Welcome to an intro to double integrals
in polar coordinates.
Sometimes a double integral is much easier to evaluate
using polar coordinates.
This is especially true if the region of integration
can be easily defined using a polar equation.
So because we're going to be converting
from rectangular to polar coordinates,
it is important that we remember the following equations.
So you may want to copy these into your notes.
If f of x is continuous on a polar region defined
as given here, then we can rewrite a double integral
in rectangular form into polar form as we see here.
And there's a couple important things to notice here.
First, our function f of x y is going to be converted
into a polar equation as we see here.
The way we do this is we replace x with r cosine theta,
and y with r sine theta.
And sometimes the function is written like this
to represent polar form,
and sometimes it's just written as f of r theta.
And there's one more very important thing to notice,
differential A is replaced with rdrd theta,
so there's an extra factor of r in the integrand
when we convert from rectangular to polar form.
And we'll take a look at why that is in just a minute.
Let's take a look at these two graphs below.
Notice in rectangular form, we were determining the volume
of these boxes where the bases were squares.
Well in polar form, we still want to determine the volume
of these boxes, but now the bases aren't quite rectangular,
so it's going to take a modification of the formula
to determine differential A.
Let's take a look at that.
So again in rectangular form, we determined that
differential A was equal to dx times dy
because this represented the area of a base,
which was a square or a rectangle.
Now if we take a look at the base of one of those boxes
in polar form, it's going to look more like this.
So delta theta would be the change in the angle,
as we see here, this shorter radius would be r1,
and the longer radius would be r2, so to determine
the length of this piece here and this piece here,
we'd actually have to use the arc length formula,
which states that arc length s is equal to r times theta.
So this length here would be r1 theta,
and this should be r2 theta.
And the length of this segment here would be delta r,
which we can set equal to r2 minus r1.
Now of course as we let the number of boxes
approach infinity, r1 is going to be approximately equal to r2,
which we'll let equal r.
So from here we can say that delta A, or the change
in the area, would be approximately equal to
r times delta theta, again because these are all
approximately equal to each other, times delta r.
So now you can see where this extra factor of r comes from.
Go back up here and look at differential A is equal to
rdrd theta, well here's the extra factor of r,
and then again as the number of boxes approaches infinity,
delta theta is approximately equal to differential theta,
and delta r is approximately equal to differential r.
So hopefully this rough outline helps you understand why
when converting to polar form
we do have this extra factor of r in the integrand.
Let's go and take a look at an example.
We want to determine the volume of the solid below the surface
given by f of xy above the xy plane
over the region bounded by x squared plus y squared
equals one and x squared plus y squared equals four.
Well the first thing we should recognize here is that
these two are circles, one with radius one
and another with radius two, as we see graphed here in red.
So our region of integration
is the region between these two circles.
This is a great example of when it'd be easier to
evaluate this using polar coordinates.
And here you can see the solid, and I've traced
the region of integration in red in the xy plane.
So let's see if we can set this up.
The first thing we need to do is rewrite the function
as a polar equation, let's first write this as
four minus the quantity x squared plus y squared.
The reason that's helpful is remember that
x squared plus y squared is equal to r squared,
so we can rewrite this in polar form
as four minus r squared.
Now see if we can set up our double integral.
Our function is now four minus r squared,
remember now we're going to have an extra factor of r,
and the ndrd theta.
Remember r stands for the radius, well this circle here
would be r equals one, and this circle here
would be r equals two.
So the limits of integration for r would just be
from one to two, and then because we want
one complete revolution around the circle, the limits
of integration of theta will be from zero to two pi.
Let's go and distribute the r.
So four r minus r to the third, drd theta.
Let's go ahead and finish this on the next slide.
So four times r squared over two minus r to the fourth
over four, let's go ahead and clean this up a little bit,
so we'll have two r squared minus
one-fourth r to the fourth.
So when r is two it looks like we'll have eight minus,
this is going to be 16 divided by four, that's four,
and then when r is one we'll have two minus one-fourth.
So four minus one and three-fourths or seven-fourths,
that's going to give us nine-fourths.
Now [speaking indistinctly] theta, that's going to be
nine-fourths theta, so when theta is two pi,
we'll have two pi and then minus zero,
give us nine halves pi.
So the volume of the solid over this region in orange
would be nine halves pi.
Let's take a look at one more example.
Here we want to determine the volume of this function
over the region x squared plus y squared
is less than or equal to four in the first octant.
So we have a circle with radius two, and we only want it
in the first octant, so it's going to be this region here.
Now that we have a region of integration, first thing
we should do is convert this to a polar equation.
So if we rewrite this as the square root of nine
minus the quantity x squared plus y squared,
we can replace x squared plus y squared with r squared.
So our double integral, the function is going to be
the square root of nine minus r squared, and then
don't forget our extra factor of r and then drd theta.
So let's determine what r would be.
Well this is a circle with radius two,
so r is going to be from zero to two.
But now because we're in the first octant,
theta is only going to be from zero to pi over two.
And now we're ready to go.
This is going to require use substitution, though.
[speaking indistinctly] have u equals nine minus r squared,
so du's going to be negative two r dr, so that tells us
that we can replace r dr with negative one half du.
So you're going to have negative one half du,
and this'll just be u to the one-half.
So we'll have negative one half, see, u to the three-halves
divided by three-halves, or times two-thirds,
and u is really nine minus r squared,
this simplifies here to negative one-third.
Let's finish this on the next slide.
So when r is equal to two, we're going to have nine minus four,
that's going to be five to the three-halves, and then when r
is zero, we're going to have nine to the three-halves.
Let's go ahead and distribute here.
Negative one-third times negative nine to the three-halves,
that's going to give us a positive nine, and we'll have minus,
and here we'd have five square root five, all over three.
This just the constant, just going to have nine minus
five squared five over three, theta, times pi over two.
Let's go ahead and get a common denominator here.
So denominator would be six, and we'd have 27 minus
five square root of five times pi.
And we'll leave it like that.
We'll take a look at some more examples in the next video.
Thank you for watching.