=(x-y)i+xj$ around the circle $r(t)=(cos (t))i +(sin(t)) j,0leq t leq 2 pi.$
I think both are same.
Both can be solved by Green's theorem.
calculus multivariable-calculus vectors vector-analysis
edited Jan 10 at 15:48
saulspatz
14.8k21329
asked Jan 10 at 15:36
sejysejy
1589
$endgroup$
Find the flux of $F=(x-y)i+xj $ across the circle $x^{2}+y^{2}=1$ in the $xy$-plane.
Find the circulation of the field $F=(x-y)i+xj$ around the circle $r(t)=(cos (t))i +(sin(t)) j,0leq t leq 2 pi.$
I think both are same.
Both can be solved by Green's theorem.
calculus multivariable-calculus vectors vector-analysis
calculus multivariable-calculus vectors vector-analysis
edited Jan 10 at 15:48
saulspatz
14.8k21329
asked Jan 10 at 15:36
sejysejy
1589
edited Jan 10 at 15:48
saulspatz
14.8k21329
asked Jan 10 at 15:36
sejysejy
1589
edited Jan 10 at 15:48
saulspatz
14.8k21329
edited Jan 10 at 15:48
saulspatz
14.8k21329
edited Jan 10 at 15:48
saulspatz
14.8k21329
14.8k21329
asked Jan 10 at 15:36
sejysejy
1589
asked Jan 10 at 15:36
sejysejy
1589
asked Jan 10 at 15:36
sejysejy
1589
1589
add a comment |
add a comment |
1 Answer
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$begingroup$
The flux of a vector field $F$ across some boundary $S$ is, if we think of the vector field as describing the movement of some fluid, the volume of that fluid crossing $S$ in a given unit time. It is given, in 2 dimensions, by the flux integral $$int_SFcdot mathbf{n}ds,$$
where $mathbf{n}: (x,y)mapsto mathbf{n}(x,y)$ is a function defined on $S$ at gives the (outward) unit normal vector to $S$ for any $(x,y)$ in $S$.
In this case, $mathbf{n}(x,y) = (x,y)$, so our integral is just
$$int_S(x-y)x + xyds = int_Sx^2ds.$$
To calculate this, we parameterise the circle using your parameterisation from your second question, so $x = cos(t)$, $y = sin(t)$, and $ds = |(-sin(t),cos(t)|dt = dt$, so our integral is
$$intlimits_0^{2pi}cos^2(t)dt = pi.$$
For your second question, however, matters are different: Intuitively, the circulation is how much of the vector field is going around the circle, rather than across it. It is given by
$$int_SFcdot ds = int_SF(s)cdot s'(t)dt.$$
In this case, we again use the same parameterisation of our circle, so our integral is
$$intlimits_0^{2pi}left(array{cos(t)-sin(t)\cos(t)}right)cdotleft(array{-sin(t)\cos(t)}right)dt = intlimits_0^{2pi}1-cos(t)sin(t)dt = 2pi.$$
So no, the questions are not both the same.
answered Jan 10 at 15:59
user3482749user3482749
4,266919
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The flux of a vector field $F$ across some boundary $S$ is, if we think of the vector field as describing the movement of some fluid, the volume of that fluid crossing $S$ in a given unit time. It is given, in 2 dimensions, by the flux integral $$int_SFcdot mathbf{n}ds,$$
where $mathbf{n}: (x,y)mapsto mathbf{n}(x,y)$ is a function defined on $S$ at gives the (outward) unit normal vector to $S$ for any $(x,y)$ in $S$.
In this case, $mathbf{n}(x,y) = (x,y)$, so our integral is just
$$int_S(x-y)x + xyds = int_Sx^2ds.$$
To calculate this, we parameterise the circle using your parameterisation from your second question, so $x = cos(t)$, $y = sin(t)$, and $ds = |(-sin(t),cos(t)|dt = dt$, so our integral is
$$intlimits_0^{2pi}cos^2(t)dt = pi.$$
For your second question, however, matters are different: Intuitively, the circulation is how much of the vector field is going around the circle, rather than across it. It is given by
$$int_SFcdot ds = int_SF(s)cdot s'(t)dt.$$
In this case, we again use the same parameterisation of our circle, so our integral is
$$intlimits_0^{2pi}left(array{cos(t)-sin(t)\cos(t)}right)cdotleft(array{-sin(t)\cos(t)}right)dt = intlimits_0^{2pi}1-cos(t)sin(t)dt = 2pi.$$
So no, the questions are not both the same.
answered Jan 10 at 15:59
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
The flux of a vector field $F$ across some boundary $S$ is, if we think of the vector field as describing the movement of some fluid, the volume of that fluid crossing $S$ in a given unit time. It is given, in 2 dimensions, by the flux integral $$int_SFcdot mathbf{n}ds,$$
where $mathbf{n}: (x,y)mapsto mathbf{n}(x,y)$ is a function defined on $S$ at gives the (outward) unit normal vector to $S$ for any $(x,y)$ in $S$.
In this case, $mathbf{n}(x,y) = (x,y)$, so our integral is just
$$int_S(x-y)x + xyds = int_Sx^2ds.$$
To calculate this, we parameterise the circle using your parameterisation from your second question, so $x = cos(t)$, $y = sin(t)$, and $ds = |(-sin(t),cos(t)|dt = dt$, so our integral is
$$intlimits_0^{2pi}cos^2(t)dt = pi.$$
For your second question, however, matters are different: Intuitively, the circulation is how much of the vector field is going around the circle, rather than across it. It is given by
$$int_SFcdot ds = int_SF(s)cdot s'(t)dt.$$
In this case, we again use the same parameterisation of our circle, so our integral is
$$intlimits_0^{2pi}left(array{cos(t)-sin(t)\cos(t)}right)cdotleft(array{-sin(t)\cos(t)}right)dt = intlimits_0^{2pi}1-cos(t)sin(t)dt = 2pi.$$
So no, the questions are not both the same.
answered Jan 10 at 15:59
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
The flux of a vector field $F$ across some boundary $S$ is, if we think of the vector field as describing the movement of some fluid, the volume of that fluid crossing $S$ in a given unit time. It is given, in 2 dimensions, by the flux integral $$int_SFcdot mathbf{n}ds,$$
where $mathbf{n}: (x,y)mapsto mathbf{n}(x,y)$ is a function defined on $S$ at gives the (outward) unit normal vector to $S$ for any $(x,y)$ in $S$.
In this case, $mathbf{n}(x,y) = (x,y)$, so our integral is just
$$int_S(x-y)x + xyds = int_Sx^2ds.$$
To calculate this, we parameterise the circle using your parameterisation from your second question, so $x = cos(t)$, $y = sin(t)$, and $ds = |(-sin(t),cos(t)|dt = dt$, so our integral is
$$intlimits_0^{2pi}cos^2(t)dt = pi.$$
For your second question, however, matters are different: Intuitively, the circulation is how much of the vector field is going around the circle, rather than across it. It is given by
$$int_SFcdot ds = int_SF(s)cdot s'(t)dt.$$
In this case, we again use the same parameterisation of our circle, so our integral is
$$intlimits_0^{2pi}left(array{cos(t)-sin(t)\cos(t)}right)cdotleft(array{-sin(t)\cos(t)}right)dt = intlimits_0^{2pi}1-cos(t)sin(t)dt = 2pi.$$
So no, the questions are not both the same.
answered Jan 10 at 15:59
user3482749user3482749
4,266919
$endgroup$
The flux of a vector field $F$ across some boundary $S$ is, if we think of the vector field as describing the movement of some fluid, the volume of that fluid crossing $S$ in a given unit time. It is given, in 2 dimensions, by the flux integral $$int_SFcdot mathbf{n}ds,$$
where $mathbf{n}: (x,y)mapsto mathbf{n}(x,y)$ is a function defined on $S$ at gives the (outward) unit normal vector to $S$ for any $(x,y)$ in $S$.
In this case, $mathbf{n}(x,y) = (x,y)$, so our integral is just
$$int_S(x-y)x + xyds = int_Sx^2ds.$$
To calculate this, we parameterise the circle using your parameterisation from your second question, so $x = cos(t)$, $y = sin(t)$, and $ds = |(-sin(t),cos(t)|dt = dt$, so our integral is
$$intlimits_0^{2pi}cos^2(t)dt = pi.$$
For your second question, however, matters are different: Intuitively, the circulation is how much of the vector field is going around the circle, rather than across it. It is given by
$$int_SFcdot ds = int_SF(s)cdot s'(t)dt.$$
In this case, we again use the same parameterisation of our circle, so our integral is
$$intlimits_0^{2pi}left(array{cos(t)-sin(t)\cos(t)}right)cdotleft(array{-sin(t)\cos(t)}right)dt = intlimits_0^{2pi}1-cos(t)sin(t)dt = 2pi.$$
So no, the questions are not both the same.
answered Jan 10 at 15:59
user3482749user3482749
4,266919
answered Jan 10 at 15:59
user3482749user3482749
4,266919
answered Jan 10 at 15:59
user3482749user3482749
4,266919
answered Jan 10 at 15:59
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
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