Mixing
find the simple closed curve of $F(x,y) = (y^3-6y)i + (6x-x^3)j$ using Green's Theorem which will have the...
$begingroup$
$F(x,y) = (y^3-6y)i + (6x-x^3)j$
a. Using Green's Theorem, find the simple closed curve C for which the integral
$ ∳F cdot dr $ (with positive orientation) will have the largest positive value.
b. Compute this largest possible value.
I'm quite certain that this is just $ iint Nx-My $ $dA$ but I do not know how to find the bounds in this scenario for both integrals. Though I'm also sure that this problem can also be done using just $ ∳ F cdot dr $ as there is an equation given for F.
integration multivariable-calculus greens-theorem
edited Jan 31 at 3:36
anomaly
17.8k42666
asked Jan 31 at 1:38
user63266user63266
92
$endgroup$
add a comment |
$begingroup$
$F(x,y) = (y^3-6y)i + (6x-x^3)j$
a. Using Green's Theorem, find the simple closed curve C for which the integral
$ ∳F cdot dr $ (with positive orientation) will have the largest positive value.
b. Compute this largest possible value.
I'm quite certain that this is just $ iint Nx-My $ $dA$ but I do not know how to find the bounds in this scenario for both integrals. Though I'm also sure that this problem can also be done using just $ ∳ F cdot dr $ as there is an equation given for F.
integration multivariable-calculus greens-theorem
edited Jan 31 at 3:36
anomaly
17.8k42666
asked Jan 31 at 1:38
user63266user63266
92
$endgroup$
add a comment |
$begingroup$
$F(x,y) = (y^3-6y)i + (6x-x^3)j$
a. Using Green's Theorem, find the simple closed curve C for which the integral
$ ∳F cdot dr $ (with positive orientation) will have the largest positive value.
b. Compute this largest possible value.
I'm quite certain that this is just $ iint Nx-My $ $dA$ but I do not know how to find the bounds in this scenario for both integrals. Though I'm also sure that this problem can also be done using just $ ∳ F cdot dr $ as there is an equation given for F.
integration multivariable-calculus greens-theorem
edited Jan 31 at 3:36
anomaly
17.8k42666
asked Jan 31 at 1:38
user63266user63266
92
$endgroup$
$F(x,y) = (y^3-6y)i + (6x-x^3)j$
a. Using Green's Theorem, find the simple closed curve C for which the integral
$ ∳F cdot dr $ (with positive orientation) will have the largest positive value.
b. Compute this largest possible value.
I'm quite certain that this is just $ iint Nx-My $ $dA$ but I do not know how to find the bounds in this scenario for both integrals. Though I'm also sure that this problem can also be done using just $ ∳ F cdot dr $ as there is an equation given for F.
integration multivariable-calculus greens-theorem
integration multivariable-calculus greens-theorem
edited Jan 31 at 3:36
anomaly
17.8k42666
asked Jan 31 at 1:38
user63266user63266
92
edited Jan 31 at 3:36
anomaly
17.8k42666
asked Jan 31 at 1:38
user63266user63266
92
edited Jan 31 at 3:36
anomaly
17.8k42666
edited Jan 31 at 3:36
anomaly
17.8k42666
edited Jan 31 at 3:36
anomaly
17.8k42666
17.8k42666
asked Jan 31 at 1:38
user63266user63266
92
asked Jan 31 at 1:38
user63266user63266
92
asked Jan 31 at 1:38
user63266user63266
92
92
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are correct that if we integrate around a closed curve $C$ that bounds a region $Omega$, then
$$
oint F cdot dr = iint_{Omega} (N_x - M_y) dA.
$$
Here, we have
$$
(N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). tag{1}
$$ In order to maximize $iint_{Omega} (N_x - M_y) dA$, we want $Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) geq 0$ if and only if $x^2 + y^2 leq 4$, i.e. $(N_x - M_y)(x,y) geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
You are correct that if we integrate around a closed curve $C$ that bounds a region $Omega$, then
$$
oint F cdot dr = iint_{Omega} (N_x - M_y) dA.
$$
Here, we have
$$
(N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). tag{1}
$$ In order to maximize $iint_{Omega} (N_x - M_y) dA$, we want $Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) geq 0$ if and only if $x^2 + y^2 leq 4$, i.e. $(N_x - M_y)(x,y) geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
add a comment |
$begingroup$
You are correct that if we integrate around a closed curve $C$ that bounds a region $Omega$, then
$$
oint F cdot dr = iint_{Omega} (N_x - M_y) dA.
$$
Here, we have
$$
(N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). tag{1}
$$ In order to maximize $iint_{Omega} (N_x - M_y) dA$, we want $Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) geq 0$ if and only if $x^2 + y^2 leq 4$, i.e. $(N_x - M_y)(x,y) geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
add a comment |
$begingroup$
You are correct that if we integrate around a closed curve $C$ that bounds a region $Omega$, then
$$
oint F cdot dr = iint_{Omega} (N_x - M_y) dA.
$$
Here, we have
$$
(N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). tag{1}
$$ In order to maximize $iint_{Omega} (N_x - M_y) dA$, we want $Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) geq 0$ if and only if $x^2 + y^2 leq 4$, i.e. $(N_x - M_y)(x,y) geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
$endgroup$
You are correct that if we integrate around a closed curve $C$ that bounds a region $Omega$, then
$$
oint F cdot dr = iint_{Omega} (N_x - M_y) dA.
$$
Here, we have
$$
(N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). tag{1}
$$ In order to maximize $iint_{Omega} (N_x - M_y) dA$, we want $Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) geq 0$ if and only if $x^2 + y^2 leq 4$, i.e. $(N_x - M_y)(x,y) geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
edited Jan 31 at 3:27
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
answered Jan 31 at 2:03
Jordan GreenJordan Green
1,146410
1,146410
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
add a comment |
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
So that would mean that the bounds would be from 0 to 2 for both integrals, correct?
$endgroup$
– user63266
Jan 31 at 3:07
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ pm sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates.
$endgroup$
– Jordan Green
Jan 31 at 3:10
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively?
$endgroup$
– user63266
Jan 31 at 3:24
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 pi$ (or, alternatively, any real numbers with a difference of $2 pi$).
$endgroup$
– Jordan Green
Jan 31 at 3:27
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
$begingroup$
Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help!
$endgroup$
– user63266
Jan 31 at 3:29
add a comment |
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