Evaluate the line integral of a vector field around a square Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Notation Question with Line Integrals over Vector FieldLine Integrals FT usage on this strange vector field: so what are the exact conditions?Flux of vector field across surface SLine integral of vector field whose curl=0Closed line integral of conservative field not zero?(RESOLVED) Nonsense circulation in a conservative vector fieldHow to calculate line integral over an ellipse with vector field undefined inside the ellipse?Line Integral of Vector Field without parametrization?Is there a specific notation to denote the potential function of a conservative vector field?Calculate line integral in a vector field
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Evaluate the line integral of a vector field around a square
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Notation Question with Line Integrals over Vector FieldLine Integrals FT usage on this strange vector field: so what are the exact conditions?Flux of vector field across surface SLine integral of vector field whose curl=0Closed line integral of conservative field not zero?(RESOLVED) Nonsense circulation in a conservative vector fieldHow to calculate line integral over an ellipse with vector field undefined inside the ellipse?Line Integral of Vector Field without parametrization?Is there a specific notation to denote the potential function of a conservative vector field?Calculate line integral in a vector field
$begingroup$
I am asking this question because I believe the answer sheet I was given has an incorrect solution.
The task is to evaluate (by hand!) the line integral of the vector field $mathbfF(x,y) = x^2y^2 mathbfhati + x^3y mathbfhatj$ over the square given by the vertices (0,0), (1,0), (1,1), (0,1) in the counterclockwise direction. This vector field is not conservative by the way.
The answer I was given is as follows:
Now the part I believe to be incorrect is the parametrization of the third curve $mathcalC_3$. I think it is wrong due to the direction: the given parametrization is the curve going up instead of down.
Since we are going in the counterclockwise direction, I believe the parametrization should be $mathbfr(t) = 1-tmathbfhati + mathbfhatj$ giving:
$$int_mathcalC_3 x^2y^2 dx + x^3ydy = int^1_0(1-t)^2dt=int^1_0 1-2t+t^2=frac13$$
Hopefully someone can confirm my suspicion or tell me why I am wrong, thank you
vector-fields line-integrals
$endgroup$
add a comment |
$begingroup$
I am asking this question because I believe the answer sheet I was given has an incorrect solution.
The task is to evaluate (by hand!) the line integral of the vector field $mathbfF(x,y) = x^2y^2 mathbfhati + x^3y mathbfhatj$ over the square given by the vertices (0,0), (1,0), (1,1), (0,1) in the counterclockwise direction. This vector field is not conservative by the way.
The answer I was given is as follows:
Now the part I believe to be incorrect is the parametrization of the third curve $mathcalC_3$. I think it is wrong due to the direction: the given parametrization is the curve going up instead of down.
Since we are going in the counterclockwise direction, I believe the parametrization should be $mathbfr(t) = 1-tmathbfhati + mathbfhatj$ giving:
$$int_mathcalC_3 x^2y^2 dx + x^3ydy = int^1_0(1-t)^2dt=int^1_0 1-2t+t^2=frac13$$
Hopefully someone can confirm my suspicion or tell me why I am wrong, thank you
vector-fields line-integrals
$endgroup$
add a comment |
$begingroup$
I am asking this question because I believe the answer sheet I was given has an incorrect solution.
The task is to evaluate (by hand!) the line integral of the vector field $mathbfF(x,y) = x^2y^2 mathbfhati + x^3y mathbfhatj$ over the square given by the vertices (0,0), (1,0), (1,1), (0,1) in the counterclockwise direction. This vector field is not conservative by the way.
The answer I was given is as follows:
Now the part I believe to be incorrect is the parametrization of the third curve $mathcalC_3$. I think it is wrong due to the direction: the given parametrization is the curve going up instead of down.
Since we are going in the counterclockwise direction, I believe the parametrization should be $mathbfr(t) = 1-tmathbfhati + mathbfhatj$ giving:
$$int_mathcalC_3 x^2y^2 dx + x^3ydy = int^1_0(1-t)^2dt=int^1_0 1-2t+t^2=frac13$$
Hopefully someone can confirm my suspicion or tell me why I am wrong, thank you
vector-fields line-integrals
$endgroup$
I am asking this question because I believe the answer sheet I was given has an incorrect solution.
The task is to evaluate (by hand!) the line integral of the vector field $mathbfF(x,y) = x^2y^2 mathbfhati + x^3y mathbfhatj$ over the square given by the vertices (0,0), (1,0), (1,1), (0,1) in the counterclockwise direction. This vector field is not conservative by the way.
The answer I was given is as follows:
Now the part I believe to be incorrect is the parametrization of the third curve $mathcalC_3$. I think it is wrong due to the direction: the given parametrization is the curve going up instead of down.
Since we are going in the counterclockwise direction, I believe the parametrization should be $mathbfr(t) = 1-tmathbfhati + mathbfhatj$ giving:
$$int_mathcalC_3 x^2y^2 dx + x^3ydy = int^1_0(1-t)^2dt=int^1_0 1-2t+t^2=frac13$$
Hopefully someone can confirm my suspicion or tell me why I am wrong, thank you
vector-fields line-integrals
vector-fields line-integrals
edited Mar 25 at 16:37
gt6989b
35.8k22557
35.8k22557
asked Mar 25 at 16:36
RooseRoose
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102
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$begingroup$
Notice that you forgot to parameterize $rm dx$. By choosing $x(t) = 1-t$, you would then have $rm dx = -rm dt$, giving you the same answer as the solution.
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$begingroup$
Notice that you forgot to parameterize $rm dx$. By choosing $x(t) = 1-t$, you would then have $rm dx = -rm dt$, giving you the same answer as the solution.
$endgroup$
add a comment |
$begingroup$
Notice that you forgot to parameterize $rm dx$. By choosing $x(t) = 1-t$, you would then have $rm dx = -rm dt$, giving you the same answer as the solution.
$endgroup$
add a comment |
$begingroup$
Notice that you forgot to parameterize $rm dx$. By choosing $x(t) = 1-t$, you would then have $rm dx = -rm dt$, giving you the same answer as the solution.
$endgroup$
Notice that you forgot to parameterize $rm dx$. By choosing $x(t) = 1-t$, you would then have $rm dx = -rm dt$, giving you the same answer as the solution.
answered Mar 25 at 16:42
![](https://i.stack.imgur.com/qVmp1.jpg?s=32&g=1)
![](https://i.stack.imgur.com/qVmp1.jpg?s=32&g=1)
mvarblemvarble
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