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Studying for a test? Prepare with these 8 lessons on Applications of definite integrals.
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Disk method rotating around vertical line

Video transcript
Let's do another example, and this time we're going to rotate our function around a vertical line that is not the y-axis. And if we do that-- so we're going to rotate y is equal to x squared minus 1-- or at least this part of it-- we're going to rotate it around the vertical line x is equal to negative 2. And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disc method. So what I want to do is construct some discs. So that's one of the discs right over here. It's going to have some depth, and that depth is going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disc is going to be the area as a function of y times the depth of the disc times dy. And then we just have to integrate it over the interval that we care about, and we're doing it all in terms of y. And in this case, we're going to integrate from y is equal to-- well, this is going to hit-- this y-intercept right over here is y is equal to negative 1. And let's go all the way to y is equal to, let's say y is equal to 3 right over here. So from y equal negative 1 to y equals 3. And that's going to give us the volume of our upside-down gumdrop-type-looking thing. So the key here, so that we can start evaluating the double integral, is to just figure out what the area of each of these discs are as a function of y. And we know that area as a function of y is just going to be pi times radius as a function of y squared. So the real key is, what is the radius as a function of y for any one of these y's? So what is the radius as a function of y? So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. If you add 1 to both sides-- and I'm going to swap sides, so you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And you get x is equal to the principal root of the square root of y plus 1. So this we can write as x-- or we can even write it as f of y if we want-- f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance here at any point? Well, this distance-- let me make it very clear. So it's going to be our total distance in the horizontal direction. So this first part as we're-- and I'm going to do it in a different color so we can see. So this part right over here is just going to be the value of the function. It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius as a function of y is going to be equal to the square root of y plus 1. This essentially will give you one of these x values when you're sitting on this curve, this x as a function of y, it'll give you one of these x values. And then from that, you add another 2, so plus 2. Another way of thinking about it, you get an x value here, and from that x value you subtract out x is equal to negative 2. And when you subtract x is equal to negative 2, you're adding 2 here. But hopefully, this makes intuitive sense. This is the x value-- let me do this in a better color-- this right over here, this distance right over here, is the x value you get when you just evaluate the function of y. But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation. Once again, if you just take a given y right over there, you evaluate the y, you get an x value. That x value will just give you this distance. If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius. So our radius as a function of y is this thing right over here. So substituting back into this, we can now write our definite integral for our volume. The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy. So I can write the pi out here-- we've done this multiple times-- times radius squared-- so it's going to be square root of y plus 1 plus 2 squared-- that's our radius-- times dy. So we've set up the definite integral. And now we just have to evaluate this thing. And I'll save that for the next video. And I encourage you to try this out on your own.