For us to find the area
of the surface of the semi cone Z equals
the square root of the quantity X squared
plus Y squared that lies between Z equals three
and Z equals eight.
If the function of two variables, G of X comma Y,
has continuous first order partial derivatives
on the closed region R in the XY-plane,
the surface area can be determined
using this double integral here.
Before we set up this double integral though,
let's look at this graph in three dimensions.
The semi cone is graphed in blue,
the lower plane is Z equals three
and this plane is Z equals eight.
Our goal is to find the surface area
of the cone that lies between these two planes.
If we look down on the XY-plane
we can view the region R, the region of integration.
The region R is the region bounded between
these two circles where we have this larger circle
and we have the smaller circle.
And again, this is from projecting the cone
onto to XY-plane between Z equals three
and Z equals eight.
So going back to our work, let's show how we can find
the equations of these two concentric circles
that form the region of integration.
Remember, these circles are formed from the intersection
of the two planes in the cone.
So if we substitute three for Z into the equation
for the cone we can determine the equation
for the smaller circle and if we substitute eight for Z
we can determine the equation of a larger circle.
Let's go ahead and show this.
If we substitute three for Z in the equation of the cone
we have three equals the square root
of the quantity X squared plus Y squared
and now to undo to square root we square
both sides of the equation which gives us
nine equals X squared plus Y squared
which we should recognize as a circle
centered at the origin with a radius of three,
which is the small circle.
For the larger circle we substitute eight for Z
which gives us eight equals the square root
of the quantity X squared plus Y squared.
And again, to undo the square root we square
both sides of the equation which gives us 64
equals X squared plus Y squared
which is a circle centered at the origin
with a radius of eight which gives us the large circle.
And because the region of integration
is the area between two circles
it'll be easier to evaluate the double integral
using polar coordinates rather than rectangular coordinates.
Remember, in polar coordinates differential A
is equal to R, DR, D theta.
For the next step let's find the partial derivatives
required in the surface area formula.
Let's first write Z as Z equals the quantity
X squared plus Y squared raised to the 1/2 power.
Now we'll find the partial derivative of Z
with respect to X.
To do this we differentiate Z with respect to X
treating Y as a constant and this will require
the chain rule.
We will have 1/2 times the quantity X squared
plus Y squared raised to the power of 1/2 minus one,
which is negative 1/2, times the derivative
of X squared plus Y squared with respect to X
which is 2X.
Let's simplify this.
Two divided by two simplifies to one.
Let's write this as X divided by,
let's move this down to the denominator
which will change the sign of the exponent
and also write it as a square root.
So we have X divided by the square root of the quantity
X squared plus Y squared.
And again, this is the partial with respect to X.
And I will find the partial of Z
with respect to Y by differentiating
Z with respect to Y, now treating X as a constant.
So we'll have 1/2 times the quantity
X squared plus Y squared raised to the negative 1/2 power
times the derivative of X squared plus Y squared
with respect to Y which is 2Y.
Simplifying, the partial of Z with respect to Y
is equal to, again, this simplifies.
Here we have Y divided by the square root
of X squared plus Y squared.
And now we have the information we need
to set up the double integral for the surface area.
The surface area is equal to the double integral
over the region R of the square root
of one plus the square of X divided by the square root
of the quantity X squared plus Y squared
plus the square of Y divided by the square root
of the quantity X squared plus Y squared, differential A.
Let's continue on the next slide.
For the next step let's simply the erratic end.
We have one plus square and this fraction
we get X squared divided by the quantity
X squared plus Y squared, squaring the denominator
undoes the square root and then here we have
plus Y squared divided by the quantity
X squared plus Y squared.
We do have a common denominator so we can go ahead
and add these two fractions.
We have one plus adding the numerator.
So we have X squared plus Y squared
over the common denominator of X squared plus Y squared.
Notice how this simplifies nicely to one.
So this simplifies to the double integral
over the region R of the square root of,
again, this is one, one plus one is two,
so we have the square root of two differential A.
This simplified nicely, but because
the region of integration is the region between
these two circles, let's evaluate the double integral
using polar coordinates.
Now again, remember, for polar coordinates
differential A is equal to R, DR, D theta.
So we have the double integral
and then we have square root two, RDRD theta.
Now let's go back and look at the region of integration
to determine the limits of integration for R and theta.
Well because the radius of the small circle is three
and the radius of the large circle is eight.
The limits of integration for R will be from three to eight.
And the limits of integration for theta
will be all the way around the circle
from zero to two pi.
The limits of integration for R
are from three to eight and for theta
are from zero to two pi.
Let's evaluate this on the next slide.
We first integrate with respect to R,
remember this is R to the first
which gives us the integral from zero to two pi.
Now square root two times R squared divided by two
or square root two divided by two R squared.
And now we find big F of B minus big F of A.
Which will give us square root two
divided by two times one R is eight, we have
eight squared minus, when R is three we have three squared.
Eight squared is 64, three squared is nine.
64 minus nine is equal to 55 which gives us
the integral from zero to two pi
of 55 square root two divided by two D theta.
And now we integrate this back to theta
which gives us 55 square root two
divided by two theta.
And now we find big F of B minus big F of A
which is 55 square root two divided by two
times the quantity two pi minus zero
which is equal to 55 square root two
divided by two times two pi.
The twos simplify which gives us
an exact surface area of 55 square root two pi.
And because this is surface area
this should be square units.
And this is approximately 244.3586.
So if we go back to our graph one last time
we now know the surface area of the cone
bounded by these two planes is exactly
55 square root two pi square units.
Which again is approximately 244.3586.
I hope you found this helpful.