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Suppose we have a region $G$ that is bounded by the straight lines $x=a$, $x=b$, $y=c$ and by an arc $y = f(x)$ (which lies above $y=c$) where $a leq x leq b$.
If $f$, $P(x,y)$ and $Q(x,y)$ are all continuously differentiable in a region which contains the region $G$ and its boundary, denoted as $partial G$, then:

$$iint_G Bigg(fracpartial Qpartial x-fracpartial Ppartial yBigg) dA = oint_partial G Big(P cdot dx + Q cdot dyBig)$$

We can prove this lemma by showing the following integrals, which are taken by splitting up the above equality, are as such:
beginequation
iint_G Bigg(fracpartial Qpartial xBigg) cdot dA = oint_partial GQ cdot dy
endequation
beginequation
-iint_G Bigg(fracpartial Ppartial yBigg) cdot dA = oint_partial GP cdot dx
endequation
First, we shall prove the bottom equation. First we need to manipulate the double integral using the bounds given and then apply the Fundamental Theorem of Calculus. This will give us the following:
beginalign
-iint_G Bigg(fracpartial Ppartial yBigg) cdot dA &= - int_a^b int_c^f(x) fracpartial Ppartial y cdot dy cdot dx labeleqn:27 \
& = - int_a^b Big(P[x,f(x)] - P[x,c]Big) cdot dx labeleqn:28
endalign
Letting $oversetrightharpoonupC_3$ be the same as $y = f(x)$ and $oversetrightharpoonupC_1$ be the same as $y=c$ we can rewrite the line above as:
$$int_oversetrightharpoonupC_3P(x,y) cdot dx + int_oversetrightharpoonupC_1P(x,y) cdot dx$$

How does: $- int_a^b Big(P[x,f(x)] - P[x,c]Big) cdot dx = int_oversetrightharpoonupC_3P(x,y) cdot dx + int_oversetrightharpoonupC_1P(x,y) cdot dx$ ???

Apologies I couldn't get the anti clockwise arrow on the integrals for the contours.