- #1

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- 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

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,

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?

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?

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