Welcome to a second example
of surface integrals
where the surface is given explicitly.
So if our surface is given by z equals g of x,y
and the following conditions are met
we going to be writing the double integral
of f of x, y, z, integrated respects to s
as a double integral over the region r
in the x, y plane
by replacing z with g of x,y
and then replacing differential s
with this extra integrating factor
differential a
where differential a would be dx, dy, or dy, dx.
And in part one we also discussed
where this jacobian or extra integrating factor came from.
Let's take a look at our example.
We want to integrate the function f of x, y, z
over the given surface in the first octant.
So we can write this double integral over the surface s
as a double integral over the region r in the x, y plane.
Where our function will become f of x comma y comma
g of x, y,
where g of x, y is equal to z.
So we have four minus two x minus two y
and then we have our jacobian or extra integrating factor
of one plus
the partial derivative of g with respects to x squared
so that's going to be negative two squared
plus the partial derivative of g
with respects to y squared
that's also going to be negative two.
Here we'll integrate with respects to y first
and then x.
Now this problem to determine the limits of integration
for y and x we're going to have to look
at the x, y trace of the given surface.
So if we set z equal to zero
and move the y to the left side
we would have two y equals negative two x plus four
go ahead and divide everything by two.
So we have y equals negative x plus two.
So our x, y trace would have a y intercept of two
and a slope of negative one so it's going to pass through
point two zero.
So this x, y trace should help us determine
the limits of integration
for y and x.
So for y,
it's going to start at zero and go up to the line
where the line must be expressed as a function of x
so it'll be from zero all the way to negative x plus two
or just two minus x.
And then for x,
this integration will be from zero to two.
Let's go ahead and continue this on the next page.
So here we have the square root of one plus four plus four.
That's going to be the square root of nine
so we have an extra factor of three.
Now let's go back and take a look at our function f.
F of x, y, z which is equal to x times y.
So this problem we don't even need to use
the z coordinates so we have xy dy dx.
Well first integrate respects to y
so we have x, y squared over two.
Replacing y with two minus x
we're going to have x times the quantity
two minus x squared all over two
and then when y is zero this will be zero.
So we multiply this out
we're going to have four minus four x plus x squared
and then we still have to multiply it by x.
So we'll have four x
minus four x squared
plus x to the third,
let's go ahead and factor the one half out
so this'll be three-halves.
So here we'll have four times x squared over two
or two x squared minus four thirds x cubed
plus x to the fourth over four.
So when x is two we'll have four times two that's eight.
Minus here we'll have 8 times 4 that's 32 thirds.
And here we're going to have 16 divided by 4 that's 4.
Next to zero we have zero.
So we have three halves times.
So it'll be 12 minus 32 thirds.
That gives us four thirds.
And this simplifies nicely so we have a value of two.
That'll do it for this video.
In the next video we'll take a look
at when the surface is defined parametrically.
Thank you for watching.