# 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.

• Your integral is $I = \iint 2(x-y)\,dx\,dy$. You can see limits from picture: $$I = \int\limits_1^2 dx \int\limits_{y=2x}^{y=4} 2(x-y)\,dy$$ – Michael Galuza Aug 3 '15 at 17:06
• @MichaelGaluza Why 2x? Why not x? – lolinda Aug 3 '15 at 17:12
• Because points $(1,2)$ and $(2,4)$ lies on it. – Michael Galuza Aug 3 '15 at 17:13
• @MichaelGaluza Does it come out to -2/3. Can you verify this? – lolinda Aug 3 '15 at 17:15
• @lolinda For concept of Multiple Integration and Vector Calculus please watch these lectures google.co.in/… – Taylor Ted Aug 3 '15 at 17:17

$$\vec{F} = (e^{\sin{x}}+y^2, x^2+y^2) = (M,N)$$ 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*}

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

• Please don't do that, use latex – Michael Galuza Aug 3 '15 at 16:42
• @MichaelGaluza oh yes sure i will do once i learn about it – Taylor Ted Aug 3 '15 at 16:44
• Your reputation is over 500, and you have not one badge. You must use latex. It's not a forum for posting pictures. – Michael Galuza Aug 3 '15 at 16:49
• @MichaelGaluza Sorry i will edit my question at once – Taylor Ted Aug 3 '15 at 16:50
• @MichaelGaluza Can you help me with this?Can you help me find the limits of integration too? And I saw that in my book that whenever it goes clockwise we have to do -C to get positive orientation. Am I right? – lolinda Aug 3 '15 at 16:51