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

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Here we've graphed the function y is equal to the square root of x. And we're going to create a solid of revolution, but we're not going to do it by rotating this around the x or the y-axis. Instead, we're going to rotate it around another somewhat arbitrary line. And in this case, I will rotate it around the line y is equal to 1. So let's say that this right over here is the line y equals 1. So this is what we're going to rotate it around, y equals 1. So the first thing we want to do is visualize what we're doing. And we actually care about the interval. So let's say that the interval is between this point right over here where the two points intersect. And let's say between that and x is equal to 4. So let's say that this right over here, so this is x equals 4. So it's going to be just right over here. So this is the interval that we're rotating, and we're going to rotate it around y equals 1, not around the x-axis. So what would our figure look like? Well, we're going to rotate it around like this. Going to rotate it around something like this. And so your figure is going to look, I guess you could call it, it will almost look like a cone head, if you view it on the side, or something like a bullet, but not quite looking like a bullet. But it would be a shape that looks something like that. So hopefully, we can visualize this. We've done this several times already, attempting to visualize these shapes. But let's think about how we could actually figure out the volume of this solid of revolution. Well, let's just think about it disk by disk. So construct a disk right over here. And we've done this many, many, many times already where we essentially want the volume of each of these disks. And then we're going to take this for each of these x's in our interval, and then we're going to take the sum of the volumes of all the disks to find the volume. We're going to stack them all up, or I guess put them next to each other in this case, and find the volume of our entire figure. So to figure out the volume of each disk, we just have to figure out the area of its face. And I'll do that in purple. We just have to figure out the area of the face and multiply it by the depth. So what's the area of this face? Well, it's going to be pi times the radius of the face squared. Now, what's the radius of the face here? Well, it's not just square root of x. Square root of x would tell us the distance between the x-axis and our function. It's now square root of x minus 1. This length right over here. The square root of x minus 1 for any given x in our interval. So it's going to be equal to square root of-- let me do that same orange color just to make it clear where I'm getting it from. It's going to be equal to square root of x minus 1. It's essentially our function minus what we are rotating around. That gives us the radius. And so this will give us the area of each of our faces. And then we just multiply that times the depth. The depth, of course, is dx. We've done that multiple times. So times dx. This is the thing that we want to sum up over our interval. And our interval-- let's see, this point right over here where the square root of x is equal to 1, that's just going to be x equals 1. This is just x equals 1 over here. And we said that we would do this all the way until x equals 4. This was kind of our end of our interval. So this is until x is equal to-- let me be careful. This is the x-axis right over here. All the way until x equals 4. So we get confused. This is the x-axis. So we're going from x equals 1 to x equals 4. And we're essentially taking this area, one way to think about it, between our square root of x and 1, and we're rotating it around 1. So we're using these little disks right over here. So the interval between 1 and 4. And so this is going to give us the volume of our solid of revolution. And so now we just have to evaluate this definite integral. So let's give a shot at it. So this is going to be equal to the integral from 1 to 4. And we can factor out-- or we could put the pi outside the integral sign. And then we can expand this out. So square root of x squared. Now, what we're going to do is expand this binomial so the square root of x minus 1 times the square root of x minus 1. Square root of x times square root of x is x. Square root of x times negative 1 is negative square root of x. Negative 1 times square root of x is another negative square root of x. And then negative 1 times negative 1 is equal to positive 1. So this part right over here will simplify to x minus 2 square roots of x. This is minus 2 square roots of x. And then you have plus 1. And then all of that times dx. So this is going to be equal to-- let's put our pi out there-- the antiderivative of this business. So that the antiderivative of x is x squared over 2. The antiderivative-- so minus 2 times the antiderivative of square root of x. Well, square root of x is just x to the 1/2 power, so we increment the power by 1. So it's going to be x to the 3/2 power times 2/3. So times 2/3. Or we could say it's going to be times 2/3x to the 3/2. Let me make it clear. This negative 2 right here is this negative 2. And this expression right over here is the antiderivative of square root of x. And you could verify this. If you take the derivative of this, 3/2 times 2/3 is 1. Decrement the 3/2, you get x to the 1/2. Now let's take the antiderivative of 1. Well, that's just going to be equal to x. And we're going to evaluate this from 1 to 4. So we're in the home stretch. So this is going to be equal to pi times-- well, first let's do it at 4. So you have 4 squared-- I'll write it all out. 4 squared over 2 minus-- what's 4 to the 3-- well, let me do this part first. So minus 4/3. I know it's 4 to the 3/2, so 4 to the 1/2 is 2, and then you raise that to the third power, you get 8. So times 8 plus 4. And then you subtract out all of this stuff evaluated at 1. So 1 squared over 2, so that's just going to be minus 1/2. And then we're subtracting this now. Actually, I don't want to skip too many steps. So it's going to be this. This is when we evaluate it at 4. And then we're going to subtract when we evaluate this whole thing at 1. So when we evaluate it at 1-- let me do this in this green color. That's not the green color I thought I was going to do. So when you evaluate it at 1, you get 1 squared over 2, which is 1/2. And then you get 1 to the 3/2, which is just 1. So this becomes minus 4/3. And then you have plus 1. And so let's simplify this. And I'll do it all in the same color now. This is going to be equal to pi times-- well, 4 squared over 2, that is 16/2, which is equal to 8. And then you have 4 times 8, which is 32, over 3. So minus 32/3, plus 4. And then you have minus 1/2-- we're just distributing the negative-- plus 4/3, and then you have minus 1. And now we just have to add up a bunch of fractions to simplify this thing. So what do we get? This is going to be equal to pi times-- let's see, our least common multiple looks like 6. We're going to put everything over a denominator of 6. And so 8 is the same thing as 48/6. 32/3 is the same thing as 64/6, so minus 64/6. 4 is the same thing as 24/6. 1/2 is the same thing as 3/6, so this is negative 3/6, the same thing as negative 1/2. 4/3 is the same thing as 8/6. And then negative 1 is the same thing as negative 6/6. A little bit of arithmetic here and let's see what we get. So if we take 48 and we subtract 64 from 48, we get negative 16. Is that right? Yes, we get negative 16. And then if we add 24 to negative 16, you get positive 8. Positive 8 minus 3 is 5. 5 plus 8 is 13. 13 minus 6 is 7. So this whole numerator simplifies to 7. So we get 7 pi over 6 as our volume. And let me just verify I did that right. So this is going to be negative 16. We get to positive 8, positive 8 plus positive 8 is 16. 16 minus 9 is-- yes, it is indeed 7. So we get 7 pi over 6, and we're done. We figured out our volume of this little sideways cone-looking thing.