Welcome to an example of
a change of two variables for a double integral
using the Jacobian.
In the first video, we showed that when converting
to a new coordinate system, we often have
an extra integrating factor
in the new coordinate system,
and we can determine that integrating factor
using the Jacobian, as we see here.
For example, if we convert a double integral
over the region R in terms of X and Y
into the UV coordinate system,
as defined here,
we convert the function into a function in terms of U and V,
and then differential A is equal to the Jacobian
times DUDV, or DVDU.
This Jacobian gives us the extra integrating factor
in the new coordinate system.
Let's go and take a look at an example.
Here we have a given double integral over the region R,
and we're going to redefine the integrating region
based upon letting X equal two times U
and Y equaling 4V plus 2U.
Let's first sketch the given region of integration.
We can see here that X is on the close interval
from zero to two,
so here's X equals zero,
and here's X equal two.
And Y is on the close interval from X to X plus four.
Let's graph Y equals X, that's a line with a slope of one
passing through the origin.
That would be this line here.
And notice that Y is above this line.
And let's go ahead and graph Y equals X plus four,
and Y is below that line.
We have a Y-intercept of four
and another slope of one.
The given region of integration is this region here.
And now, we're going to redefine the integrating region
based upon X equals 2U and Y equals 4V plus 2U.
Let's go ahead and list the vertices of this region
in terms of X and Y.
We have the origin, point zero, zero,
then we have zero, four,
then we have two, two,
and then we have two, six.
Now, I want to determine what the vertices would be
in the UV coordinate system.
Let's go ahead and solve this equation for U.
We would have U equals X divided by two.
Let's go ahead and solve this for V,
so we'd first subtract 2U, and then divide by four.
If we sub in X over two for U,
we would have Y minus X divided by four.
So, we'll use these two equations to find the vertices
in the new coordinate system.
Determining U, we'll just divide X by two,
so this would be zero, this would still be zero.
This would be one, and this would be one.
And then for the new V coordinate,
we'll have Y minus X divided by four.
Zero minus zero divided by four is still zero.
Four minus zero is four, divided by four, that's one.
Here we're going to have two minus two.
That's zero divided by a four, we have zero.
Here we have six minus two, that's four,
divided by four, that's one.
Notice how the new region of integration is much simpler.
We have zero, zero,
zero, one,
one, one,
and one, zero.
Our new region of integration is just a one-by-one square,
which is going to simplify our limits of integration.
Before we set up the new double integral,
let's go ahead and find the Jacobian, or integrating factor,
when converting to the UV coordinate system.
Here's our Jacobian.
Let's start by determining our partial derivatives.
The partial derivative of X with respects to U,
that's just going to be two.
The partial derivative of X with respects to V,
that's going to be zero.
And now, we'll have the partial of Y with respects to U.
That's going to give us two.
And then, partial derivative of Y with respects to V.
That's going to give us four.
Our Jacobian is going to be equal to this
two-by-two determinant,
so we'll have two, zero,
two, four.
So, we'll have eight minus zero, which is equal to eight.
Converting to the new coordinate system,
we'll have an extra factor of eight
in our double integral.
Let's go ahead and set this up.
We're going to rewrite the given double integral
in the UV coordinate system.
The given function was X times Y,
which will be 2U times
4V plus 2U.
And then, the Jacobian was equal to eight,
so we have an extra factor of eight.
Now, let's go back and take a look
at our region of integration again.
Both intervals for U and V are from zero to one.
Let's integrate first with respects to V,
from zero to one,
and then with respects to U,
from zero to one.
And we could've easily have switched
this order of integration.
Let's go ahead and evaluate this on the next page.
If we distribute here, we would have 8UV
plus 4U-squared
times eight.
So, we can see there's a common factor of four here
as well as this eight,
so we're going to go ahead and factor out 32.
That's going to leave us with 2UV
plus U-squared,
DVDU.
Integrate in respects to V will have
2U times V-squared over two.
The twos will simplify out,
so we'll UV-squared plus U-squared V.
Remember, we're substituting in for V,
so when V is one,
we'll have U plus U-squared,
and then when V is zero, these'll both be zero.
We'll have one-half U-squared
plus one-third U to the third.
We're going to have 32 times,
when U is one we'll have one-half plus one-third,
and then when U is zero, these'll both be zero.
So, if 32 times, this'll be five-sixths,
and this simplifies to 80-thirds.
Just remember when converting to a new coordinate system,
we do have to make sure that we calculate
the Jacobian
to determine the extra integrating factor.
I hope you found this example helpful.
Thank you for watching.