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Luigi0210
 one year ago
Need a little help setting up an integral:
Luigi0210
 one year ago
Need a little help setting up an integral:

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Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Consider \(\large \int_{C} (x^2+y^3)dx+3xy^2dy \) 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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3could you plot the region ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3\(\Large \oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x}  \frac{\partial L}{\partial y}\right)\, dx\, dy\)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1something really weird happens with that integrand thingy doesn't it?

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1nah I got the same thing for the integrand

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3oh, thats right, it becomes 0

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0That's why I got confused.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I never applied green's theorem before... but if I were to read this like a "normal" integral we are looking at nothing

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.1but @hartnn as I said I'm not familar with green's theorem are there any conditions that the integrand must satisfy?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3"If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there ..." ~Wiki

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.0So using Green's Theorem would technically give me the same answer as if I solved it using line integrals?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3thats right...can we find the answer using line integral and verify it to be =0 ... ?

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3we 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
 one year ago
Best ResponseYou've already chosen the best response.3hmm.. i was treating this as double integral :P

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0Would it need to be split into two single integrals?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3we'll have to use parametric equations for x and y x= f(t) y= g(t)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3I am aware of the linear parameterization, but thinking how do we include y= x^2 there...

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3x= t y= t^2 dx =dt dy = 2t dt

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3plug all those in the original integral

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \int [(t)^2+(t^2)^3 (1)+3(t)(t^2)^2)(2t)] dt\)?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3yes, this integral we need to find along, t = 0 to t =2

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3and i highly doubt you'll get a 0, because all terms are positive...

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0I'm looking in the book and it says something about using the line segments in \(r(t)=(1t)r_{0}+tr_{1} \) where t is from 0 to 1 to make a parametric equation?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Here 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
 one year ago
Best ResponseYou've already chosen the best response.3so using green's theorem, getting a 0 is indeed correct. we were trying to find the line integral, (not using green's theorem)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3along 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
 one year ago
Best ResponseYou've already chosen the best response.3if you get 0, you would have verified the answer we got from green's theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436985910945:dw 3 line integrals are needed... looks a bit painful hmm

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3oh yes, i took 0,2 instead of 2,0 :P
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