Is the Calculation of the Vector Line Integral Over a Square Correct?

[WMDhamnekar]

Homework Statement

$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy,$where c is the boundary of the unit square,oriented clockwise.

Relevant Equations

No relevant equation

Author's answer:
Recognizing that this integral is simply a vector line integral of the vector field $F=(x^2−y^2)i+(x^2+y^2)j$ over the closed, simple curve c given by the edge of the unit square, one sees that $(x^2−y^2)dx+(x^2+y^2)dy=F\cdot ds$
is just a differentiable 1-form. The process here would be, then, the parameterize the unit square perimeter by time, and integrate under the parameterization: We get $\begin{equation}c(t)=\begin{cases}(0,t), 0≤t≤1\\ (t−1,1) ,1≤t≤2 \\ (1,3−t), 2≤t≤3\\ (4−t,0) ,3≤t≤4. \end{cases} \end{equation}$

as our clockwise parameterization, beginning and ending at the origin. To understand the switch to the parameterization, we highlight the first “piece”: Along the left-side edge of the unit square only, the parameterization is the path c1, going from (0, 0) to (0, 1) and parameterized by t in the y-direction only. We get

$\begin{align*}\displaystyle\int_{c_1} F\cdot ds &= \displaystyle\int_{c_1} (x^2-y^2)dx + (x^2+y^2)dy \\ &= \displaystyle\int_0^1 F_1 (x(t),y(t)) x'(t) dt + F_2 (x(t), y(t))y'(t) dt \\ &=\displaystyle\int_0^1 ((0)^2 -(t)^2 )(0dt) + ((0)^2 +(t)^2 )(1dt) \\ &=\displaystyle\int_0^1 t^2 dt = \frac{t^3}{3}\big{|}_0^1 =\frac13 \end{align*}$

Hence on the four pieces (so once around the square), we get

$\begin{align*}\displaystyle\oint_c F \cdot ds &= \displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy \\ &=\displaystyle\int_0^1 t^2 dt \displaystyle\int_1^2 ((t-1)^2 -1^2) dt + \displaystyle\int_2^3 (1^2 - (3-t)^2 )dt + \displaystyle\int_3^4 (4-t)^2 dt\\ &= \displaystyle\int_0^1 t^2 dt +\displaystyle\int_1^2 (t^2 - 2t )dt + \displaystyle\int_2^3 (10 -6t +t^2 )dt +\displaystyle\int_3^4 (16 - 8t +t^2)dt\\ &= \frac13 + \left ( \frac{t^3}{3}- t^2\right ) \big{|}_1^2 + \left( 10t - 3t^2 +\frac{t^3}{3}\right)\big{|}_2^3 +\left( 16t - 4t^2 +\frac{t^3}{3}\right) \big{|}_3^4 \\ &= \frac13 +\left( \frac83 -4 -\frac13 +1\right) +\left( 30 -27 +9 -20 +12 - \frac83 \right) + \left( 64-64 +\frac{64}{3}-48 + 36 -9\right) \\ &= \frac13 -\frac23 + \frac43 +\frac13 = \frac43 \end{align*}$

My answer:

Here, c is the boundary of the unit square oriented clockwise of the regionR={(x,y):0≤x≤1,0≤y≤1}
By Green's theorem $P(x,y)=(x^2−y^2),Q(x,y)=(x^2+y^2)$we have
$\begin{align*} \displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2 )dy &= \displaystyle\iint\limits_R \left( \frac{\partial{Q}}{\partial{x}} - \frac{\partial{P}}{\partial{y}}\right)d A\\ &= \displaystyle\iint\limits_R (2x+2y)dA =2 \end{align*}$

Whose answer is correct? My answer or author's answer?

 

[PeroK]

Whose answer is correct? My answer or author's answer?

Neither. Green's Theorem involves an anticlockwise orientation, so the answer should be $-2$.

 

[Orodruin]

Neither. Green's Theorem involves an anticlockwise orientation, so the answer should be $-2$.

Which is also easily obtained from Stokes’ theorem (of which Green’s theorem is a special case):
$$
d((x^2-y^2) dx + (x^2 + y^2) dy) = -2y\, dy\wedge dx + 2x \, dx \wedge dy = -2(x + y) dy\wedge dx
$$
Anti-clockwise rotation means $dy\wedge dx$ is correct surface orientation to get the correct sign.

 

[WMDhamnekar]

Neither. Green's Theorem involves an anticlockwise orientation, so the answer should be $-2$.

Would you tell me where the author is wrong?

Would you tell me the correct upper and lower integral limits of the area for my answer?

 

[PeroK]

Would you tell me where the author is wrong?

It's a mess. I'll let you look for the error.

Would you tell me the correct upper and lower integral limits of the area for my answer?

You have:
$$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy$$$$= \int_0^1 (0 + y^2) dy + \int_0^1(x^2 - 1)dx - \int_0^1(1 + y^2)dy - \int_0^1 (x^2 - 0)dx$$Note that the $x^2$ and $y^2$ terms cancel out, leaving:
$$= \int_0^1(- 1)dx - \int_0^1(1)dy = -2$$

 

[PeroK]

Note that, more generally,for the given path:
$$\displaystyle\oint_c (f(x) - g(y))dx + (f(x)+g(y))dy$$$$= f(0) + g(0) - f(1) - g(1)$$

 

[WMDhamnekar]

It's a mess. I'll let you look for the error.

You have:
$$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy$$$$= \int_0^1 (0 + y^2) dy + \int_0^1(x^2 - 1)dx - \int_0^1(1 + y^2)dy - \int_0^1 (x^2 - 0)dx$$Note that the $x^2$ and $y^2$ terms cancel out, leaving:
$$= \int_0^1(- 1)dx - \int_0^1(1)dy = -2$$

I hope the following presentation of answer would be correct

$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy=\displaystyle\int_0^1 \int_1^0 (2x +2y)dA =-2$

 

[WMDhamnekar]

It's a mess. I'll let you look for the error.

You have:
$$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy$$$$= \int_0^1 (0 + y^2) dy + \int_0^1(x^2 - 1)dx - \int_0^1(1 + y^2)dy - \int_0^1 (x^2 - 0)dx$$Note that the $x^2$ and $y^2$ terms cancel out, leaving:
$$= \int_0^1(- 1)dx - \int_0^1(1)dy = -2$$

How shall I know that you have used here green theorem? I used in my answer Green's theorem.

 

[PeroK]

How shall I know that you have used here green theorem? I used in my answer Green's theorem.

I didn't use Green's theorem. I just did the four line integrals separately. That seemed the simplest approach.

 

[WMDhamnekar]

Which is also easily obtained from Stokes’ theorem (of which Green’s theorem is a special case):
$$
d((x^2-y^2) dx + (x^2 + y^2) dy) = -2y\, dy\wedge dx + 2x \, dx \wedge dy = -2(x + y) dy\wedge dx
$$
Anti-clockwise rotation means $dy\wedge dx$ is correct surface orientation to get the correct sign.

What is the meaning of $dy \wedge dx?$

 

[PeroK]

I hope the following presentation of answer would be correct

$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy=\displaystyle\int_0^1 \int_1^0 (2x +2y)dA =-2$

Yes, but it would seem more logical to write:
$$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy=-\displaystyle\int_0^1 \int_0^1 (2x +2y)dA =-2 $$

 

[Orodruin]

What is the meaning of $dy \wedge dx?$

That is a 2-form.

Generally, Stokes’ theorem (not to be confused with the curl theorem — which is often also called Stokes’ theorem and is a special case, just as Green’s theorem and the divergence theorem) states that if $\omega$ is a $p$-form and $\Omega$ a $p+1$-dimensional region then
$$
\oint_{\partial\Omega} \omega = \int_\Omega d\omega.
$$

 

[WWGD]

Homework Statement:: $\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy,$where c is the boundary of the unit square,oriented clockwise.
Relevant Equations:: No relevant equation

Author's answer:
Recognizing that this integral is simply a vector line integral of the vector field $F=(x^2−y^2)i+(x^2+y^2)j$ over the closed, simple curve c given by the edge of the unit square, one sees that $(x^2−y^2)dx+(x^2+y^2)dy=F\cdot ds$
is just a differentiable 1-form. The process here would be, then, the parameterize the unit square perimeter by time, and integrate under the parameterization: We get $\begin{equation}c(t)=\begin{cases}(0,t), 0≤t≤1\\ (t−1,1) ,1≤t≤2 \\ (1,3−t), 2≤t≤3\\ (4−t,0) ,3≤t≤4. \end{cases} \end{equation}$

as our clockwise parameterization, beginning and ending at the origin. To understand the switch to the parameterization, we highlight the first “piece”: Along the left-side edge of the unit square only, the parameterization is the path c1, going from (0, 0) to (0, 1) and parameterized by t in the y-direction only. We get

$\begin{align*}\displaystyle\int_{c_1} F\cdot ds &= \displaystyle\int_{c_1} (x^2-y^2)dx + (x^2+y^2)dy \\ &= \displaystyle\int_0^1 F_1 (x(t),y(t)) x'(t) dt + F_2 (x(t), y(t))y'(t) dt \\ &=\displaystyle\int_0^1 ((0)^2 -(t)^2 )(0dt) + ((0)^2 +(t)^2 )(1dt) \\ &=\displaystyle\int_0^1 t^2 dt = \frac{t^3}{3}\big{|}_0^1 =\frac13 \end{align*}$

Hence on the four pieces (so once around the square), we get

$\begin{align*}\displaystyle\oint_c F \cdot ds &= \displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy \\ &=\displaystyle\int_0^1 t^2 dt \displaystyle\int_1^2 ((t-1)^2 -1^2) dt + \displaystyle\int_2^3 (1^2 - (3-t)^2 )dt + \displaystyle\int_3^4 (4-t)^2 dt\\ &= \displaystyle\int_0^1 t^2 dt +\displaystyle\int_1^2 (t^2 - 2t )dt + \displaystyle\int_2^3 (10 -6t +t^2 )dt +\displaystyle\int_3^4 (16 - 8t +t^2)dt\\ &= \frac13 + \left ( \frac{t^3}{3}- t^2\right ) \big{|}_1^2 + \left( 10t - 3t^2 +\frac{t^3}{3}\right)\big{|}_2^3 +\left( 16t - 4t^2 +\frac{t^3}{3}\right) \big{|}_3^4 \\ &= \frac13 +\left( \frac83 -4 -\frac13 +1\right) +\left( 30 -27 +9 -20 +12 - \frac83 \right) + \left( 64-64 +\frac{64}{3}-48 + 36 -9\right) \\ &= \frac13 -\frac23 + \frac43 +\frac13 = \frac43 \end{align*}$

My answer:

Here, c is the boundary of the unit square oriented clockwise of the regionR={(x,y):0≤x≤1,0≤y≤1}
By Green's theorem $P(x,y)=(x^2−y^2),Q(x,y)=(x^2+y^2)$we have
$\begin{align*} \displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2 )dy &= \displaystyle\iint\limits_R \left( \frac{\partial{Q}}{\partial{x}} - \frac{\partial{P}}{\partial{y}}\right)d A\\ &= \displaystyle\iint\limits_R (2x+2y)dA =2 \end{align*}$

Whose answer is correct? My answer or author's answer?

If I may, you're missing a + sign in the second term of the second line of your math .