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

• WMDhamnekar
In summary: Omega## is the boundary of ##\Omega##. In the special case of ##p=1##, the boundary of a 1-dimensional region is a set of points. For example, if ##\Omega## is a 1-dimensional region, then ##\partial \Omega = \{p_1, p_2, \dots, p_n\}## is a set of points. If we add a direction to each point, we get a 1-dimensional oriented region, and ##\partial \Omega## is now a 1-form (a differential form of degree 1). The notation for this is ##\partial \Omega = \sum_{
WMDhamnekar
MHB
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
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 ##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}
##

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*} ##

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*}##

Last edited:
WMDhamnekar said:
Neither. Green's Theorem involves an anticlockwise orientation, so the answer should be ##-2##.

PeroK said:
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.

PeroK said:
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?

WMDhamnekar said:
Would you tell me where the author is wrong?
It's a mess. I'll let you look for the error.
WMDhamnekar said:
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:
$$= \int_0^1(- 1)dx - \int_0^1(1)dy = -2$$

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

PeroK said:
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 ##

PeroK said:
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.

WMDhamnekar said:
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
Orodruin said:
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?##

WMDhamnekar said:
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:
$$\displaystyle\oint_c (x^2-y^2)dx + (x^2+y^2)dy=-\displaystyle\int_0^1 \int_0^1 (2x +2y)dA =-2$$

WMDhamnekar
WMDhamnekar said:
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
$$\oint_{\partial\Omega} \omega = \int_\Omega d\omega.$$

WMDhamnekar
WMDhamnekar said:
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

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 ##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}
##

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*} ##

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*}##

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

## 1. What is a line integral?

A line integral is a mathematical concept used in multivariable calculus to calculate the total value of a function along a given curve or path. It takes into account the direction of the curve and the values of the function at each point along the curve.

## 2. What is the purpose of evaluating a line integral?

Evaluating a line integral allows us to calculate the total value of a function along a given curve, which can be useful in many real-world applications such as calculating work done by a force along a specific path or determining the average temperature along a certain trajectory.

## 3. How is a line integral calculated?

A line integral is calculated by breaking the curve into small segments, approximating the value of the function at each point, and then summing up all these approximations. As the number of segments approaches infinity, the approximation becomes more accurate and the line integral can be calculated more precisely.

## 4. What is the difference between a line integral and a regular integral?

A regular integral calculates the area under a curve in two dimensions, while a line integral calculates the value of a function along a curve in three dimensions. In other words, a regular integral is a special case of a line integral where the curve is a straight line.

## 5. What are some real-world applications of line integrals?

Line integrals have many applications in physics, engineering, and other fields. They can be used to calculate work done by a force, electric or magnetic fields along a path, fluid flow along a certain trajectory, and many other physical quantities. They are also used in computer graphics to render three-dimensional objects and in economics to calculate consumer surplus.

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