 In this video, we provide the solution to question number eight for practice exam number two for math 1220, in which case we're asked to set up and simplify the integral that will compute the surface area generated by the surface formed by revolving the arc y equals the square root of 1 minus x squared from x equals 0 to x equals 1 around the axis y equals 1. Okay, so this is a surface area type problem. Let's make sure we have an understanding of what this thing is doing here. So the curve right there, y equals the square root of 1 minus x squared, we can very quickly visualize that this function is actually the graph of a semicircle of radius one, but we're only going from zero to one. Let me actually just put in the portion that we're interested in here. So we get this just a quarter of the circle like so. And we're supposed to rotate this around the axis y equals one. So that's going to come right here. And so we're going to take this this arc and rotate it around the line y equals one. And so you'll get some type of like funnel like shape surface there, because you'll get like some other part over here, again, spin that thing around. And so we're trying to look for the surface area of that region. So it'll be helpful for us to consider what is our typical cross section, something like that you get a diagonal like so. It is like a trapezoid. We're going to integrate this thing with respect to ds. So remember when it comes to the surface area, the basic formula, recall is a two pi times the integral of your radius, the radius of your rotation there ds, for which with the ds you do have the option to integrate it with respect to x or y. And so we can choose which one ever seems most advantageous to us as the bounds are given in terms of the x coordinates and also the function is a function with input variable x. It seems like integrating with respect to x will probably be the best choice here, but you actually could integrate this with respect to y. You do have that option here. So let's let's first consider doing with respect to x. Now if you're going to do with respect to x here, you're going to of course replace. Well, let's actually do with this radius here first, because the radius is then this distance right here. How far are you from the axis? And so if you take a typical point on the curve here, x comma y, this here is y equals one, this is y. So the distance would be one minus y. And so that's important to note here. If we take that radius, since we're going to integrate with respect to x, we'll go from zero to one here. Our radius is one minus y. We'll come back to that one in a second. And then we have to take the square root for the ds here. We take the square root of one plus y prime squared dx. So we're going to make some, we're going to have to transition some things over here. For example, we need this to be an x, but we have a formula here. That's not so bad. We're just going to substitute that in there. We do need to calculate the derivative of our y function. That is this one right here. So let's do that to the side right here, y prime. Since it's a square root, when you take the derivative of square root, that actually puts the square root in the denominator. And then you get a coefficient of two in there as well. That's just a consequence of the power rule. But then by the chain rule, we have to take the inner derivative, the derivative of one minus x squared, which is negative two x. The two's cancel. And then of course, when you square it, it won't make much of a difference in the negative sign either. So we're going to plug that in right here. Doing so, we end up with, oh, I forgot my two pi from before. So we're going to have two pi times the integral here. So this thing, I'm going to leave this as a one minus y for just a little bit longer. This thing right here, you have one plus, then you're going to have to fit your whole square root in there. Excuse me, the inside the square root, you have to fit the y prime. We're going to square that. So the numerator when you square it is going to become an x squared. The denominator, since it's a self-square root, when you square it, you're going to get one minus x squared. There is a dx, of course, over here as well. And so then with that in play, let's try to simplify this thing. And the instructions do require we try to simplify this. So for example, inside the square root, I'm going to find a common denominator. So we get two pi times the integral from zero to one. Again, I'm leaving it as one minus y. This is mostly for spatial purposes at the moment. That one is going to become a one minus x squared. We have an x squared in there and the denominator becomes one minus x squared. So for which then in the numerator, you'll notice these x squareds cancel out, which is kind of nice. The numerator then becomes just one. So you take the square root of one, which is just one. So when you rewrite that one more time, two pi integral from zero to one, you're going to have in the denominator, the square root of one minus x squared. That's nice dx. Now in the numerator, we have the one minus y. So let's consider that for a moment. You get one minus the square root of one minus x squared, like so. And so with this one, I notice that because we have the square root of one minus x squared here and here, if I break it up into two fractions, I could simplify. I mean, you could insert this into the answer right here. I'd say that's sufficiently simplified, but I'll do one more step here. The surface area is going to be two pi times the integral from zero to one, for which if you take the first fraction, one over the square root of one minus x squared, the second fraction will just be a negative one. It just simplifies like that. So if you were to integrate this thing with respect to x, you would get the following expression right here. It gives you your answer. Now, if you want to integrate this with respect to y, one thing we have to do is we have to solve this equation for x, for which honestly, if you go through the algebra that's square both sides, move some things around, you're going to end up with x equals the square root of one minus y squared. Kind of a nice symmetry of the circle. You get the exact same equation to represent this exact same curve. And notice with regard to the y-coordinate, you're going to go from y equals zero up to y equals one. And so then when we put that into our integral there, two pi times the integral from zero to one here, our radius is still going to be one minus y, so you get one minus y like so. Then we have to take the square root of one plus dx over dy squared dy, but be aware that because of the exact same formula that we had before, by basically the exact same calculation, if I were to look at dx over dy, dy there, you're going to still end up with a y this time over the square root of one minus y squared. And then when you insert that into the square root, it'll simplify in the exact same manner that we saw before, but now we just have y's instead of x's there. So that's going to simplify to be the square root of one minus y squared dy. And honestly that would then be the answer that you're looking for. You can put that up, put this up here in the box, put that box on the line or something like that. That would give us the correct antiderivative status to say the correct integral there. And so if you wanted to do that instead using y as opposed to x, you could do that. There's something a little bit nice about it. But at the same time, the hiccup is that this thing was set up for x. If you wanted to in terms of y, you'd have to solve for x, which really wasn't too painful in this situation.