We want to evaluate the given double integral
over the region R, where R is the region 25 x squared
plus four y squared, less or equal to 100.
We're going to evaluate this double integral
by performing a change of variables,
sometimes called a transformation.
But before we do this, let's take a closer look
at the region R.
The region R is defined by the inequality 25 x squared
plus four y squared, less or equal to 100.
We should recognize this is an ellipse,
but to graph the ellipse we would want the right side
to be equal to one, so let's divide everything by 100.
Simplifying, we'd have x squared divided by four
plus y squared divided by 25, less or equal to one.
So we have an ellipse centered at the origin
where the vertical axis is the major axis
and there's a length of 10 units as we see here.
The minor axis has a length of four units, as we see here.
So the area of this ellipse is the region R.
Now we're going to perform a change of variables.
To simplify the region of integration,
we're going to simplify this to the region
of integration that's a circle.
We can do this by performing a change of variables.
Notice how if you'll let x equal two u,
and y equal five v, if we perform substitution
into the inequality we would have two u squared
divided by four plus five v squared divided by 25,
less or equal to one.
And this simplified very nicely to u squared
plus v squared, less or equal to one
where this region would be a circle centered
at the origin with a radius of one,
which I've already graphed here.
This is going to be the region S in the uv plane.
Now that we know the region of integration for the uv plane,
we still have to write f of x, comma, y
as a function of u and v.
And then we're performing a change of variables.
We're going to have this extra factor in integrated function
which is called the Jacobian.
Then we have the udv.
Notice how differential A in the xy plane
is equal to this extra factor, the Jacobian, the udv,
in the uv plane.
Now let's work on determining f of u, comma, v,
we're writing the integrated function as a function
of u and v.
But before we do that, now we know that region S
in the uv plane is bounded by u squared plus v squared
less or equal to one.
When we use the transformation or change of variables
of x equals two u and y equals five v.
Notice how f of x comma y, the integrated function here
is equal to x squared which means f of u, comma, v
would be equal to two u squared since x is equal
to two u.
So f of u, comma, v is equal to four u squared.
Next we need to find the Jacobian,
and for review, remember the Jacobian is equal
to the value of this two by two determinant.
So to find the elements in the determinant we need
to find the partial derivatives using our equations
for x and y.
So the partial of x with respect to u is going to be two.
The partial of x with respect to v would be zero.
The partial of y with respect to u would be zero.
And the partial of y with respect to v would be five.
The Jacobian of x and y with respect to u and v
is equal to the two by two determinant
where the first row would have elements two and zero.
The second row would have elements of zero and five.
That value of this two by two determinant is equal
to this product minus this product
which would be 10 minus zero, which equals 10.
Going back to our previous slide,
notice how we use the absolute value of the Jacobian.
The absolute value of the Jacobian of x and y
with respect to u and v would be equal
to the absolute value of 10, which equals 10.
Now we have all the pieces we need in order to write
the given double integral in terms of u and v.
The given double integral is going to be equal
to the double integral over the region S of f of u,
comma, v which is four u squared,
times the absolute value of the Jacobian,
which is 10.
Then we have the udv.
Now we need to find that limit of integration
for u and v.
Notice, in this case because the region of integration
is a circle, it's actually going to be easier
to evaluate this if we perform another change of variables
and write this double integral in polar form.
So let's work on doing this.
Let's re-write this as double integral over the region S
of this would be 40 squared.
And we have the udv.
On the uv plane, this would be the u-axis
and this would be the v-axis.
So on the uv plane we can say that u is equal to r
cosine theta, and v is equal to r sine theta.
Now writing the double integral using polar coordinates,
we're going to have the double integral.
40 u squared is going to be 40 and then u is equal
to r cosine theta.
So u squared would be r squared, cosine squared theta.
Then remember when using polar coordinates,
we'll have an extra factor of r,
which would actually be the Jacobian.
Then we have drd theta.
Now we need you to find the limits of integration
for r and theta using this region of integration.
The limits of integration for r would be from zero to one.
The limits of integration for theta would be from
zero to two pi all the way around the circle.
Let's go ahead and simplify the intergrated function.
We've got a double integral over the region S
using polar coordinates.
We'd have 40 r cubed cosine squared theta drd theta.
Now this integral on the next slide and now we integrate
with respect to r, treating theta as a constant.
So the anti-derivative is going to be 40 cosine squared
times r to the fourth divided by four
which would be 10 r to the fourth cosine squared theta.
Now you define big f of b minus big f of a
by performing substitution for r.
So when r is equal to one we're going to have
10 cosine squared theta.
When r is equal to zero we just have zero.
So the new integrated function is 10 cosine squared theta.
Before we integrate with respect to theta, though
we'll have to use a power producing formula
for cosine squared theta.
Remember, cosine squared theta is equal to one-half
times one plus cosine two theta.
Let's perform this substitution for cosine squared theta.
Well 10 times one-half is five.
This is going to be five the quantity
one plus cosine two theta.
Let's go ahead and distribute.
So we have five plus five cosine two theta.
Integrating with respect to theta,
the anti-derivative of five is going to be five theta.
Now we integrate five cosine two theta we do have
to perform u substitution where u equals two theta.
Du equals two d theta.
So dividing both sides by two notice how one-half du
is equal to d theta.
So we're going to have an extra factor of one-half
when integrating five cosine two theta.
So the anti-derivative is going to be plus five halves
sine two theta.
Again, that's because we can think of this
as five times one-half cosine udu.
Now evaluating here, when theta's equal to two pi
we'll have five times two pi plus five halves
times the sine of two times two pi, that's four pi.
And when theta's equal to zero we have five times zero
plus five half times sine zero.
Well, sine four pi is zero.
Five times zero is zero and sine zero is zero.
So this simplifies nicely to 10 pi
which would be approximately 31.1416.
I hope you found this helpful.