# Evaluate the line integral

• WMDhamnekar

#### 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 ##\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*} ##

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

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.

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?

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:
$$= \int_0^1(- 1)dx - \int_0^1(1)dy = -2$$

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

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 ##

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.

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?##

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
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
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 ##\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*} ##