Today, we're going to be talking about how to evaluate a double integral by converting it to polar coordinates and in this particular problem
we've been [given] the double integral of e to the negative x squared minus y squared and we've been told that [D] as
indicated by [d] here we're taking the double integral of this region that
D is Bounded by X equals the square root of 4 [minus] y squared and the y?
axis so in order to convert this to polar coordinates
We're going to need to deal with [a] couple of different components of this double integral
The first is the function itself e to the negative x squared minus y squared we're going to need to change that to polar coordinates
We're going to need to change this Da here to polar coordinates
[and] we're going to need to give limits of integration in terms of polar coordinates
And that means limits of integration for R and limits of integration for theta we're going to be converting everything in
Terms of x and y to things in terms of R and theta so first let's talk about our region here
We've been told that the region for which we're finding the area is bounded by this equation here for x and the y [axis]
[so] let's see if we can go ahead and sketch this the first thing we want to do here is change
[this] equation for [x] into something maybe more recognizable
So if we take that [equation] for [x] and we square both [sides] on the left hand side
We'll get x squared on the right hand side the square root sign will go away
And we'll be left with 4 minus y squared if we add y squared to both [sides] we'll get x [squared] plus
Y squared equals 4 and we can clearly see that this is the equation of a circle centered at the origin
With Radius 2 now if we go ahead and draw a very simple graph of that
we can draw it small because we don't need a lot of
Detail here, but if we draw a sketch of that circle
it's centered at the origin and it has radius 2 because 2 is the square root here of
4 so if we say [that] that's our circle, and that this is 2 here
And this is 2 here, and this is an xy coordinate system
We're sketching it on the Cartesian coordinate system [now]. Here's our circle the important thing to remember though
Is that the region we were given was we're told D is bounded by I [x] equals positive
Root 4 minus y squared and this is really important because if we try to solve this equation
Again for x remember we squared both sides if we try to solve this equation
Again for x what will actually end up with we'll subtract y squared from both sides and we'll get x squared equals
4 minus y squared and then if we take the square root of both sides we'll get x equals
Positive or negative square root of 4 minus y squared well the original
equation we were given as a bound [for] d our region was not positive or negative 4 minus y squared it was just
Positive 4 minus y squared what that tells us is that we're dealing with only [x-values] that are positive
So we're not dealing with the entire circle
We're only dealing with the right-hand side of the circle this half circle where x values are positive
So we're actually looking at this region
And that makes sense because we've also been told that the region is bounded by the y
Axis which we can see here the y axis slices this circle in half
So this region to the right of the y axis is actually
the region that we're going to find the area for and the one that we're
Interested in now in order to get limits of integration for this region in terms of R and theta in polar coordinate system
Let's pretend that this is no longer a cartesian coordinate system. So let's just scratch out this x and y here
Let's now pretend that this is a polar coordinate system
If you'll remember in a polar coordinate system
Instead of x and y, we have R and theta r
Is always the distance of the point away [from] the origin so at the origin R is equal to 0?
but if I move out in any Direction
Here to the edge of my circle now the value of R is 2 because my radius is 2 and it doesn't matter where I?
Move here in this circle the radius is always going to be 2 so the lowest value R
Can attain in this polar coordinate system is 0 where it's right here at the origin the largest value it can attain
Given this bounded region here that we've drawn in purple is
2 so we know [that] our limits of integration for R are going to be 0 to 2 so we can go ahead and actually
start writing out our
double integral here
So our limits of integration are going to be zero to two four are now four theta remember that in a polar coordinate system
theta starts here
It's the angle theta starts here at zero and moves
Counterclockwise around this polar coordinate system up here is pi over [two]. This is Pi
this is 3 Pi over 2 all the way back to
2 pi where you've gone all the way around [the] circle [if] we were looking for the area of the entire
Circle in other words if we were given plus or minus root
4 minus y squared then our limits of integration for theta would be 0 to 2 pi, but in this case
we're only
interested in this half circle on the right here now if we start with the Angle 0 and we want to give the
Area for this region here that our limits of integration for theta would be 0 to pi over
2 but then we'd still [have] to account for this other quadrant down here and our limits of integration
Would be 3 pi over 2 to 2 pI and we'd have to [have] two separate integrals it gets real messy
so what we want to do is we want to start from this point down here and
Work our way up, so we can give one
Continuous interval for theta and the way that we're going to do that is by calling instead of calling this 3 Pi over 2
We'll just call it negative Pi over 2 because notice that the angle positive Pi over 2 is right here
This is 90 degrees from the Angle 0 moving in the positive Direction here
We get 2 Pi over 2 if we move 90 degrees here in the opposite Direction
It's moving the same distance in terms of Angle removing an angle of Pi over 2 but in the negative Direction
So we're going to say that theta starts at negative Pi over 2 down here and moves its way all the way up [here] to
Positive Pi over 2 so we're going to call those limits of integration
negative Pi over 2 to positive Pi
Over 2 and those will be [our] limits of integration
For theta now in terms of the actual function itself e to the negative x squared minus y squared?
Remember that we have a conversion formula to convert
Cartesian coordinates to polar Coordinates which is R squared equals x squared plus y [squared]?
You'll often see this conversion formula paired with the other two formulas x equals R
Cosine theta and y equals R
Sine theta we use these three formulas together to convert from Cartesian to polar the only one
We really need is this first one
But if we convert this equation here if we factor out a negative from the exponent we get e to the negative
X squared plus y squared if we just factor out that negative and now you see we have x [squared] plus y squared
Here we have x squared plus y squared in our conversion equation so we can make a substitution and call this e
To the Negative R
Squared and that's how we'll convert our function here from Cartesian coordinates to polar coordinates
Now all we need to deal with is the da and whenever you're replacing da or this can also be written
Dy/dx when you're replacing either one what you want to do is replace it with [R] dr
D-Theta, [and] that's always the case if you have the Da here or the Dy/Dx you want to replace it with R dr
D-Theta and make sure to include this extra R
Now in fact, that's all we need to do to set up our polar double integral at this point
We just need to evaluate it in the way that we'll do that is with U substitution
We'll set U equal to R. Squared because we have r squared here in our exponent
We'll take the derivative of U to get d u the derivative of R squared is to our
Dr. And now if we [just] divide both sides by [two] what we'll get is our dr
Is equal to 1/2 d U and the reason that we only divide it by [two]?
Instead of [two] R is because notice that we've solved here for r Dr. And we have r dr
In our equation so we can substitute [1/2] to you for R dr
And cancel that are in one step so let's go ahead and rewrite our integral here
We've got negative Pi over 2 to positive Pi over 2 of the integral from 0 to 2
Here we have e to the negative U and then rDR
We're going to substitute [1/2] you let's go ahead and put this 1/2 in the front and call this d you and then d theta
Now we have something
We can actually integrate so we'll say negative pi over 2 to [pi] over 2 when we take the integral of this
Remember with e to the u or e to the negative u we're just dividing by the coefficient on the U so
We're really [just] dividing by negative 1 here what we end up with is negative [1/2] e
To the negative U we need to go ahead and back substitute for U and get negative R
Squared here, and we'll be evaluating that on the interval 0 to 2 and then of [course] we have here our d theta
When we plug in our limits of integration?
for R
0 and 2 here what we'll get is negative 1/2 we plug in our upper limit of integration first [-]
2 squared is 4 [so] we get e to the negative 4 then we'll subtract whatever we get when we plug in 0
Well here, we get 0 squared. Which is [just] 0 negative 0 is still just 0 e to the 0 [is] 1?
1 [times] negative 1/2 is a negative 1/2 so minus a negative 1/2 is just plus
1/2 here and then we have D theta now
This is simplified enough that we can easily integrate with respect to theta we'll go ahead [and] do that
We'll get negative 1/2 E to the negative 4 times theta plus 1/2 times
theta and we'll be evaluating that on negative Pi over 2 to positive Pi over 2
We'll plug in our upper limit of integration first Pi over 2 we'll get negative Pi over
4 [E] to [the] Negative 4 because we'll multiply this pi over 2 here
We'll plug it in for theta here, and when we combine that with
Negative 1/2 we get negative Pi over 4 e to the negative 4 when we plug it in here?
We'll get plus Pi over 4 so [plus] Pi over
4
Then we'll subtract whatever we get when we plug in negative pi over 2 so we'll get minus and then here
Plugging in negative Pi over 2 negative Pi over 2 times a negative 1/2 is going to give us a positive
Pi over 4 E to [the] negative four and then plugging it in here
We're going to get minus pi over four now if we combine like terms
You'll notice we have negative pi over four e to the negative four minus pi over four e to the negative four?
That's a minus two pI over
Four e to the negative four we have a plus pi over four minus
A negative pi over four is the same as pi over 4 plus [pi] over four so we end up with plus two pI over
Four now if we factor out a 2 pi over four we'll get two pi over four times
[well] multiply first by this term so we can lead inside the parentheses with a positive
So we'll say two pi over four times one, then we'll go to this term and we'll say minus e to the negative
Four now if we just reduce this to pi over four fraction. We'll get pi over
[two] times [one] minus e to the negative four you're welcome to leave it like that
Or you can always say Pi over 2 times 1 minus 1
over E to the 4 you can move that e to the negative 4 to the denominator and
the exponent changes sign from a negative to a positive
So either way either of these last two are acceptable for a final answer
But that's how you can evaluate a double integral by converting it from Cartesian coordinates to polar coordinates
So I hope you found that video helpful if you did like this video down below and subscribe
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