1
$\begingroup$

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.

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Please don't do that, use latex $\endgroup$ – Michael Galuza Aug 3 '15 at 16:42
  • $\begingroup$ @MichaelGaluza oh yes sure i will do once i learn about it $\endgroup$ – Taylor Ted Aug 3 '15 at 16:44
  • $\begingroup$ Your reputation is over 500, and you have not one badge. You must use latex. It's not a forum for posting pictures. $\endgroup$ – Michael Galuza Aug 3 '15 at 16:49
  • $\begingroup$ @MichaelGaluza Sorry i will edit my question at once $\endgroup$ – Taylor Ted Aug 3 '15 at 16:50
  • $\begingroup$ @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? $\endgroup$ – lolinda Aug 3 '15 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.