 In this video, we're going to find the area of the surface generated by rotating the curve y equals e to the x from x equals 0 to x equals 1 about the x axis. So if you were to try to look at this picture real quickly, what it looks like is the following. We have our exponential curve y equals e to the x and it looks something like this, y equals e to the x. So we're going to go from x equals 0 to x equals 1, like so. So we have 0, 1 as a point and 1 e as a second point. And we want to take this curve and rotate it around the x axis, like so. So drawing a typical radius will look something like this. Notice we're just looking for the y coordinate of the function. To find the area of the surface of revolution, we're going to use our formula s equals the integral of 2 pi r ds. Now let's be careful as we're going through this, right? We have to decide do we want to integrate with respect to x or y. We know the bounds if we're going to integrate with respect to y. x is going to sit between 0 and 1, like we were told. y, we can see, will range from 1 to e, which is perfectly acceptable. The radius is equal to y, so that's kind of nice, but we can also describe y in terms of x. So I'm not really leaning one way or the other. Do be aware though if we take a dx approach, then our ds will look like the square root of 1 plus y prime squared dx, which would look more like, because in this situation, y is e to the x, we would get 1 plus e to the 2x dx. So that's something to deal with. But notice if the radius is an e to the x, you'd have an e to the x with this together, and I think that actually looks really good for us. Let's take a look at that. If you integrate this with respect to x, you'll go from 0 to 1, 2 pi e to the x. Take the square root of 1 plus e to the 2x dx, or if you prefer, it might be better to look at this as e to the x squared dx. We'll see in a second why that actually might be a good choice. This is an integral that we could do. We could do this one because I think we could first start off with a u substitution, u to be e to the x. That way du equals e to the x dx, like so. And then as you switch from x to u, you'll go from 0 to 1, which you would get, let's see, when x is 0, or u would equal 1. When x is 1, you're going to get u is equal to e. And so you get the following integral. So if you do this u substitution, you'd end up with 1 to e to pi. And the e dx dx would come together just to be du, and you're going to get the square root of 1 plus u squared du. Now I want to mention to you that this integral we just set up using u, this was the integral we would get if we had set it up with respect to y. So the thing is, in this situation, if you try to set this thing up with respect to x or y, you're going to gravitate towards this integral anyways. Like I said, this is the integral you get when you set it up with respect to y. But this is okay because we can actually calculate this thing just fine. If we were to calculate this, well, we're going to have to do another substitution, this time a trigonometric substitution. If we set u to equal tangent theta, then du equals secant squared theta d theta, and then the square root of 1 plus u squared will equal secant theta. As we switch the bounds one more time, as you go from u to theta, when u equals 1, you want tangent, when is tangent equal to 1? This happens at pi force. And then the last one when u equals e, you're going to get arc tangent of e, which I'm just going to call that alpha for short as we go forward here. So making that substitution, you're going to get 2 pi integrate from pi force to alpha. You're going to get the square root of 1 plus u squared, which is a secant theta. 2 is going to become a secant squared theta d theta, which we've seen this puppy before. Secant cubed is not a super easy one to do, but we have done it before. So we're just going to make that statement right here. You're going to get 2 pi times 1 half times tangent theta secant theta. And then add to that the natural log of that value of secant theta plus tangent theta. And we're going to go from alpha to pi force. So plug in these things in here. Do notice that 2 pi times 1 half, of course, is just equal to pi. Plugging in alpha, we're going to get tangent of alpha. Remember alpha is the square, or alpha, like we saw before is arc tangent of e. So tangent of alpha is going to equal e. We're going to get secant of alpha. We're going to have to come back for that one. Plus the natural log of the absolute value of secant of alpha plus e. That's the first bit. Then we're going to subtract tangent of pi halves. Sorry, pi force. We already know that one is going to equal 1. So I'm just going to replace that with a 1 times secant of pi force. We'll have to come back to that one in just a second. And then we're also subtracting the natural log of secant of pi force plus 1. Now secant of pi force isn't so bad because after all secant of pi force is just going to equal 1 over cosine of pi force. Cosine of pi force, a lot of us remember is root 2 over 2. But this is the same thing as 1 over the square root of 2, which means secant of pi force is the square root of 2. So we can use that in these calculations above. Secant of alpha takes a little bit more effort, but not beyond our reach. If we think of the associated right triangle here, where the angle is associated to alpha. Since we know that tangent of alpha equals e over 1, we get e and 1. Then the other side will be the square root of 1 plus e squared. Secant is going to be the hypotenuse over the adjacent side. So we see that secant of alpha is the square root of 1 plus e squared, square root of 1 plus e squared, like so. And so putting this all together, we end up with pi times e times the square root of 1 plus e squared. Plus the natural log of e plus the square root of 1 plus e squared. I would love to say there's some nice simplification there, but not one that seems obvious. You get minus the square root of 2 and then minus the natural log of 1 plus the square root of 2, like so. In which case we then see that we get this solution, which is massively complicated. This becomes the square root of 22.940. And so we get this estimate right here, accurate to four decimal places. Now one thing I want to mention is that if you've been paying attention to our examples with surface area, with surface area, even though many of them are able to be calculated like we did this one, much easier than arc length, we still are usually ending up with these like massively complicated irrational numbers, which we have no appreciation of how big they are until we estimate them anyways. So since we're sort of ending the calculation within approximation anyways, it sort of begs the question, could we save some time and just estimated this using say Simpsons rule from the very get go. And in practice, that is again how one would often do these type of things, arc length and surface area. You're probably not going to have much benefit of doing this exactly because you want an approximation anyway. So what I'm saying is in practice, numerical approximation is typically sufficient for these type of situations. Now of course, with a homework question, it might want you to practice these integration techniques so an exact answer might be required. But be aware that again in practice, numerical approximations become very much king in situations like we're in right now.