Luigi0210
  • Luigi0210
Need a little help setting up an integral:
Mathematics
jamiebookeater
  • jamiebookeater
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Luigi0210
  • Luigi0210
Consider \(\large \int_{C} (x^2+y^3)dx+3xy^2dy \) where C is the parabola \(y=x^2\) from (2,4) to (0,0) followed by the line segment from (0, 0) to (2, 0) and another line from (2,0) to (2, 4) Evaluate the integral using Green's Theorem
Luigi0210
  • Luigi0210
hartnn
  • hartnn
could you plot the region ?

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hartnn
  • hartnn
\(\Large \oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dx\, dy\)
myininaya
  • myininaya
something really weird happens with that integrand thingy doesn't it?
Luigi0210
  • Luigi0210
Yes, I end up getting 0 when taking the partials, unless I did that wrong... I had set it from y: 0 -> 2x and x: 0 -> 2
myininaya
  • myininaya
nah I got the same thing for the integrand
hartnn
  • hartnn
oh, thats right, it becomes 0
Luigi0210
  • Luigi0210
That's why I got confused.
myininaya
  • myininaya
I never applied green's theorem before... but if I were to read this like a "normal" integral we are looking at nothing
myininaya
  • myininaya
you know like when I think about this integral I think about something like: \[\int\limits_{1}^{2}0 dx=0\] this 0 because |dw:1436983532812:dw|
myininaya
  • myininaya
but @hartnn as I said I'm not familar with green's theorem are there any conditions that the integrand must satisfy?
hartnn
  • hartnn
"If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there ..." ~Wiki
hartnn
  • hartnn
so i guess, there are no conditions other than that you should be able to find partial derivatives, in this case, you could find it...so you can apply
Luigi0210
  • Luigi0210
So using Green's Theorem would technically give me the same answer as if I solved it using line integrals?
hartnn
  • hartnn
thats right...can we find the answer using line integral and verify it to be =0 ... ?
Luigi0210
  • Luigi0210
I don't know how to set up the integral as a line integral though, not used to the form :/ Could you help me set it up? :P
hartnn
  • hartnn
|dw:1436984142535:dw|
hartnn
  • hartnn
|dw:1436984249338:dw|
hartnn
  • hartnn
|dw:1436984422686:dw|
hartnn
  • hartnn
|dw:1436984491305:dw|
hartnn
  • hartnn
we need to find the bounds of the double integration. limits for x : from curve to vertical line so, x = + sqrt y to x =2 limits for y: from 0 to 2
hartnn
  • hartnn
hmm.. i was treating this as double integral :P
Luigi0210
  • Luigi0210
Yes, the bounds make sense, just don't know how to handle it when it's split into dx and dy instead of something like \(\large \int_{C} y^ ds \)
Luigi0210
  • Luigi0210
Would it need to be split into two single integrals?
Luigi0210
  • Luigi0210
*ds not s
hartnn
  • hartnn
we'll have to use parametric equations for x and y x= f(t) y= g(t)
hartnn
  • hartnn
I am aware of the linear parameterization, but thinking how do we include y= x^2 there...
hartnn
  • hartnn
x= t y= t^2 dx =dt dy = 2t dt
hartnn
  • hartnn
plug all those in the original integral
Luigi0210
  • Luigi0210
\(\large \int [(t)^2+(t^2)^3 (1)+3(t)(t^2)^2)(2t)] dt\)?
hartnn
  • hartnn
yes, this integral we need to find along, t = 0 to t =2
hartnn
  • hartnn
and i highly doubt you'll get a 0, because all terms are positive...
hartnn
  • hartnn
Luigi0210
  • Luigi0210
I'm looking in the book and it says something about using the line segments in \(r(t)=(1-t)r_{0}+tr_{1} \) where t is from 0 to 1 to make a parametric equation?
ganeshie8
  • ganeshie8
Here curl is 0, that means the given vector field is conservative. Which is same as saying the work done is path independent, consequently any closed loop line integral is 0.
hartnn
  • hartnn
so using green's theorem, getting a 0 is indeed correct. we were trying to find the line integral, (not using green's theorem)
hartnn
  • hartnn
along the curve, y= x^2, we took, x =t, y = t^2 and we'll get a certain integral value. now along the line x= 0 to x= 2, again try to find the integral. in this case, y =0, dy =0 similarly along the line y= 0 to y=4, again try to find the integral, in this case, x= 2, dx =0 the addition of all these 3 integrals should be = 0
hartnn
  • hartnn
if you get 0, you would have verified the answer we got from green's theorem
ganeshie8
  • ganeshie8
|dw:1436985910945:dw| 3 line integrals are needed... looks a bit painful hmm
hartnn
  • hartnn
oh yes, i took 0,2 instead of 2,0 :P
hartnn
  • hartnn
|dw:1436986201469:dw|
Luigi0210
  • Luigi0210
Thanks guys!

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