Where we left off in the last video, we were finding the

surface area of a torus, or a doughnut shape.

And we were doing it by taking a surface integral.

And in order to take a surface integral, we had to find the

partial of our parameterization with respect to s, and the

partial with respect to t, and now we're ready to take

the cross product.

And then we can take the magnitude of the cross product.

And then we can actually take this double integral and

figure out the surface area.

So let's just do it step by step.

Here we could take the cross product, which is not

a non-hairy operation.

This is why you don't see many surface integrals actually get

done, or many examples done.

Let's take the cross product of these two fellows.

So the partial of r with respect to s, crossed with--

in magenta-- the partial of r with respect to t.

This will be a little bit of review of cross

products for you.

You might remember this is going to be equal

to the determinant.

I'm going to write the unit vectors up here.

The first row is i, j, and k.

And then the next 2 rows are going to be-- let me do that in

that yellow color-- the next 2 rows are going to be the

components of these guys.

So let me copy and paste them.

You have that right there.

Copy and paste.

Put that guy right there.

Then you have this fellow right there.

Copy and paste.

Put him right there.

And then you got this guy right here.

This'll save us some time.

Copy and paste.

Put him right there.

Then the last row is going to be this guy's components.

Copy and paste.

Put him right here.

Almost done.

This guy-- copy and paste.

Put him right there.

Make sure we know that these are separate terms.

And finally, we don't have to copy and paste it, but just

since we did for all of the other terms, I'll do it

for that 0, as well.

So the cross product of these is literally the determinant

of this matrix right here.

And so, just as a bit of a refresher of taking

determinants, this is going to be i times the subdeterminant

right here, if you cross out this column and that row.

So it's going to be equal to i-- you're not used to seeing

the unit vector written first, but we can switch the order

later-- times i times the submatrix right here.

If you cross out this column and that row.

So it's going to be this term times 0-- which is just

0-- minus this term times that term.

So minus this term times this term- the negative signs are

going to cancel out, so this'll be positive.

So it's just going to be i times this term times this

term, without a negative sign right there.

So i times this term, which is a cosine of s.

It's really that term times that term, minus that term

times that term, but the negatives cancel out.

That times that is 0.

So that's how we can do this.

It's a cosine of s times b plus a cosine of s-- I'll just all

switch to the same color-- sine of t.

So we've got our i term for the cross product.

Now it's going to be minus j-- remember when you take the

determinant, you actually have this, kind of, you have to

checker board of switching sines.

So now it's going to be minus j times-- and you cross out that

row and that column-- and it's going to be this term times

this term-- which is just 0-- minus this term

times this term.

And once again, when you have-- oh, sorry.

When you cross out this column and that row.

So it's going to be that guy times that guy, minus

this guy times this guy.

So it's going to be minus this guy times this guy-- so it's

going to be-- let me do it in yellow.

So the negative times negative that guy, b plus a cosine of s

cosine of t times this guy, a cosine of s.

We'll clean it up in a little bit.

Well, we'll clean this up, and you see this negative and that

negative will cancel out.

We're just multiplying everything.

And then finally, the k term.

So plus-- I'll go to the next line-- plus k times-- cross out

that row, that column-- it's going to be that times that,

minus that times that.

So that looks like a kind of a beastly thing.

But I think if we take it step by step, it

shouldn't be too bad.

So that times that.

The negatives are going to cancel out.

So this term right here is going to be a sine

of t, sine of s.

And then this term right here is b plus a

cosine of s sine of t.

So that's that times that-- and the negatives canceled out,

that's why I didn't put any negatives here-- minus

this times this.

So this times this is going to be a negative number.

But if you take the negative of it, it's going to

be a positive value.

So it's going to be plus that a cosine of t

sine of s times that.

Times b plus a cosine of s cosine of t.

Now you see why you don't see many examples of surface

integrals being done.

Let's see if we can clean this up a little bit, especially if

we can clean up this last term a bit.

So let's see what we can do to simplify it.

So our first term.

So let's just multiply it out, I guess is the

easiest way to do it.

Actually, the easiest first step would just be factor out

the b plus a cosine of s.

Because that's in every term. b plus a cosine of s.

b plus a cosine of s.

So let's just factor that out.

So this whole crazy thing can be written as b plus a cosine

of s-- so we factored it out-- times--.

I'll put in some brackets here, so you don't multiply

times every component.

So the i component, when you factor this guy out, is going

to be a cosine of s sine of t.

Let me write it in green.

So it's going to be a cosine of s sine of t times i-- you're

not used to seeing the i before, so I'm going to write

the i here-- and then plus--.

We're factoring this guy out, so you're just going to be

left with cosine of t, a cosine of s.

Or we can write it as a cosine of s cosine of t-- that's that

right there, just putting it in the same order as that--

times the unit vector j.

And then when we factored this guy out-- so we're not going

to see that or that anymore.

When you factor that out, we can multiply this

out, and what do we get?

So in green, I'll write again.

So if you multiply sine of t times this thing over here--

because that's all that we have left after we factor out this

thing-- we get a sine of s, sine squared of t, right?

We have sine of t times sine of t.

So that's that over there.

Plus-- what do we have over here?

We have a sine of s times cosine squared of t.

And all of that times the k unit vector.

And so things are looking a little bit more simplified,

but you might see something jump out at you.

You have a sine squared and a cosine squared.

So somehow, if I can just make that just sine squared plus

cosine squared of t, those will simplify to 1.

And we can.

And this term right here, we can-- if we just focus on that

term-- and this is all kind of algebraic manipulation.

If we just focus on that term, this term right here can be

rewritten as a sine of s-- if we factor that out-- times sine

squared of t plus cosine squared of t times

our unit vector, k.

Right?

I just factored out an a sine of s from both of these terms.

And this is our most fundamental trig identity

from the unit circle.

This is equal to 1.

So this last term simplifies to a sine of s times k.

So, so far we've gotten pretty far.

We were able to figure out the cross product of these 2, I

guess, partial derivatives of the vector valued,

or our original parameterization there.

We were able to figure out what this thing right here, before

we take the magnitude of it, it translates to this

thing right here.

Let me rewrite it-- well, I don't need to rewrite it.

You know it.

Well, I'll rewrite it.

So that's equal to-- I'll rewrite it neatly and we'll use

this in the next video-- b plus a cosine of s times open

bracket a cosine of s sine of t times i plus-- switch back to

the blue-- plus a cosine of s cosine of t times j plus--

switch back to the blue-- this thing-- plus-- this simplified

nicely-- a sine of s times k.

Times the unit vector k.

This right here is this expression right there.

And I'll finish this video, since I'm already

over 10 minutes.

And in the next video, we're going to take

the magnitude of it.

And then, if we have time, actually take

this double integral.

And we'll all be done.

We'll figure out the surface area of this torus.