Welcome to a lesson on change of two
variables for double integrals.
This video will introduce the Jacobian
for a transformation of two variables.
We just finished studying double integrals
in rectangular and polar form as well as triple integrals
in rectangular, cylindrical, and spherical form.
We know that when converting to a different
coordinate system there are extra factors in the integrand.
For example, for double integrals in polar form
we had an extra factor of r in the integrand,
as well as in cylindrical form for triple integrals.
And we also had the extra factors of rho squared sin phi
when converting from rectangular to circle form
using triple integrals.
So, we're going to develop a formula that will tell us,
when we switch coordinate systems, what the extra factors
in the integrand would be.
Often called the integrating factor.
Let's say we redefine x and y into the uv
coordinate system where x is equal to g of uv,
and y is equal to h of uv.
If we have a double integral in rectangular form,
and we want to convert it to the new coordinate system
defined by u and v, we want to be able to determine
what differential A would be equal to.
Remember that differential A represents the area of a small
parallelogram in the new uv coordinate system.
Now, here we are calling it a parallelogram,
but you can also think of it as
a rectangle if you needed to.
So, now if we let vector r have components x comma y,
in the new coordinate system vector r would have the
components g of uv,
and h of uv.
So, the area of a small parallelogram
would be determined by
the partial derivative of r with respects to u,
times delta u,
and the partial derivative of r with respects to v,
times delta v.
Remember, if we have two vectors we can determine the area
of the parallelogram formed by them by determining the
magnitude of the cross product of the two vectors.
So, in our case we'd have the magnitude of the partial
derivative of r with respects to u, times delta u,
crossed with the partial derivative of r
with respects to v, times delta v.
This would be equal to the magnitude of the partial of r
with respects to u, crossed with the partial derivative of r
with respects to v, times delta u, times delta v.
Now, at this point, let's go ahead and take a closer look
at this cross product here.
This would give us a three by three determinant
where the first row would be i, j, k.
And now for the second row.
Because r is defined using u and v in the new coordinate
system, this would give us the partial derivative of x
with respects to u, and the partial derivative of y
with respects to u.
The k component would be zero.
And then for the third row we have the partial derivative
of x with respects to v,
partial derivative of y with respects to v, and then zero.
Because of this third column being zero
the result is only going to be a k component
where we'll have the two by two determinant formed by
the partial of x with respects to u,
partial of y with respects to u,
partial of x respects to v,
and then partial of y with respects to v, times k.
Then we go back up here.
Remember, our ultimate goal was to determine
the magnitude of this cross product.
So, the magnitude of the cross product, now, would be equal
to the partial of x with respects to u,
times the partial of y with respects to v,
minus the partial of x with respects to v,
times the partial derivative of y with respects to u.
So, putting these pieces together we can now say
that differential A, the small change in area,
is going to be equal to the absolute value of
partial of x with respects to u,
times the partial of y with respects to v,
minus the partial of x respects to v,
times the partial of y with respects to u,
times differential u, times differential v.
Let's go back and take a quick look.
Again, we show that all of this was
equal to the difference of this product.
And then here's our differential u and our differential v,
which gives us the approximate area of a small parallelogram
which would give us differential A.
This right here is called the Jacobian,
and this calculation here is going to give us the extra
factors in our integrand when we convert differential A
to our new coordinate system in terms of u and v.
And the difference of this product, called the Jacobian,
is often written using this notation here.
So, let's go ahead and summarize.
The Jacobian of transformation of two variables
is given by this formula here, as we just arrived.
So, when converting a double integral,
where r and s are regions in the xy and uv plane,
related by the given equations here,
we can determine the extra integrating factor
using this formula here.
Now, let's go ahead and look at an example where we convert
a rectangular double integral to polar form.
And when we calculate the Jacobian we should get
the extra factor of r in our integrand.
Let's go ahead and see if we do.
So, in this case, x of r theta would be equal to
r cosine theta,
and y of r theta would be equal to r sin theta.
So, we want to verify that the Jacobian
would be equal to r in this case.
Let's go ahead and determine our partial derivatives.
Partial of x with respects to r
is going to give us cosine theta.
And the partial of x with respects to theta
would give us negative r sin theta.
Then the partial of y with respects to r
would give us sin theta.
Then the partial of y with respects to theta,
that would give us r cosine theta.
Now, let's go ahead and calculate the Jacobian.
So, the first row will be cosine theta,
negative r sin theta.
The second row will be sin theta,
r cosine theta.
So, we're going to have r cosine squared theta,
minus negative r sin squared theta.
We have a common factor of r, so we'll have r and then
cosine squared theta, plus sin squared theta,
which, of course, is equal to one.
So, this verifies that we're converting to polar form.
We have an extra factor of r in the integrand.
We'll take a look at another example in the second video.
Thank you for watching.