INSTRUCTOR: Calculate the surface integral, the double integral over the region S of the function x minus y, where S is the cylinder x squared plus y squared equals 1, with 0 less than or equal to z less than or equal to 2, including the circular top and bottom.

Now, notice that this double integral over the region S of x minus y is actually composed of three separate surface areas. So I have the double integral. I skipped my dS. That is equal the double integral over the region S1 of x minus y, dS1, plus-- well, you get the picture.

There are three sides. There's the sides of the cylinder. They're the side of the cylinder. There's the bottom base, and there's the top base-- those circular top and bottom dimensions. So we're actually going to write this as three separate surface areas, and we will sum them at the end.

So let's start with what I'm going to call side 1, which is the side of the cylinder. In order to find that integral, we need to parameterize these sides. So this would be r of u comma v equal to angle bracket, cosine u comma sine u comma v with 0 less than or equal to u less than or equal to 2 pi and 0 less than or equal to v less than or equal to 2.

For this parameterization, t sub u is angle bracket, negative sine u comma cosine u comma 0, end angle bracket. And t sub v is equal to angle bracket, 0 comma 0 comma 1.

t sub u crossed with t sub v then is equal to angle bracket, cosine u comma negative sine u comma 0. And the magnitude of tu cross tv is then equal to 1 because of the Pythagorean identity.

So this means that the double integral over the region S1 of the function x minus y is equal to the integral from 0 to 2 with respect to v of the integral from 0 to 2 pi with respect to u of the function cosine u minus sine u.

The inner integral is going to be equal to integral-- well, this will become the integral from 0 to 2 of the inner antiderivative becomes sine u plus cosine u if I went here from 0 to 2 pi.

And this is equal to 0, actually. So this integral ends up being the integral from 0 to 2 with respect to v of 0. And that is equal to 0. So the surface area with this integral and this surface is equal to 0 on the sides.

Now let's work on the-- what did I say S sub 2 was? The bottom base. Let's do the bottom base next. So for S sub 2, our bottom base, our parameterization can be r of u comma v equal to angle bracket, u cosine v comma u sine v comma 0.

Now, you might notice the changes here. So in the first setup, the first surface, there's a fixed radius of 1 for the sides of the cylinder and a varying height from 0 to 2. Well, here, we have a fixed height for the bottom base of 0 and a varying radius because we're beginning at the center of our circle and working our way outwards to find that surface.

Now, that is going to be restricted. 0 less than or equal to u less than or equal to 1. And 0 less than or equal to v less than or equal to 2 pi. Because again, our radius going from the inside to the out, 0 to 1, and going in a complete circle of 0 to 2 pi.

Now let's find t sub u. t sub u is equal to angle bracket, cosine v comma sine v comma 0, close angle bracket.

tv is equal to angle bracket, negative u sine v comma u cosine v comma 0 so that tu cross tv is equal to angle bracket, 0 comma 0 comma u, close angle bracket. And the magnitude of tu cross tv is then equal to u.

Our integral is going to be very similar. So the double integral over surface 2, S sub 2, of the function x minus y will be equal to integral from 0 to 1 with respect to u of the integral from 0 to 2 pi with respect to v of the function u cosine v minus u sine of v.

Now, the inner integral. Evaluating that means our whole expression, this surface area of the surface, is integral from 0 to 1 of u multiplied by sine v plus cosine v, evaluated from 0 to 2 pi. And that is still with respect to u.

And this is, again, equal to 0. So we have the integral from 0 to 1, du of the function 0, so that the area of that surface with respect to our integral of the function x minus y is equal to 0.

Now, our third surface-- this is going to be the top of this cylinder-- is going to be almost identical to this one. It's going to have the same parameterization except that the third component will be equal to 2 because we're at the top of this. Height is equal to 2. And that makes tu and tv the same, which means tu cross tv is the same, which means the magnitude of tu cross tv is the same.

And it actually means our whole integral is going to be identical. So that third surface, the top of this, is equal to 0. The surface area of that is equal to 0, which means the double integral over the surface S of the function x minus y is equal to 0 plus 0 plus 0, which is equal to 0.