Example of calculating a surface integral part 3

In the last couple of videos we've been slowly moving towards our goal of figuring out the surface area of this torus. And we did it by evaluating a surface integral, and in order to evaluate a surface integral we had to take the parameterization-- take its partial with respect to s and with respect to t. We did that in the first video. Then we had to take its cross product. We did that in the second video. Now, we're ready to take the magnitude of the cross product. And then we can evaluate it inside of a double integral and we will have solved or we would have computed an actual surface integral-- something you see very few times in your education career. So this is kind of exciting. So this was the cross product right here. Now, let's take the magnitude of this thing. And you might remember, the magnitude of any vector is kind of a Pythagorean theorem. And in this case it's going to be kind of the distance formula to the Pythagorean theorem n 3 dimensions. So the magnitude-- this is equal to, just as a reminder, is equal to this right here. It's equal to the partial of r with respect to s cross with the partial of r with respect to t. Let me copy it and paste it. That is equal to that right there. Put an equal sign. These two quantities are equal. Now we want to figure out the magnitude. So if we want to take the magnitude of this thing, that's going to be equal to-- well, this is just a scalar that's multiply everything. So let's just write the scalar out there. So b plus a cosine of s times the magnitude of this thing right here. And the magnitude of this thing right here is going to be the sum, of-- you can imagine, it's the square root of this thing dotted with itself. Or you could say it's the sum of the squares of each of these terms to the 1/2 power. So let me write it like that. Let me write the sum of the squares. So if you square this you get a squared cosine squared of s, sine squared of t. That's that term. Plus-- let me color code it. That's that term. I'll do the magenta. Plus that term squared. Plus a squared cosine squared of s, cosine squared of t. That's that term. And then finally-- I'll do another color-- this term squared. So plus a squared sine squared of s. And it's going to be all of this business to the 1/2 power. This right here is the same thing as the magnitude of this right here. This is just a scalar that's multiplying by both of these terms. So let's see if we can do anything interesting here. If this can be simplified in any way. We have a squared cosine squared of s. We have an a squared cosine squared of s here, so let's factor that out from both of these terms and see what happens. I'm just going to rewrite this second part. So this is going to be a squared cosine squared of s times sine squared of t-- put a parentheses-- plus cosine-- oh, I want to do it in that magenta color, not orange. Plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1/2 power. Now what is this? Well, we have sine squared of t plus cosine squared of t. That's nice. That's equal to 1, the most basic of trig identities. So this expression right here simplifies to a squared cosine squared of s plus this over here: a squared sine squared of s. And all of that to the 1/2 power. You might immediately recognize you can factor out an a squared. This is equal to a squared times cosine squared of s plus sine squared of s. And all of that to the 1/2 power. I'm just focusing on this term right here. I'll write this in a second. But once again, cosine squared plus sine squared of anything is going to be equal 1 as long as it's the same anything it's equal to 1. So this term is a squared to the 1/2 power. Or the square root of a squared, which is just going to be equal to a. So all of this-- all that crazy business right here just simplifies, all of that just simplifies to a. So this cross product here simplifies to this times a, which is a pretty neat and clean simplification. So let me rewrite this. That simplifies, it simplifies to a times that. And what's that? a times b, so it's ab. ab plus a squared cosine of s. So already, we've gotten pretty far and it's nice when you do something so beastly and eventually it gets to something reasonably simple. And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region from s-- over the region over with the surface is defined. So s going from 0 to 2 pi and t going from 0 to 2 pi. Over this region. So we want to integrate this over that region. So that region we're going to vary s from 0 to 2 pi. So ds. And then we're going to vary t from 0 to 2 pi-- dt. And this is what we're evaluating. We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization. So this is what we can put in there. Things are getting simple all of a sudden, or simpler. ab plus a squared cosine of s. And what is this equal to? So this is going to be equal to-- well, we just take antiderivative of the inside with respect to s. So the antiderivative-- so let me do the outside of our integral. So we're still going to have to deal with the 0 to 2 pi and our dt right here. But the antiderivative with respect to s right here is going to be-- ab is just a constant, so it's going to be abs plus-- what's the antiderivative of cosine of s? It's sine of s. So plus a squared sine of s. And we're going to evaluate it from 0 to 2 pi. And what is this going to be equal to? Let's put our boundaries out again or the t integral that we're going to have to do in a second-- 0 the 2 pi d t. When you put 2 pi here you're going to get ab times 2 pi or 2 pi ab. So you're going to have 2 pi ab plus a squared sine of 2 pi. Sine of 2 pi is 0, so there's not going to be any term there. And then minus 0 times ab, which is 0. And then you're going to have minus a squared sine of 0, which is also 0. So all of the other terms are all 0's. So that's what we're left with it, it simplified nicely. So now we just have to take the antiderivative of this with respect to t. And this is a constant in t, so this is going to be equal to-- take the antiderivative with respect to t-- 2 pi abt and we need to evaluate that from 0 to 2 pi, which is equal to-- so we put 2 pi in there. You have a 2 pi for t, it'll be a 2 pi times 2 pi ab. Or we should say, 2 pi squared times ab minus 0 times this thing. Well, that's just going to be 0, so we don't even have to write it down. So we're done. This is the surface area of the torus. This is exciting. It just kind of snuck up on us. This is equal to 4 pi squared ab, which is kind of a neat formula because it's very neat and clean. You know, it has a 2 pi, which is kind of the diameter of a circle. We're squaring it, which kind of makes sense because we're taking the product of-- you can kind of imagine the product of these 2 circles. I'm speaking in very abstract, general terms, but that kind of feels good. And then we're taking just the product of those two radiuses, remember. Let me just copy this thing down here. Actually, let me copy this thing because this is our new-- this is our exciting result. Let me copy this. So copy. So all of this work that we did simplified to this, which is exciting. We now know that if you have a torus where the radius of the cross section is a, and the radius from the center of the torus to the middle of the cross sections is b. That the surface area of that torus is going to be 4 pi squared times a times b. Which I think is a pretty neat outcome.