# Application of Green’s theorem | MIT 18.02SC Multivariable Calculus, Fall 2010

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.

## 8 thoughts on “Application of Green’s theorem | MIT 18.02SC Multivariable Calculus, Fall 2010”

1. jazerazo says:

Thanks a lot.

2. shylildude says:

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

3. Sci Fi Animator says:

Thanks for all your great videos!

4. Alberto Palacios says:

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.

5. boanice223 says:

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

6. Li Jie says:

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

7. Salena Chaudhry says:

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

8. Bhavuk Garg says:

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