Is this a particular case of the Stoke's Theorem? How to prove the equality? I'm doing my Calculus III homework and I'm stuck in a question. It seems to be a particular case of the Stoke's Theorem but I'm not sure. The problem is:
Be $B$ a triangle with vertex $(0,0)$, $(1,0)$ and $(1,1)$; $\gamma$ being the border of the triangle in the anticlockwise direction, verify that:
$$\int_{\gamma}Pdx+Qdy=\iint_B \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$$
Where $P$ and $Q$ are supposed to be of class $C^1$ in a open space $\Omega$ containing $B$.
The problem does not say about the Stokes Theorem and the suggestion is to solve both sides and compare results. But I'm stuck in the double integral part. Here is my current resolution:
I parameterized this right triangle in the following way:
$$\gamma (t)=\begin{cases} \gamma _{ 1 }=(t,0),0\le t\le 1 \\ \gamma _{ 2 }=(1,t-1),1\le t\le 2 \\ \gamma _{ 3 }=(3-t,3-t),2\le t\le 3 \end{cases}$$
Knowing that $$\int_{\gamma}Pdx+Qdy=\int_a^b \left[P\frac{dx}{dt}+Q\frac{dy}{dt}\right]dt$$
the first side of the equation can be written as:
$$\int_0^1P(t,0)dt+\int_1^2Q(1,t-1)dt-\int_2^3P(3-t,3-t)-\int_2^3Q(3-t,3-t)dt$$
Then I'm stuck by here. How could I simplify this equation in order to equate to other side?
Also, there is some tip about how I develop the right hand side of the equation? I mean, how do I do the double integral of this partial thing? I'm stuck in the following:
$$\iint_B \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy=\int_0^1 \int_0^x \frac{\partial Q}{\partial x}dxdy-\int_0^1 \int_y^1 \frac{\partial P}{\partial y}dxdy$$
The limits were chosen using the constraints of the triangle.
Is there a tip to develop both sides of the equation and prove the equality? Is there some mistake in my resolution so far? 
Thank you in advance! I really need tips to solve this problem.
 A: The relation can be shown separately for the cases 


*

*(1) when $Q=0$, and $P$ arbitrary, 

*(2) when $P=0$, and $Q$ arbitrary.


(It depends linearly on the $1$-form $P\; dx + Q\;dy$, so it is enough to show it for $P\; dx+0\; dy$ and $0\; dx+Q\; dy$.) 
(Observe that (2) follows from (1) by changing the orientation, i.e. using the map
$T:(x,y)\to(y,x)$, if we would show (1) for a general triangle. This is because $T$ maps the volume element $dx\wedge dy$ into $dy\wedge dx=-dx\wedge dy$. Our triangle is not general, but it is clear what should be transposed in the computations from one case to the other one.)
So let us show (1). So $Q=0$. I will write $P'_y$ for the differential of $P$ w.r.t. $y$. We start the computation, use Fubini for the integral on $B$:
$$
\begin{aligned}
\iint_B -P'_y\; dx\; dy
&=
-\int_0^1dx\int_0^xP'_y\; dy
\\
&=
-\int_0^1(P(x,x)-P(x,0))\; dx
\\
&=
\int_0^1P(x,0)\; x'\cdot dx +
\int_0^1P(1,x)\; 1'\cdot dx +
\int_1^0P(x,x)\; x'\cdot dx 
\\
&=
\int_{\gamma_1}P\; dx +
\int_{\gamma_2}P\; dx +
\int_{\gamma_3}P\; dx
\\
&=
\int_{\gamma=\gamma_1\cup\gamma_2\cup\gamma_3=\partial B}P\; dx\ .
\end{aligned}
$$
Recall here the following definition. Let $P=P(x,y)$ be a function defined in some domain $U\subseteq\Bbb R^2$, let $\gamma:I=[a,b]\to U$ be a function. We take it to be of the shape $\gamma(t)=(x(t),y(t))$, $t$ being a parameter in $I$. Then we have by definition
$$
\int_\gamma P\; dx:=\int_a^b P(x(t), y(t))\; x'(t)\; dt\ .
$$
In the above case, the paths $\gamma_1,\gamma_2,\gamma_3$ are "paths" defined on the "intervals" from $0$ to $1$, from $0$ to $1$, and from $1$ to $0$ (for this last one exchange extremities, and the sign), given by $\gamma_1(x)=(x,0)$, $\gamma_2(x)=(1,x)$, $\gamma_3(x)=(x,x)$.
The same idea of computation (with less details) works for the case $(2)$, we use again Fubini to rewrite the integral on $B$:
$$
\begin{aligned}
\iint_B Q'_x\; dx\; dy
&=
\int_0^1dy \int_y^1 Q'_x\; dx
\\
&=
\int_0^1 (Q(1,y)-Q(y,y))\; dy
\\
&=
\int_0^1 Q(y,0)\; 0'\cdot dy +
\int_0^1 Q(1,y)\; y'\cdot dy +
\int_1^0 Q(y,y)\; y'\cdot dx 
\\
&=
\int_{\gamma_1}Q\; dy +
\int_{\gamma_2}Q\; dy +
\int_{\gamma_3}Q\; dy
\\
&=
\int_{\gamma=\gamma_1\cup\gamma_2\cup\gamma_3=\partial B}Q\; dy\ .
\end{aligned}
$$
