# Surface integral example part 2: Calculating the surface differential | Khan Academy

Now that we have our parametrization right
over here, let's get down to the business of actually evaluating
this surface integral.
It's a little bit involved, but we'll
try to do it step by step.
And so the first thing I'm going to do
is figure out what d sigma is in terms of s and t,
in terms of our parameters.
So we can turn this whole thing into a double integral
with respect to-- or a double integral in the s, t plane.
And remember, d sigma right over here,
it's just a little chunk of the surface.
It's a little area of the surface right over there.
And we saw in previous videos, the ones
where we learned what a surface integral even
is, we saw that d sigma right over there,
it is equivalent to the magnitude of the cross product
of the partial of our parametrization with respect
to one parameter crossed with the parametrization
with respect to the other parameter
times the differentials of each of the parameters.
So this is what we're going to use right here.
It's a pretty simple looking statement, but as we'll see,
taking cross products tend to get a little bit hairy.
Especially cross products of three dimensional vectors.
But we'll do it step by step.
But before we even take the cross product,
we first have to take the partial of this with respect
to s, and then the partial of this with respect to t.
So first, let's take the partial with respect to s--
the partial of r with respect to s.
So right over here, all the stuff with the t in it,
you can just view that as a constant.
So cosine of t isn't going to change.
The partial of-- or the derivative of cosine of s
with respect to s is negative sine of s.
So this is going to be equal to-- I'll
put the negative out front-- negative cosine of t sine of s.
I'm going to keep everything that has a t involved purple.
Sine of s.
And let me make-- I don't know.
I'll make the vectors orange-- i, and then you we'll plus
and once again, we'll take the derivative with respect to s,
cosine of t is just a constant, derivative
of sine of s with respect to s is cosine of s.
So it's going to be plus cosine of t cosine of s j.
And then plus the derivative of this with respect to s.
Well, this is just a constant.
The derivative of 5 with respect to s would just be 0.
This is the same thing.
It's just a constant with-- this does not
change with respect to s.
So it does-- so our partial here with respect to s is just 0.
So we could write even 0k.
Let me write 0-- I'll just write 0k right over there.
And that's nice to see because that'll
make our cross product a little bit more straightforward.
Now, let's take the partial with respect to t.
all right.
And we get-- so the derivative of this with respect
to t-- now, cosine of s is the constant.
Derivative of cosine t with respect to t
is negative sine of t.
So this is going to be negative sine of t cosine of s i--
I'll do it in that-- I'll use this blue-- i, and then
plus-- now derivative of this with respect to t.
Derivative of cosine of t is negative sine of t.
So once again, so now we have minus sine of t sine of s--
my hand is already hurting from this.
This is a painful problem.
j plus derivative of sine of t with respect to t.
We're taking the partial with respect to t.
It's just cosine of t.
So plus cosine of t.
And now times the k unit vector.
Now we're ready to take the cross product of these two
characters right over here.
And to take the cross products-- let me write this down.
So we're going to say, we're going
to take the cross product of that with that.
Is going to be equal to-- and I'm
going to set up this huge matrix.
Or, it's really just a three by three matrix
but it's going to be huge because it's
going to take up a lot of space to have
to write down all this stuff.
So maybe I'll take up about that much space
so that I have space to work in.
And I'll write my unit vectors up here.
i, j, k.
Or at least this is how I like to remember
how to take cross products of three dimensional vectors.
Take the determinant of this three
by three matrix-- the first row are just
our unit vectors-- the second row is the first vector
that I'm taking the cross product of.
So this is going to be negative.
I'm just going to rewrite this right over here.
So it's going to be negative cosine of t sine of s.
And then you have cosine of t cosine of s.
And then you have 0 which will hopefully
simplify our calculations.
And then you have the next vector, that's the third row.
Negative sine of t cosine of s.
And I encourage you to do this on your own
if you already know where this is going.
It's good practice.
Even if you have to watch this whole thing
to see how it's done, try to then do it again on your own
because this is one of those things
you really got to do yourself to really have it sit in.
Negative sine of t sine of s.
And then finally, cosine of t.
So let's take the determinant now.
So first we'll think about our i component.
Are i component, you would essentially
ignore this column-- the first column in the first row--
and then take the determinant of this submatrix right over here.
So it's going to be i-- so this is
going to be equal to i times something.
I'll put the something in parentheses there.
Normally you see the something in front of the i
but you can swap them there.
So let me just-- so it's going to be i times something.
Ignore this column, that row.
This determinant is cosine of t cosine of s times cosine of t,
which is going to be cosine squared of t.
Let me write it a little neater.
Cosine squared of t cosine of s.
And then from that we need to subtract 0 times this.
But that's just going to be 0 so we're just left with that.
Now we're going to do the j component.
But you probably remember the checkerboard thing
when you have to evaluate three by three matrices.
Positive, negative, positive.
So you will have a negative-- you write a negative j.
A negative coefficient, I guess, in front of the j times
something.
And so you ignore j's column, j's row.
And so you have negative cosine of t sine
of s times cosine of t.
Well, that's going to be negative cosine
squared of t times sine of s.
Let me make sure I'm doing that right.
Ignore that and that.
It's going to be that times that.
So yep.
Negative cosine sine of s minus 0 times that.
And so that's just going to be 0, so we can ignore it.
And you have a negative times a negative here
so they can both become positive.
And then finally, you have the k component.
And once again, you can go back to positive there.
Positive, negative, positive on the coefficients.
That's just evaluating a three by three matrix.
And then you have plus k times-- and now this
might get a little bit more involved
because we won't have the 0 to help us out.
Ignore this row, ignore this column,
take the determinant of this sub two by two.
You have negative cosine t sine s times
negative sine of t sine s.
Well, that's going to be-- the negatives cancel out.
So it's going to be cosine of t sine
of t times sine squared of s.
And then, from that we're going to subtract
the product of these two things.
But that product is going to be negative
so you subtract a negative-- that's
the same thing as adding a positive.
So plus-- and you're going to have cosine t sine t again--
plus cosine t-- let me scroll to the right a little bit--
plus cosine t sine t again.
And that's times cosine squared of s.
Now this is already looking pretty hairy
but it already looks like a simplification there.
And that's where the colors are helpful.
Actually, now I have trouble doing math
in anything other than multiple pastel colors
because this actually makes it much easier
to see some patterns.
And so what we can do over here is
we can factor out the cosine t sine t.
And so this is equal to cosine t sine t times sine
squared s plus cosine squared s.
And this we know the definition of the unit circle.
This is just going to be equal to 1.
So that was a significant simplification.
And so now we get our cross product
being equal to-- let me just rewrite it all.
Our cross product.
r sub s crossed with r sub t is going to be equal to cosine
squared t cosine s times our i unit vector plus cosine squared
t cosine sine of s times our j unit vector
plus-- all we have left because this is just 1-- cosine t sine
t times our k unit vector.
So that was pretty good but we're still not done.
We need to figure out the magnitude of this thing.
Remember, d sigma simplified to the magnitude
of this thing times ds dt.
So let's figure out what the magnitude of that is.
And this is really the home stretch
so I'm really crossing my fingers
that I don't make any careless mistakes now.
So the magnitude of all of this business
is going to be equal to the square root--
and I'm just going to square each of these terms
and then add them up-- the square root
of the sum of the squares of each of those terms.
So the square of this is going to be--
cosine squared squared is cosine to the fourth-- cosine
to the fourth t cosine squared of s plus cosine
to the fourth t sine squared s-- and I already
see a pattern jumping out-- plus cosine
squared t sine squared t.
Now, the first pattern I see is this--
is just this first part right over here.
We can factor out a cosine to the fourth t.
Then we get something like this happening again.
So let's do that.
So these first two terms are equal to cosine
to the fourth t times cosine squared s plus sine squared s.
Which, once again, we know is just 1.
So this whole expression has simplified
to cosine to the fourth t plus cosine squared t sine
squared t.
Now we can attempt to simplify this again
because this term and this term both
have a cosine squared t in them.
So let's factor those out.
So this is going to be equal to-- everything
I'm doing is under the radical sign.
So this is going to be equal to a cosine squared
t times cosine squared t.
And when you factor out a cosine squared t here,
you just have a plus a sine squared t.
And that's nice because that, once again, simplified to 1.
All of this is under the radical sign.
Maybe I'll keep drawing the radical signs here
to make it clear that all this is still