Welcome to an example on how
to find the surface area of a function of two variables.
In this example, we're asked to find the are of the surface
of the paraboloid given by z equals x squared plus y squared
that lies below the plane z equals 16.
So our goal is to find the surface area
of this blue paraboloid that lies
below the plane z equals 16 which is graphed here in red.
In our previous example, we found the surface area
of a parametric surface given by r of u comma v.
But in this example, we're trying to find the surface area
of a surface given by z equals g of x comma y
over the region R in the xy-plane.
So the surface area is equal to the double integral
over the surface S integrated with respect to S
which equals the double integral over the region R.
Where, again, the region R would be the region
in the xy-plane of the square root of one plus
the partial derivative of g with respect to x squared plus
the partial derivative of g with respect to y squared,
differential A.
So going back to our example, notice how if we project
the paraboloid onto the xy-plane we get this circle.
Where because the equation of this red plane is z equals 16
if we replace z with 16 in the equation of the paraboloid,
we get 16 equals x squared plus y squared.
Which gives us this region, R, on the xy-plane
which is a circle centered at the origin
with a radius of four.
And looking at our formula here, g of x comma y is equal
to z which equals x squared plus y squared.
Now let's go ahead and find our partial derivatives.
So the partial derivative of g with respect to x is equal
to the derivative of x squared plus y squared with respect
to x which would be two x.
And the partial derivative of g with respect to y
would be the derivative of x squared plus y squared
with respect to y which would be two y.
So if the surface area, which equals the double integral
over the surface S, differential S is equal
to the double integral over the region R
on the xy-plane of the square root
of one plus the partial with respect to x squared
which would be two x to the second plus the partial
with respect to y squared which would be two y squared,
differential A.
Let's go ahead and simplify the exponents.
So we have the double integral over the region R
of the square root of one plus ...
Well this is going to be four x squared plus four y squared.
Let's factor out the four.
So we'd have four times the quantity x squared
plus y squared, differential A.
So because the region R is a circle,
let's write this using polar coordinates.
Where differential A is equal to r, dr, d theta
and x squared plus y squared equals r squared.
So in polar coordinates we'd have the double integral
over the region R of the square root of one plus,
this would be four r squared.
And again, differential A is equal to r, dr, d theta.
And now using the region R here will determine the limits
of integration.
The limits of integration with respect to r would be
from zero to four.
The limits of integration for theta would be
from zero all the way around to two pi.
Let's go ahead and evaluate this
to determine the surface area on the next slide.
Well to perform u substitution,
we'll let u be equal to one plus four r squared.
So differential u is equal to eight r, differential r.
So let's divide both sides by eight.
So we have one eighth differential u equals r dr.
Which means in terms of u, all of this would be ...
Well the square root of one plus four r squared
would be the square root of u which would be u
to the one half and then r dr is equal to one eighth du.
So now if we integrate, with respect
to u we'd have one eighth times u
to the three halves times two thirds.
Which means with respect to r we'd have one eighth
times two thirds times the quantity one plus four r squared
raised to three halves.
We can simplify here.
There's one, two, and two and four two's in eight.
So here we have one twelfth.
So we have one twelfth times the integral from zero
to two pi.
Now performing substitution for r when r is four,
we'd have the quantity one plus four times four squared
to the three halves minus when r is zero,
we'd have one plus four times zero squared
to the three halves, d theta.
Simplifying again, we have one twelfth times the integral
from zero to two pi.
Here we're going to have one plus 64.
That's 65 to the three halves minus here we'll have one
to the three halves which is one.
This is just a constant.
So we can write this as one twelfth times the quantity 65
to the three halves minus one times the integral
from zero to two pi, d theta or if we want one d theta.
So the antiderivative with respect to theta here
would just be theta.
So we have one twelfth times 65 to the three halves
minus one.
Then we have the antiderivative which is theta.
Performing substitution, we'll just have one twelfth
times the quantity 65 to the three halves minus one
times, we just have two pi minus zero.
So the two and the twelfth simplify.
So the surface area is exactly pi divided
by six times the quantity 65 to the three halves minus one.
Which as a decimal would be approximately 273.8666.
And because it's the surface area,
this would be square units which again is the surface area
of this paraboloid below the red plane z equals 16.
I hope you found this helpful.