Evaluating Line Integrals using Green's Theorem I am currently learning about Green's Theorem, Curl and Divergence, and I came across a problem:
Given a two dimensional vector field: $$ F=\langle e^{\sin{x}}+y^2, x^2+y^2 \rangle$$
And then I am also given that there is a oriented curve $C$ such that it starts at point $(1,2)$ and moves along a line segment to the point $(1,4)$, then moves along another line segment to the point $(2,4)$, and then moves one more time along a third line segment back to $(1,2)$.
How do I calculate $$\int_C F\,dr?$$
My thoughts for this were that we could parameterize the movement of $C$? I would like to solve this using Green's Theorem if possible. But, I am very vague on this and I would like some explained help on this concept, since I will be having a test in the near future.
 A: $$\vec{F} = (e^{\sin{x}}+y^2, x^2+y^2) = (M,N)$$
From Green's theorem $$\int_C{F}\;\mathrm{d}r = \iint_R \text{curl }\vec{F} \; \mathrm{d}A$$
Since the curl of $\vec{F}$ is $N_x - M_y$
\begin{align*}
\iint_R \text{curl }\vec{F} \; \mathrm{d}A &= \iint_R 2x-2y\; \mathrm{d}A\\
&= 2\iint_R x-y\; \mathrm{d}A
\end{align*}
To convert to iterated integral we need to know the bounds of $y$ and $x$. If we fix some $x$ notice that $y$ ranges from the line that passes through $(1,2)$ and $(2,4)$ (the line $y = 2x$) to $y = 4$. $x$ ranges from $1$ to $2$. So now we have the iterated integral:
\begin{align*}
2\int_1^2\int_{2x}^{4} x-y \;\mathrm{d}y \;\mathrm{d}x &= 2\int_1^2\left(xy-\frac{y^2}{2}\bigg|_{2x}^4\right)\;\mathrm{d}x \\
&= 2\int_1^2\left(4x-8\right)-(2x^2-2x^2)\;\mathrm{d}x \\
&= 2\int_1^2 4x-8\; \mathrm{d}x \\
&=-4
\end{align*}
A: Using Green's Theorem. Now you have to evaluate double integral over region i have drawn instead of parametrization of line 3 times. Hope i have helped you
We have to compute 
$\int F.dr $ over 3 lines given in your question
So we use Greens Theorem to avoid parametrization of line three times.Given our vector field
M= $e^{sinx} + y^{2}$
N=$ x^{2} + y^{2}$
So using Green's Theorem we have
$\int F.dr $ over 3 lines given in your question = $ \iint\limits_{R} \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\mathrm{d}A $, where R is region of Triangle having vertices as $(1,2)$ $(1,4)$ and $(2,4)$
So we putting values of partials we get,
$ \iint\limits_{R} 2(x-y) \mathrm{d}A $
Now i hope you can take from here.It is simple double integral
EDIT
We have our double integral as 
$ \iint\limits_{R} 2(x-y)dydx $ .We have limits as , $y$ goes from $2x$ to $4$ and $x$ goes from $1$ to $2$. When you write equations for three lines, from there you will find your y limits as i have already mentioned
