Welcome to the first video on
surface integrals where the surface is given explicitly.
For a surface S given by z equals g of x, y,
that is continuous and differentiable over a region R
in the xy-plane, we can evaluate this double integral
of f of x, y, z integrated over the surface S
as the double integral as we see here over the region
R in the x, y plane, and there are some important
things we should notice here.
First, we're going to replace z with g of x, y,
and second, differential S is replaced with this
extra integrating factor here,
and then differential a written as dxdy or dydx.
And notice in the previous video,
when we were talking about surface area,
we introduced a surface integral here
where f of x, y, z was equal to one,
but now they integrated as a function other than one.
Let's take a look at where this integrating factor
or this jacobian comes from.
So in the previous video we showed how to determine the
surface area of a parametized surface using this
double integral here on the right.
Well if we have z equals g of x, y,
then our vector valued function for the surface
would be x, y, g of x, y, and so the participative
of r with respects to x would be here
and the participative of y would be here.
So if we determine the integrating factor here,
the magnitude of the cross product,
this would be the cross product of our two partial
derivatives and then the magnitude does give us a jacobian
or the extra integrating factor that
we saw on the previous screen.
Let's go and take a look at our first example
by looking at an application.
Here we have a roof given by the graph of g of x, y,
over the following region.
If the density of the roof is given by f of x, y, z,
determine the mass of the roof.
So if we integrate the density function,
over the surface g of x, y, this will give us the mass.
So let's take a look at a graph of actually what's happening
in this situation.
The surface of the roof would be the purple plane
and the given region is bounded by these yellow planes.
So if we integrate the density function over this purple
surface it will give us the mass of the roof.
So let's go ahead and set this up.
The mass is going to be equal to the double integral
of f of x, y, z, our density function,
integrated over the surface s, which is going to be equal
to the double integral over the region r
of f of x, y, g of x, y, which is 25, plus,
I'm going to write this as 1/2x plus 1/2y
times your integrating factor of the square root
of one plus the partial derivative of g with respect
to x squared, and here's g, so the partial derivative
with respects to x would be 1/2,
so we're going to have 1/2 squared,
plus the partial derivative of g with respects to y squared.
That'll be 1/2 squared as well.
And because the integrating region r is rectangular,
the order of integration doesn't matter.
Let's go ahead and go with respects to x first
and then y, so limits to integration with respect to x
will be from zero to 40, and for y
they'll be from zero to 20.
Now let's take a look at this integrating factor here.
1/2 squared is 1/4, so we have one plus 1/4 plus 1/4,
or 4/4 plus 1/4 plus 1/4 underneath the square root,
which is going to be the square root of 6/4,
and the square root of four is equal to two,
so we have square root of six divided by two.
Let's go ahead and factor that out.
And now for our function f, we'll have 150
minus two times z, but z is 25 plus 1/2x plus 1/2y.
Now let's go ahead and clear these parentheses here,
so we can think of distributing a negative two.
So I'll have 150 minus 50.
That's 100.
And then we'll have minus x, minus y.
Let's continue this on the next slide.
Now we'll integrate with respects to x
treating y as a constant, so we'll have 100x minus x
squared over two, minus xy, and we'll evaluate this
when x is equal to 40 and when x is equal to zero.
Notice when x is zero, everything here is going to be zero
so we can just replace x with 40,
so we'll have 100 times 40 minus 40 squared divided by two
minus, this will be 40y.
Well 1,000 times 40, that's going to be 4,000,
minus, this is going to be 1600 divided by two,
which is 800, so 4,000 minus 800 is 3,200, minus 40y.
And now we'll integrate with respects to y.
So 3,200y minus 40 times y squared divided by two,
which will be 20y squared.
Now we'll evaluate this.
So we'll first replace y with 20.
We'll have 3,200 times 20 minus 20 times 20 squared,
and again notice when y is equal to zero these
will both be zero.
Now to save some time I've already evaluated this.
This here is equal to 56,000.
So square root six divided by two times 56,000
is approximately equal to 68,585.7.
So this would be the mass of that roof,
and if this was the metric system this
would be in kilograms.
So again this would be the mass of this roof in purple,
using the given density function.
And we'll take a look at several more examples of surface
integrals over the next several videos.
Thank you for watching.