Welcome to a second example of converting a double integral in rectangular form to polar form.
As we discussed in a previous video.
To convert a double integral in rectangular form to polar form.
We have to convert the function f of x, y
into a function in terms of r and theta.
And differential A is replaced with
an extra factor of r, dr, d theta.
So we can't forget about this extra factor of r
when converting to polar form.
Let's go and take a look at our second example.
Let's see if we can determine the region of integration.
We're integrating first with respect to y which means y has to be greater than or equal to zero and
less than or equal to the square root of x minus x squared.
So y has to be greater than or equal to zero.
And
less or equal to the square root of
x minus x squared.
And the outer integral
we're integrating with respect to x which means x has to be greater than or equal to zero and
less than or equal to one.
Let's see if we can determined the region of integration from this.
Well we know x is positive and y is positive so we're dealing with the first quadrant.
So let's focus on determining
the graph of y is less than or equal to the square root of
x minus x squared.
One thing we could do is square both sides.
That would give us y squared
is less than or equal to x minus x squared.
If we add x squared to both sides we have x squared plus
y squared
less than or equal to x.
So we might wanna type this into our graphing calculator but we could also try to do this by hand.
By making a T table.
So were gonna go ahead and see if we can graph x squared plus y squared
equals x.
By making a T table.
Well if x is zero.
Course we have zero on both sides so y would have to be zero.
And if x was one.
We have a one on both sides y would have to be zero.
Let's focus on what would happen if x was one half.
Well if x is one half we have the equation one fourth
plus
y squared
must equal one half.
So if we subtract one fourth on both sides we have y squared equals
one fourth.
And then y would have to be equal to one half.
So from this you might be able to determine that we have to graph
that looks like this.
Here's the point
zero, zero.
Here's the point one half, one half.
Here's the point one, zero.
So because y has to be less than this semi circle.
We would shade below
meaning
this would be our region of integration.
Let's see if we can start to set up our double integral from here.
I'm gonna go ahead and label
this curve
it was a curve y
equals
the square root of
x
minus x squared.
So now we have our function f of x, y equals x times y.
Let's see if we can determine f of r, theta.
Meaning convert
this function into polar form.
Member x is equal to r cosine theta.
And y is equal to r sine theta.
So f of r, theta.
Would be r squared cosine theta
sine theta.
So we can go to our double integral in polar form.
Here's f of r, theta.
So we have r squared.
Cosine theta, sine theta.
And don't forget we have an extra factor of r
and then dr, d theta.
So now we need to determine the limits of integration.
First respect to r.
And then with respect to theta.
And so this one it's gonna be a little bit tricky there's a couple ways are going about doing it.
If we don't recognize the polar equation
that we create.
This semi circle here.
You go ahead and take this curve here.
And rewrite it as a polar equation.
So let's go ahead and do that we would have
y equals
the square root
of x minus x squared.
And y is equal to r sine theta.
We have the square root
again x is r cosine theta.
And we have minus r squared.
Cosine squared theta.
Now if we squared both sides of this equation we're gonna have r squared
sine squared theta.
Would equal r cosine theta.
Minus r squared cosine squared theta.
Let's go and add r squared cosine squared to both sides
that will give us r squared
sine squared theta.
Plus
r squared cosine squared theta.
Equals r cosine theta.
Now if we factor out r squared we're gonna have
sine squared plus cosine squared but that's equal to one.
And divide both sides by r.
You have r equals cosine theta.
So this tells us that the radius.
This tells us we could use this polar equation to trace out the radius.
So the limits of integration with respect to r would be from zero
to cosine theta.
Now let's go ahead and verify this with our graphing calculator.
To write down as I put the calculator into green mode because it is easier to read and also polar form.
And I typed in r equals cosine theta.
Let's take a look at the graph.
Notice it does generate this half circle as well as the second half in the fourth quadrant.
Let's see what theta would be.
If we press trace.
We can see that when theta is equal to zero.
So we be at this point here.
And as we increase theta.
You can see it does trace out the semi circle that we want.
And when theta is equal to 90 degrees or pi over two.
We're back at the origin.
So that works perfectly.
That verifies
our limits of integration for r.
It also tells us that theta would be from zero to pi over two.
And notice for the limits of integration for r.
There was in terms of function that had to be in terms of theta.
Let's go and finish this on the next slide.
Notice we are going to have r to the third power because of this extra factor of r.
And now we're gonna integrate with respect to r treating theta as a constant.
So we have r to the fourth over four.
Go ahead and pull out the one fourth.
Now we want to replace
r with cosine theta.
It's gonna give us
cosine to the fourth theta here plus another factor of cosine theta we'll have cosine to the fifth theta
sine theta.
Then when r is zero this should be zero.
Now we're gonna have to perform u substitution if we let u equal cosine theta.
And differential u is gonna be equal to negative sine theta
d theta.
So we're gonna replace sine theta, d theta with negative du.
So negative
du takes the place of
sine theta, d theta.
This would be u to the fifth.
So we have negative one fourth times u to the six over six.
Let's go rewrite this in terms of theta we have negative one twenty fourth.
This should be cosine to the sixth theta.
Well cosine pi over two would be zero.
So we have zero to the sixth is zero.
Minus
cosine to the zero which is one. One to the sixth is one.
So we have negative one twenty fourth
times a negative one or positive one twenty fourth.
That will do it for this second example.
I hope you found this helpful.