Need a little help setting up an integral:

- Luigi0210

Need a little help setting up an integral:

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- 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

@hartnn

- hartnn

could you plot the region ?

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## More answers

- 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

something really weird happens with that integrand thingy
doesn't it?

- 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

nah I got the same thing for the integrand

- hartnn

oh, thats right, it becomes 0

- Luigi0210

That's why I got confused.

- 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

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

but @hartnn as I said I'm not familar with green's theorem
are there any conditions that the integrand must satisfy?

- 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

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

So using Green's Theorem would technically give me the same answer as if I solved it using line integrals?

- hartnn

thats right...can we find the answer using line integral and verify it to be =0 ... ?

- 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

|dw:1436984142535:dw|

- hartnn

|dw:1436984249338:dw|

- hartnn

|dw:1436984422686:dw|

- hartnn

|dw:1436984491305:dw|

- 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

hmm.. i was treating this as double integral :P

- 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

Would it need to be split into two single integrals?

- Luigi0210

*ds not s

- hartnn

we'll have to use parametric equations for x and y
x= f(t)
y= g(t)

- hartnn

I am aware of the linear parameterization,
but thinking how do we include y= x^2 there...

- hartnn

x= t
y= t^2
dx =dt
dy = 2t dt

- hartnn

plug all those in the original integral

- Luigi0210

\(\large \int [(t)^2+(t^2)^3 (1)+3(t)(t^2)^2)(2t)] dt\)?

- hartnn

yes, this integral we need to find along,
t = 0 to t =2

- hartnn

and i highly doubt you'll get a 0, because all terms are positive...

- hartnn

@dan815 @ganeshie8

- 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

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

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

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

if you get 0, you would have verified the answer we got from green's theorem

- ganeshie8

|dw:1436985910945:dw|
3 line integrals are needed... looks a bit painful hmm

- hartnn

oh yes, i took 0,2 instead of 2,0 :P

- hartnn

|dw:1436986201469:dw|

- Luigi0210

Thanks guys!

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