Evaluating $\int_γ y^2 \;dx + x^2\; dy$ In an exercise I have to evaluate:
$$\int_γ y^2 \; dx + x^2 \; dy$$
where $γ$ is the line from $(1,0)$ to $(0,1)$. I solved line integrals before using $dS = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \cdot dt$. However I have never seen this particular notation before and I don't know what it means.
I would appreciate any help.
 A: The parametric representation of line from $(1,0)$  to $(0,1)$ is $y=-x+1, 0\leq x \leq1$, so $$\int_{\gamma}y^2dx+x^2dy=\int_1^0[(-x+1)^2-x^2]dx=0.$$
A: I imagine it can be construed as  the following:
If $\def\i{{\bf i}}\def\j{{\bf j}}\def\vdot{{\bf\cdot}}\def\r{{\bf r}}\def\ts{\textstyle}\def\/{}{\bf F}(x,y)=M(x,y)\i+N(x,y)\j$ gives a force field, the total work done by $\bf F$ on a point moving along the curve $C$  parameterized by ${\bf r}(t), a\le t\le b$ is given by the work integral:
$$\tag{1}
     \eqalign{\int_C \!\!{\bf F \vdot T}\,ds=\!\int_a^b \!\!{\bf F\/}\vdot d\r =\! \int_a^b \!{\bf F\/}\vdot \r'(t) \,dt  
       = \! \!\int_a^b \Bigl[\ts M\,{dx\over dt}+N\,{dy\over dt}   \Bigr] \,dt}.
  $$
   We abbreviate the rightmost integral above as 
$$\tag{2}\int_a^b M\,dx+N\,dy.$$
To evaluate  $\int_C {\bf F \vdot T}\,ds$, where ${\bf F}(x,y) = M\i+N\j\,$:


*

*Parameterize $C$: ${\bf r}(t) = x(t)\i+y(t)\j$, $a\le t\le b$.

*In $M$ and $N$, replace $x$ with $x(t)$ and $y$ with $y(t)$.

*From the parameterization, find $dy\over dt$ and $dx\over dt$.

*Substituting everything into the rightmost integral in $(1)$, calculate the integral (which should by now be a normal integral in $t$) $\int_C {\bf F \vdot T}\,ds=\int_a^b M {dx}+
    N{dy}$.



In your case ${\bf F}(x,y)=x^2\i+y^2\j$, and you'll have to find a parameterization of the path $\gamma$.
A: Along the line $x+y=1$ we have $dy=-dx$, so
$$\int_{\gamma}y^2dx+x^2dy=\int_1^0 \left((1-x)^2-x^2\right) dx=\int_0^1(2x-1)\,dx=\left[x^2-x\right]_0^1=0.$$
If the curve were closed, we could use Green's theorem:
$$\int_{\partial R}Ldx+Mdy=
\int_R\left(
\frac{\partial M}{\partial x}-
\frac{\partial L}{\partial y}
\right)\,dx\,dy$$
while if the $1$-form $\omega=y^2dx+x^2dy$
were the exterior derivative $\omega=df$ of a
potential (a smooth function) or $0$-form $f(x,y)$,
then the integral would be $f(0,1)-f(1,0)$
by Stokes' theorem, which generalizes
the fundamental theorem of calculus
and has the physical interpretation of
work as a change in potential energy,
as @DavidMitra points out.
However, neither of these apply in our case.
If we view $\omega$ in stead as a vector field $\mathbf{F}=(y^2,x^2)$,
then there exists a potential $f$ so that $\mathbf{F}=\nabla f$ iff
$\mathbf{F}$ is conservative, i.e., iff the scalar curl of $\mathbf{F}$, $\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}$, vanishes.
In such a case, the integral is path-independent and the result is the
difference in potential; in our case however, once again, this doesn't help,
since $\nabla\times\mathbf{F}=2(x-y)\hat{k}\ne\vec{0}$.
In our case, the path conspires with the form $\omega$ or vector field $\mathbf{F}$ to produce a vanishing result. If we integrated along line
$x+y=a\ne0,1$ from $(a,0)$ to $0,a$, we would get
$$
\int_0^a(2ax-a^2)\,dx=a\left[x^2-ax\right]_0^1=a(1-a)\ne0\,,
$$
while if we used in stead the path along the axes
from $(a,0)$ via $(0,0)$ to $(0,a)$, we would get $0$.
A: $x(t)=1-t $
$y(t)=t$  
with $0 \le t \le 1 $.  We have $dx=-dt$, $dy=dt$.
$\int_{\gamma} y^2+x^2 dy= \int_{0}^{1} t^2  (-dt)+(1-t)^2 dt=\int_{0}^{1} (1-2t) dt= 0 $
