 In this video, we're going to take a look at the arc of the parabola y equals x squared from 1, 1 to 2, 2, 4. And consider that arc being rotated around the y-axis. So we're going to rotate this thing around the y-axis. This gives us a surface of revolution, the illustration you can see in front of you. And this image is courtesy of James Stewart's calculus textbook here. We want to find the area of this curve right here. And so when we talk about the curve, we're describing this curve right here. We go from 1, 1 to 2, 4, 1, 1 right there. And so this thing is going to be rotated around the y-axis. And we want to find the surface area of this thing. So the general formula for the surface area is s equals the integral of 2 pi r ds. The ds can be adapted to integrate with respect to x or with respect to y. So we'll decide on what the best choice of ds is after we determine what is our best choice of r and also the bounds we want to consider. We can see that in this problem, x is going to range from 1 to 2. So if we integrate with respect to x, those will be the bounds we use. But we also see that y will range from 1 to 4 right here. So in terms of the bounds, we can describe the bounds in terms of x or with respect to y. So we have that freedom so far. We have the function y equals x squared, but we also have x equals the square root of y, which we could do that. So so far with the data I've collected so far, y equals x squared to me is much more preferable than x equals the square root of y. And so I'm leaning towards integrating with respect to x, but the real kicker is going to be the radius, right? What is the radius here? The radius is going to be the distance from the axis to the curve. And so taking a typical point on the curve right here, x comma y, the radius is going to be the number x, right? It's going to be the x coordinate. Now if I want to integrate with respect to y, if you take a dy approach, your r, which is x, would have to be given as the square root of y. On the other hand, if you take the dx approach, your radius, which is still x, it could just stay as x, right? And so that's going to be wonderful for us. Our radius would be the square root of x. Our function, because the ds is going to need to have a function there, you're going to have an x equals the square root of y. There's going to be a bunch of square root of y going on if you use the dy approach. If you use the dx approach, then you just have to have y equals x squared. You're going to have some monomials, no square roots. And so that's going to be the winner for us, winner, winner chicken dinner right there. So let's integrate this thing with respect to x. That means ds is going to look like the square root of one plus y prime squared dx. And let's set up our integral using that approach. Move it down here. So like the data we collected, we're going to integrate s. We're going to integrate with respect to x. So we want to go from one to two. We get two pi my radius. The radius here we found out was an x. And then we get the square root of one plus. Remember y was x squared, which means y prime is 2x. So we're going to get a four. Sorry, I'll have myself there. 2x squared dx. For which case, if we square the 2x, we're going to get a 4x there. One to two. Two pi x. Square root of one plus four x squared dx. Now, when we did this calculation, we did a very similar calculation this for arc length. We had to find the arc length of a parabola. That led itself to some type of tangent substitution. And you can notice here that because we have a square root of one plus 4x squared, this does seem to suggest that we do another 2x equals tangent theta substitution. But like I said before, because you have this radius that sits in front, there might be some alternatives like maybe integration of the parts or even better, u substitution. If we take u to be one plus 4x squared, then du would equal 8x dx, which we have a dx. We have an x. And all we need is an eight in front. We already have a two. So times that by four and then divide by four to compensate for that. We can change our integral to have the format pi force, the integral of the square root of u, so u to the one half du. That's a pretty nice integral right there. Let's also change the bounds. So as x ranges from one to two, u will range from plug these values into our function right here. When x is one, u will be five. When x is two, let's see, you get two square root of four. Four times four is 16 plus one is a 17, like so. So we'll range from five to 17. Anti-derivative of u to the one half. We're going to raise the power so we get u to the three halves, then divide by that power, which is multiplied by its reciprocal two thirds. We integrate from five to 17. And see, there's a little bit of cancellation with the coefficients. Two goes into four two times. So we're going to get a coefficient of pi sixth in the front. And then as we plug in the 17, we're going to get 17 to the three halves minus five to the three halves. There's no perfect squares there to simplify. So we get something like pi sixth times 17 root 17 minus five root five. No other simplifications we can use in this situation. We can approximate this if we want to, but we'll keep it as an exact answer right here. And so we can see that calculating surface area generally is much easier than arc length, even though it involves the arc length formula, because like we saw in this example, u substitution or other techniques of integration come into play that can dramatically simplify the calculation.