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Evaluate the line integral of a vector field around a square

0

$begingroup$

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

share|cite|improve this question

edited Mar 25 at 16:37

gt6989b

35.8k22557

asked Mar 25 at 16:36

RooseRoose

102

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this question

edited Mar 25 at 16:37

gt6989b

35.8k22557

asked Mar 25 at 16:36

RooseRoose

102

$endgroup$

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

share|cite|improve this question

edited Mar 25 at 16:37

gt6989b

35.8k22557

asked Mar 25 at 16:36

RooseRoose

102

$endgroup$

share|cite|improve this question

edited Mar 25 at 16:37

gt6989b

35.8k22557

asked Mar 25 at 16:36

RooseRoose

102

share|cite|improve this question

edited Mar 25 at 16:37

gt6989b

35.8k22557

asked Mar 25 at 16:36

RooseRoose

102

edited Mar 25 at 16:37

gt6989b

35.8k22557

edited Mar 25 at 16:37

gt6989b

35.8k22557

asked Mar 25 at 16:36

RooseRoose

102

asked Mar 25 at 16:36

RooseRoose

102

1 Answer
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oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 25 at 16:42

mvarblemvarble

182

$endgroup$

add a comment | 

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0

$begingroup$

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.

share|cite|improve this answer

answered Mar 25 at 16:42

mvarblemvarble

182

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 25 at 16:42

mvarblemvarble

182

$endgroup$

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.

share|cite|improve this answer

answered Mar 25 at 16:42

mvarblemvarble

182

$endgroup$

share|cite|improve this answer

answered Mar 25 at 16:42

mvarblemvarble

182

answered Mar 25 at 16:42

mvarblemvarble

182

answered Mar 25 at 16:42

mvarblemvarble

182