WEBVTT

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CHRISTINE BREINER: Welcome
back to recitation.

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In this video I want to show
how we can use change of

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variables to polar coordinates
in particular to evaluate an

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integral that, without the
change in variables, we don't

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have the techniques to do.

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So I'm going to show us how to
evaluate the integral from

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minus infinity to infinity of
e to the minus x squared dx.

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And I'll just point out that
if you do anything in

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probability you will see this
integral a great deal.

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So this is a distribution and
you'll see it a great deal if

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you ever do anything in
probability theory.

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But how are we going
to use polar

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coordinates to evaluate this?

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Well, the object is going to be
to introduce a little bit

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more into this integral, so that
when I actually introduce

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that in I'm going to have an r
squared term in the exponent

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and I'm going to have --of
course by the change of

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variables in the Jacobian--

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multiplied by r.

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And that's what's going
to save us.

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So let me actually just
get right to it.

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I'm going to call this integral,
this quantity,

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capital I. We'll just call this
quantity capital I. And

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then what I'm going
to do is introduce

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a couple more things.

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So let me write it down, and
then I will justify that it's

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reasonable to even look
at this thing.

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And that it will give
us something useful.

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OK, so I just took this integral
and what I've done is

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I've now taken this quantity and
I've multiplied by an e to

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the minus y squared.

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And then I'm now integrating for
minus infinity to infinity

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and dy also.

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And what I want to point out,
that maybe isn't immediately

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obvious from the way it's
written here, but will be

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immediately obvious when I write
it a different way, is

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that this quantity is
actually just the

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square of this quantity.

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So it's actually just two of
these multiplied together.

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And let me point out
why that is.

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Because this term right here can
be rewritten as e to the

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minus x squared, e to
the minus y squared.

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And then I have two integrals,
which are for minus infinity

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to infinity both.

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Sorry, that one looks
a little off kilter.

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But those two integrals, which
the bounds are minus infinity

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to infinity for both.

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And then I have a dx and a dy.

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Now the reason that it
is just I squared--

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this quantity is just I
squared-- is because if you

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notice, this first integral, the
inside integral, this is

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independent of x.

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So I can move it out.

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So let me actually come up
here and write that down.

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So I can move it out and I can
actually rewrite the integral.

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And I'll put the bounds here,
because it's high up enough

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that I can actually write it.

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e to the minus y squared.

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And then integral for minus
infinity to infinity.

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e to the minus x
squared dx dy.

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Right?

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So let's point out
again what I did.

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This quantity I circled was the
constant when I considered

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x the variable.

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Right?

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In terms of x, this is the
constant, so I can move it in

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front of that integral, which
is the dx integral.

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And so now what do I see?

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Well, this quantity is what
I've called I. And so now

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that's a constant.

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So I can move that
out to the front.

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That's a constant.

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That's I, as designated.

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And so now I have the integral
for minus infinity to infinity

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of either the minus
y squared dy.

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And then that again is another
I. Because if you notice, I

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mean, this was either the
minus x squared dx--

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and so if I keep this as either
the minus y squared dy,

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it's going to be exactly
the same value.

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So this is actually equal
to I squared.

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So the point I want to make, if
you come back over here to

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this integral, is that this
quantity I'm going to be able

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to integrate very easily.

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And when I do that, if I take
the square root, I'm going to

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get this value.

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Because I just showed this
thing in the box

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was equal to I squared.

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So let me write it down.

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On this last part of the board
I will write down that thing

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again and then we'll actually
evaluate it and

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we'll see what we get.

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So let me rewrite.

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This is the thing we
want to evaluate.

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And I'm going to remind
myself that

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that's equal to I squared.

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I squared is equal
to this quantity.

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And I mentioned at the beginning
that we're going to

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use a trick.

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We're going to change
variables into polar

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coordinates.

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And so notice that the region
we're integrating over is the

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entire xy plane.

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And in polar coordinates,
what's that going to be?

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Theta is going to run from 0
all the way around to 2 pi,

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and r is going to run
from 0 to infinity.

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And so those will be our
bounds for r and theta.

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Because if you want to get the
entire xy plane, in terms of r

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and theta, that is what
you have to do.

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And I want to mention also, what
is this quantity going to

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become in terms of
r and theta?

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Well, notice that this
is e to the minus x

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squared minus y squared.

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It's just really e to
the minus quantity x

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squared plus y squared.

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It's really e to the
minus r squared.

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So what we get when
we do a change of

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variables, is we do--

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I'll put the theta on the
inside actually, because

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that'll be easy to do first.
So the r is going

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to be on the outside.

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The theta is going to
be on the inside.

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And then e to the minus r
squared, that's a direct

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substitution.

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x squared plus y squared
equals r squared.

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And then I get r d theta d r.

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And this comes from the Jacobian
that you computed--

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I believe in lecture even--

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to show how you change from xy

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variables to r theta variables.

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And I put the d theta first
here because I wanted to

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integrate in theta first. And
the reason I want to integrate

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in theta first is
notice that--

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because nothing here
depends on theta--

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all I pick up is a 2 pi.

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I just pick up --when
I integrate--

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I get a theta.

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I evaluate it at 2 pi, and then
I evaluate it at 0 and I

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take the difference.

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And so if I do that line all
I get is-- well, I should

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probably write it in front--

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all I get is 2 pi from the
theta, and then the integral

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from 0 to infinity.

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e to the minus r squared
times r dr.

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Now this is a much easier
quantity to evaluate than e to

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the minus x squared dx.

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Because now we have an r here.

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So now it's a natural
substitution type of problem.

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And we can do it right away.

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I'm going to write it down and
then we'll check and make sure

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I didn't make a mistake.

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We should get something like e
to the minus r squared over 2

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with a negative sign in front.

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So let me make sure when I
take this derivative--

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you could just do a substitution
to check, but you

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should get exactly this
kind of thing.

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When I take the derivative of
e to the minus r squared, I

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get a negative 2r and then--

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did I do something wrong here?

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Oh yeah.

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I'm good.

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I get a negative 2r.

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And so the negatives kill off,
the 2s divide out, and I get

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my e to the minus r
squared times r.

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So that's good.

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And now I have to evaluate
it at the bounds.

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So again, this was much easier
to evaluate if I do a

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substitution and I let r--

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I think I want to let
r squared equal u or

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something like this.

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I don't even know.

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I just did the substitution
without thinking about it.

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So you can figure out what
you need to substitute.

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But this is what you get.

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And so let me actually just
evaluate at the bounds and see

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what happens.

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So as r goes to infinity, I get
e to the minus r squared.

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As r goes to infinity, e to the
minus r squared goes to 0.

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And so the first term is
0 when I evaluate.

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And then so the second term I
get is, I do 0 minus 2 pi

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times-- if I evaluate at
e to the 0 I get 1--

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so I get negative 1 over 2.

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The negative 1 comes
from e to the 0.

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And then I divide by 2
so I get my 2 there.

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So a negative and a negative
gives me a positive.

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2 divided by 2.

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And so the whole thing
equals pi.

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Whew!

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Just enough room.

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And so I just want to remind
us where we came from.

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We came from pi--

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we wanted to show was equal--
well, the pi,

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where did it come from?

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It's equal if we go all the
way back up to I squared.

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And so we wanted to show
what I was equal to.

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I then is equal to just
the square root of pi.

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So I can come back over
to where I started

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which is over here.

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And I can say this is equal
to the square root of pi.

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And again, I just want to
mention what it came down to.

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It came down to introducing
a little bit more.

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Essentially we took something
that was a single variable

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problem, we made it a
multivariable problem.

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But what that allowed us to
do is to do a change of

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coordinates from xy into the
polar coordinate system.

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And the trick here-- as you come
through-- the trick where

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it actually happens
is right here.

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Is because this is easy
to integrate now

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that I have an r here.

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If I didn't have-- my problem
initially was I didn't have an

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x when I was trying to find
an antiderivative.

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But when I do the change of
variables, I get an e to the

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minus r squared.

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And then with an r here, that's
much easier to find an

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antiderivative.

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OK.

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So I think that is
where I'll stop.