Welcome to Calculus. 
I'm Professor Ghrist. We're about to begin Lecture 36, on 
surface area. In previous lessons, we've computed areas 
of flat, two-dimensional regions, but what about surface area? 
The area of a shape that might be wavy or curve in three-dimensional space. 
In this lesson, we'll see how the surface area element is related to the arc length 
element. Let's begin with the computation of 
surface area for a very simple surface. In this case, a cone over a circle. 
Let's say of height h, where the circle has radius r. 
Now, the best way to compute this is to decompose into triangles and integrate. 
So let's take our surface area element to the sum triangle. 
Defined by a segment of the circle with angle, d theta. 
The length of the base of this triangle may be approximated by rd theta. 
What is the height? Not h, but rather L, the slant length of 
this cone. Although, expressable in terms of r and 
h, let's just use L. So our surface area element is 1 half the 
base, rd theta times the height, L. Integrating to obtain the surface area, 
we get the integral from 0 to 2pi, from 1 half rLd theta. 
That is simply pi rL. That part is easy enough. 
We're going to put this to use in determining the surface area of a surface 
of revolution, obtained by revolving a curve. 
In the x-y plane, about an axis, let's say about the x-axis. 
What is the appropriate surface area element? 
We're going to fix a value of x, and consider a vertical slice of this 
surface, obtaining a circle. We then thicken it up by dS. 
What is this surface area element? Well, this is not a cylinder. 
It is rather the tail end of some cone, a cone with slat length L and of radius r. 
Now, what we care about is not the slat length. 
But rather dL, the small change in that length. 
That is indeed the arc length element. Now, let us use what we know about the 
service area of this cone. It is pi times r times L, but how do we 
get dS? Well, of course, we apply the 
differentiation operator and we see that dS is in fact pi times, through the 
product rule, rdL plus Ldr. Now, dL we know that's the arch length 
element. What do we do with dr? 
Well, consider the right triangle with hypotenuse L and height r. 
Then, if we extend things a little bit, into d l and d r Then through similar 
triangles we can argue that dL over L is dr over r. 
Cross-multiplying we see that L times dr is equal to r times dL, unless we can 
simplify the surface area element to twice pie r dL. 
And this is wonderful because we know dL, the arc length element and we know r, 
because r is typically given in terms of y of x, where this curve is some function 
of y of x. Therefore, using what we know about the 
arc length element of such a graft. We have dS is pi times y times square 
root of 1 plus the of ydx squared dx, well, let's see how this works in a 
context of a fun problem. Consider a round ball of radius R divide 
this ball into slices of equal width. The question is which of these slices has 
the most surface area. Is it the one in the middle or maybe the 
one at the end? Well, let's set things up in terms of 
some coordinates. If this is a ball of radius r, then it 
can be considered a surface of revolution where we take the curve, y equals square 
root of r squared minus x squared, and rotate that out the x-axis. 
Therefore, we know that the surface area element is 2 pi y dL, in this case, it's 
2 pi times the square root of r squared minus x squared times the square root of 
1 plus the derivative squared, dx. We've done this derivative before. 
It's minus x over square root of r squared minus x squared. 
And so, substituting that in for the surface area element, we see something 
that looks a little frightening, but it's not so bad. 
If we put the arc length element above the common denominator r squared minus x 
squared, then indeed we see that, that denominator cancels with the r squared 
minus x squared to the left. We're left with 2pi, square root of r 
squared dx. That is 2 pi rdx. 
That is a simple service area element, but it's not only simple. 
It is independent of x. It does not matter as long as your 
thickness is dx. That means that in the question of which 
slice has the most area. None of them or rather all of them, they 
all have the same area. In fact, we can take this surface area 
element and very easily integrate it as x goes from negative r to r to obtain in 
our heads. The familiar formula for surface area of 
a ball of radius r for pi r squared. Now, that might seem like a trick or a 
curiosity, but if we slice horizontally and consider a slightly different 
setting, then we have a basis for what is called the Lambert cylindrical 
projection. In cartography, this is a solution to the 
problem of how to make a map of the world where everything has the same area as on 
the surface on the ball even though the map was flat. 
It distorts length, but not area. Let's turn to a different example. 
This time, we're going to take the curve 1 over x to the p as x goes from 1 to 
infinity and rotate that curve about the x-axis. 
This is going to give us some improper integrals. 
The question is what is the surface area? Well, let's begin by doing something a 
little different. Let's compute the volume of this solid, 
obtained if we slice by orthogonal discs. Then, what do we obtain? 
For the volume. Well, at some point x we have a volume 
element given by pi times the radius x to the minus p squared times dx. 
We can integrate this to obtain the volume. 
That gives us the integral from 1 to infinity of pi times x to the minus 2P 
dx. This is a simple integral. 
We've done this before when we did the P integrals. 
When 2P is bigger than 1, we get pi or 2P minus 1. 
We could rewrite that as saying, P is bigger than a half. 
Otherwise, when P is less than or equal to 1 half, we have a divergent integral 
and the volume is infinite. Now that we've done volume, let's do 
surface area. Surface area element is 2 pi times the 
height x to the minus p times dL. We can rewrite that dL as square root of 
1 plus the derivative negative px to the minus p minus 1 quantity squared. 
Now, what do we get when we integrate that? 
That looks like an unpleasant integral. It is an unpleasant integral. 
But, we can at least determine whether it's convergent or divergent. 
How do we do that? Recall, what matters is the leading-order 
term. We could apply the binomial expansion to 
that square root. Since when x is very large the P squared 
x to the negative 2P minus 2 is very small. 
And we obtain 2 pi x to the minus P times 1 plus some higher order terms. 
Then we can be very specific about what order those are. 
What really matters is that the leading order term is 2 pi times x to the minus 
p. And therefore, we know when the integral 
converges and when it diverges. It converges when p is strictly bigger 
than 1 and diverges otherwise. Now, let's compare that surface area to 
what we saw in the case of volume. There are different constraints on P for 
surface area. What we have is something that is finite, 
if P is strictly bigger than 1. But for volume, we have something that is 
finite, when P is strictly bigger than 1 half, and this leads us to the very 
curious result. That if your value of P is somewhere 
between 1 half and 1, then you can have an object with finite volume, but 
infinite surface area. That seems a bit counterintuitive at 
first, but it's very cool. Let's try an example, that's a bit more 
finite. Let's consider, simply, a parable y 
equals x squared, but let's rotate it about the y-axis. 
As x goes from 0 to 1, that's going to give us some bowl shaped region. 
In this case, we're going to want to slice horizontally and use the 
appropriate surface area element to pi rdL. 
What is r? Well, in this case it's equal to x. 
That is the square root of y. And in this case, what is dL? 
Well, it's 1 plus the derivative squared. Now, that derivative is what? 
We need to differentiate square root of y. 
That gives us 1 over 2 root y. Simplifying, we get a surface area 
element that is 2 pi times square root of y plus 1 4th dy. 
And now, to obtain the surface area, we integrate this surface area element as y 
goes from 0 to 1. This is not too hard of an integral. 
We get 2 pi times quantity y plus 1 quarter to the 3 halves times 2 3rds. 
Evaluating that from 0 to 1, I'll let you check that that yields pi divided by 6 
times 5 square root of 5 minus 1. And it's worth noting briefly that this 
problem can also be solved by integrating with respect to x instead of y. 
In this case, r is equal to x. And dL is square root of one plus dy, dx 
squared dx. That derivative simply 2 times x. 
I'll leave it to you to set up the integral and show that with a simple u 
substitution, it's possible to compute the exact same answer with just about the 
same amount of work. Sometimes, it's better to integrate the 
other way. Let's end with one last example, a 
catenoid. This is what happens when you rotate an 
catenary about the axis. Recall the catenary, the shape obtained 
by hanging chain or rope. Now, we know the equation for that. 
It is y is some constant 1 over kappa times the hyperbolic cosine of kappa x 
plus sum initial height y naught. In this case, what is the surface area 
element? 2 pi y square root of 1 plus dy, dx 
squared dx. It's easy to differentiate a hyperbolic 
cosine and so we obtain a surface area element. 
Well, doesn't look so nice we can simplify that square root, but in the 
end, we've got a COSH squared term and a COSH term. 
Now, this is not impossible to integrate. It is doable, but we're not going to do 
it here. What I do want to point out is that this 
is a surface that you can see, it's a wonderful example of something called a 
minimal. Surface something that minimizes the 
surface area for a fixed boundary. If you took two wire loops and dipped 
them in a soap film solution and held them apart, if you do it just right, then 
the surface that you get will be a catenoid. 
And that's kind of fun to play with. And now, you know how to compute its 
surface area. With this, we complete our applications 
of definite integrals to problems in geometry. 
But of course, there are many other applications we could explore. 
In our next lesson, we'll consider a more physical application related to work.