CHRISTINE BREINER: Welcome
back to recitation. In this video I want to show how
we can use change of variables, to polar coordinates
in particular, to evaluate an integral
that, without the change in variables, we don't
have the techniques to do. So I'm going to show us how
to evaluate the integral from minus infinity to infinity of
e to the minus x squared dx. And I'll just point out that if
you do anything in probability you will see this
integral a great deal. So this is a
distribution and you'll see it a great deal if
you ever do anything in probability theory. But how are we going to
use polar coordinates to evaluate this? Well, the object is going to be
to introduce a little bit more into this integral, so
that when I actually introduce that in I'm
going to have an r squared term in the exponent and
I'm going to have-- of course by the change of variables in
the Jacobian-- multiplied by r. And that's what's
going to save us. So let me actually
just get right to it. I'm going to call this
integral, this quantity, capital I. We'll just call
this quantity capital I. And then what I'm going to do is
introduce a couple more things. So let me write it
down, and then I will justify that it's
reasonable to even look at this thing. And that it will give
us something useful. OK, so I just took this
integral and what I've done is I've now taken this quantity
and I've multiplied by an e to the minus y squared. And then I'm now integrating
from minus infinity to infinity in dy also. And what I want to point out,
that maybe isn't immediately obvious from the way
it's written here, but will be immediately
obvious when I write it a different way, is
that this quantity is actually just the square
of this quantity. So it's actually just two of
these multiplied together. And let me point
out why that is. Because this term right here can
be rewritten as e to the minus x squared, e to the
minus y squared. And then I have two integrals,
which are from minus infinity to infinity both. Sorry, that one looks
a little off kilter. But those two integrals,
which the bounds are minus infinity to infinity for both. And then I have a dx and a dy. Now the reason that it is just I
squared-- this quantity is just I squared-- is
because if you notice, this first integral, the
inside integral, this is independent of x. So I can move it out. So let me actually come up
here and write that down. So I can move it out and I can
actually rewrite the integral. And I'll put the bounds
here, because it's high up enough that I can
actually write it. e to the minus y squared. And then integral from
minus infinity to infinity e to the minus x squared dx dy. Right? So let's point out
again what I did. This quantity I circled
was the constant when I considered x the variable. Right? In terms of x, this
is the constant, so I can move it in front
of that integral, which is the dx integral. And so now what do I see? Well, this quantity
is what I've called I. And so now that's a constant. So I can move that
out to the front. That's a constant. That's I, as designated. And so now I have the integral
for minus infinity to infinity of e to the minus y squared dy. And then that again is another
I. Because if you notice, I mean, this was e to
the minus x squared dx, and so if I keep this as e
to the minus y squared dy, it's going to be
exactly the same value. So this is actually
equal to I squared. So the point I want to make,
if you come back over here to this integral, is
that this quantity I'm going to be able to
integrate very easily. And when I do that, if
I take the square root, I'm going to get this value. Because I just showed this
thing in the box was equal to I squared. So let me write it down. On this last part of the board
I will write down that thing again and then we'll
actually evaluate it and we'll see what we get. So let me rewrite. This is the thing
we want to evaluate. And I'm going to remind myself
that that's equal to I squared. I squared is equal
to this quantity. And I mentioned at the
beginning that we're going to use a trick. We're going to change variables
into polar coordinates. And so notice that the
region we're integrating over is the entire xy-plane. And in polar coordinates,
what's that going to be? Theta is going to run from 0
all the way around to 2 pi, and r is going to run
from 0 to infinity. And so those will be our
bounds for r and theta. Because if you want to
get the entire xy-plane, in terms of r and theta,
that is what you have to do. And I want to
mention also, what is this quantity going to become
in terms of r and theta? Well, notice that this
is e to the minus x squared minus y squared. It's just really e
to the minus quantity x squared plus y squared. It's really e to
the minus r squared. So what we get when we
do a change of variables, is we do-- I'll put
the theta on the inside actually, because that'll
be easy to do first. So the r is going to
be on the outside. The theta is going
to be on the inside. And then e to the
minus r squared, that's a direct substitution. x squared plus y squared
equals r squared. And then I get r d theta dr. And this comes from the
Jacobian that you computed-- I believe in lecture
even-- to show how you change
from x, y variables to r, theta variables. And I put the d theta
first here because I wanted to integrate in theta first. And the reason I want
to integrate in theta first is notice that-- because
nothing here depends on theta-- all I pick up is a 2 pi. I just pick up-- when I
integrate, I get a theta. I evaluate it at 2 pi, and
then I evaluate it at 0 and I take the difference. And so if I do that line
all I get is-- well, I should probably
write it in front-- all I get is 2 pi from
the theta, and then the integral from
0 to infinity, e to the minus r
squared times r dr. Now this is a much
easier quantity to evaluate than e to
the minus x squared dx. Because now we have an r here. So now it's a natural
substitution type of problem. And we can do it right away. I'm going to write
it down and then we'll check and make sure
I didn't make a mistake. We should get something like e
to the minus r squared over 2 with a negative sign in front. So let me make sure when I take
this derivative-- you could just do a substitution
to check, but you should get exactly this kind of thing. When I take the derivative
of e to the minus r squared, I get a negative 2r and then--
did I do something wrong here? Oh yeah. I'm good. I get a negative 2r. And so the negatives kill
off, the 2s divide out, and I get my e to the
minus r squared times r. So that's good. And now I have to
evaluate it at the bounds. So again, this was
much easier to evaluate if I do a substitution
and I let r-- I think I want to let r squared
equal u or something like this. I don't even know. I just did the substitution
without thinking about it. So you can figure out what
you need to substitute. But this is what you get. And so let me actually
just evaluate at the bounds and see what happens. So as r goes to infinity, I
get e to the minus r squared. As r goes to infinity, e to
the minus r squared goes to 0. And so the first term
is 0 when I evaluate. And then so the
second term I get is, I do 0 minus 2 pi times--
if I evaluate at e to the 0 I get 1-- so I get
negative 1 over 2. The negative 1 comes
from e to the 0. And then I divide by
2 so I get my 2 there. So a negative and a negative
gives me a positive. 2 divided by 2. And so the whole
thing equals pi. Whew! Just enough room. And so I just want to remind
us where we came from. We came from pi-- we wanted to
show was equal-- well, the pi, where did it come from? It's equal, if we go all the
way back up, to I squared. And so we wanted to show
what I was equal to. I then is equal to just
the square root of pi. So I can come back over to where
I started, which is over here. And I can say this is equal
to the square root of pi. And again, I just want to
mention what it came down to. It came down to introducing
a little bit more. Essentially we
took something that was a single variable
problem, we made it a multivariable problem. But what that
allowed us to do is to do a change of
coordinates from x, y into the polar
coordinate system. And the trick here-- as you
come through-- the trick where it actually
happens is right here. Is because this is easy to
integrate now that I have an r here. If I didn't have--
my problem initially was I didn't have an
x when I was trying to find an antiderivative. But when I do the
change of variables, I get an e to the
minus r squared. And then with an r
here, that's much easier to find an antiderivative. OK. So I think that is
where I'll stop.
