Welcome back to recitation. In this video I’d like us to

consider the following problem. So the question is for what

simple closed curve C, which is oriented positively around

the region it encloses, does the integral over

C of minus the quantity x squared y plus 3x minus 2y dx

plus 4 y squared x minus 2x dy achieve its minimum value? So again, the thing

we want to find here, the point we want to make,

is that we have this integral and we’re allowed

to vary C. So we’re allowed to change the curve,

the simple closed curve. And we want to know for what

curve C does this integral achieve its minimum value? So why don’t you work

on that, think about it, pause the video, and

when you’re feeling like you’re ready to see how I

do it, bring the video back up. Welcome back. So again, what we would

like to do for this problem is we want to take

this quantity, which is varying in C, and

we want to figure out a way to minimize it. To find its minimum value. And actually what’s

interesting to think about, before we proceed

any further, is you might think you want to take

the smallest possible simple closed curve you can. In other words, you might want

to shrink it down to nothing. But then this integral would

be 0 and the question is, can we do better? Because we could have a

minimum value, of course, that’s negative. And so we could do

better by having a larger curve put in the right place. So I just want to point that

out, if you were thinking “I’ll just make C

not actually a curve, I just shrink it

down to nothing.” That won’t be the

best we can do. We can actually

find a curve that has an integral that is not

0, but is in fact negative. And then we want to make it

the most negative we can. So that’s the idea. So what we’re

going to do here is we’re going to use Green’s

theorem to help us. And so what we want to remember

is that if we have the integral over C of M*dx plus N*dy we want

to write that as the integral over R of N sub x

minus M sub y dx dy. So what I want to

point out, I’m just going to write down

what these are. I’m not going to take

the derivatives for you. I’m just going to show

you what they are. So in this case, with

the M and N that we have, we get exactly this

value for the integral. We get N sub x minus M sub

y is equal to x squared plus 4 y squared minus 4 dx dy. You can obviously

compute that yourself. Just take the derivative

of N with respect to x, subtract the derivative

of M with respect to y, you get a little cancellation

and you end up with this. And so now, instead of

thinking about trying to minimize this quantity

over here in terms of a curve, now we can think about trying

to minimize the quantity here in terms of a region. And the goal here

is to make the sum of all of this over

the whole region– we want to make it as negative

as we can possibly make it. So essentially

what we want to do is we want– on the

boundary of this region, we would like the

value here to be 0, and on the inside of the region

we’d like it to be negative. So let me point that

out again and just make sure we understand this. To make this quantity

as small as possible, what we would like–

let me actually just draw a little region. So say this is the region. To make this integral as small

as possible, what we want is that x squared plus

4 y squared minus 4 is negative inside the region. So if this whole quantity is

less than 0 inside the region, and we want it to, on the

boundary of the region, equal 0. And why do we want it to

equal 0 on the boundary? Well, that’s because then we’ve

gotten all the negative we could get and we haven’t

added in any positive and brought the value up. So that’s really why we want

the boundary of the region to be exactly where

this quantity equals 0. And so let’s think about,

geometrically, what describes R– oops,

that should have an s. What describes R,

where x squared plus 4 y squared minus 4

is less than 0 inside R, and that’s the same thing as–

in what we’re interested in– x squared plus 4y squared minus 4

equals 0 on the boundary of R. And if you look at this,

this is really the expression that will probably help you. This expression,

if you rewrite it, you rewrite it as x squared

plus 4y squared is equal to 4. And you see that this is

actually the equation that describes an ellipse. Maybe you see it more often

if you divide everything by 4, and so on the right-hand

side you have a 1, and your coefficients are

fractional, potentially, there. But this is exactly the

equation for an ellipse. And so the boundary of R

is an ellipse described by this equation,

but the boundary of R is actually just C. So C we now know is

exactly the curve that is carved out by this

equation on the plane, on the xy-plane. That’s an ellipse. So just to remind you

what we were trying to do. If you come back

here, we were trying to figure out a way to

minimize the certain integral over a path. And what we did was

instead of trying to look at a bunch of

paths and figure out what would minimize

that, we tried to see if Green’s

theorem would help us. So Green’s theorem allowed

us to take something that was an integral

over a path and change it to an integral over a region. And then when we look at

what we ended up with, we realize that we could

make this the most minimum if we let this be on the

region where it was negative everywhere. So we were looking

for a region where this quantity was everywhere

negative on the inside and 0 on the boundary. And that’s exactly what we did. And then we see that

we get to a point where the boundary

has this equation, x squared plus 4y

squared equals 4. We see then the

boundary’s an ellipse, and C is indeed the

boundary of the region. So we see that C is the ellipse

described by this equation. So that’s where I’ll stop.

Thanks a lot.

shouldn't it be -x^2 +4y^2 -2?

Thanks for all your great videos!

So you found an equation of an elipse; great! The fundamental question still remains, what is the importance of Green's theorem? What is the relevance? What physical or real world problems can be solved by Green's Thrm.? Or is Green's Thrm only a fun mathematicians way of spending time? Can someone make a video explaining why we should all know about Green's Thrm, please.

Am I the only one who's learning this for the first time?

Great video, unlike most of the mathematical videos on Youtube, it teaches you the way of thinking instead of stuffing algebraic calculations.

I think it's suppose to be 4y^2 – x^2 – 4.

can you explain why you want value along curve to be zero, can't it be negative also that the value along the curve is also below z axis rather than zero