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# Skallmetoden for rotasjon rundt en loddrett linje

## Video transkripsjon

I've got the function y is
equal to x minus 3 squared times x minus 1. And what I want to do is
think about rotating the part of this function that
sits right over here between x is equal
to 1 and x equals 3. And x equals 3 and x
equals 1 are clearly the zeroes of this
function right over here. And I want to take this
region and rotate it around the y-axis. And if I did that,
I'd get a shape that looks something like that. And I want to figure out
the volume of that shape. And what we're going
to do is a new method called the shell method. And the reason we're going
to use the shell method-- you might say, hey,
in the past, we've rotated things around
a vertical line before. We used the disk method. We wrote everything
as a function of y, et cetera, et cetera. We created all of these disks. We figured out the volume
of each of those disks. But the problem here is
this is hard to express as a function of y. How do you solve explicitly
for y right over here? So instead, we're going to
keep things in terms of x and have a different geometric
visualization for how we can come up with the volume. What we're going to
imagine instead-- instead of constructing disks, we're
going to construct shells. And what do I mean by a shell? So for each x at the interval,
on this kind of cut of it, we can construct a rectangle. And what happens if we were
to rotate this rectangle? So this is the rectangle
right over here. What happens if we rotate this
rectangle around the y-axis along with everything else? I'll try my best
attempt to draw it. It's going to look
something like this. This is challenging
my art skills, but I think I can handle it. So it's going to look
something not too dissimilar to that
right over there. So it looks like a
hollowed-out cylinder. I guess that's why
we call it a shell. And it's going to
have some depth. The depth is going to be dx. And the height
right over here is going to be the
value of my function. The height is f of x. In this case, f of x is x minus
3 squared times x minus 1. How do we figure out the
volume of a cylinder like this? Well, if we can figure out the
circumference of the cylinder, and then multiply that
circumference times the height of the cylinder,
we'd essentially figure out the area
of the outside surface of our cylinder. And then if we multiply the
area of the outside surface of our cylinder by that
infinitesimally small depth, then that'll give
us the volume-- I shouldn't say
cylinder-- of our shell. So let's try to do it. What is the
circumference of a shell? What is the circumference
of one shell going to be? Well, it's going to be 2 pi
times the radius of that shell. We need to express this
as a function of x. And so what is that going to be? It's going to be 2 pi. So for a given x,
what is the radius? Well, the radius right over here
is just the horizontal distance between the y-axis and that x. So that's just x. So the circumference,
in this case, is just going to
be 2 pi times x. Now, what is the height going to
be for any one of those shells? The height is
going to be f of x. This is right over here. And so what is going to be the
surface area of the outside? So let me put this in quotes--
"outside" surface area. I'm not worried about the
depth right now, the dx. I'm not worried about this
top part and the bottom part. I'm just worried about
the outside surface area. Well, the outside
surface area is just going to be the circumference
times the height. It's going to be 2
pi x times f of x. And in this situation,
in the situation we're looking at
right over here, that's going to be 2 pi x
times x minus 3 squared times x minus 1. Now, what's going
to be the volume? So the volume of the
shell-- shell volume-- is just going to be all
this business times dx. So it's going to be 2
pi x times f of x dx. And so now we're ready to
integrate over the interval. So the volume of
our entire shape is going to be the
definite integral. We're going to integrate over
all the x's in the interval, from x is equal to 1 to x
is equal to 3 of this thing. And we could take
the 2 pi out front. So we'll put 2 pi out front. And on the inside,
we have x times f of x, which in our
situation, is this business. So it's going to be x times
x minus 3 squared times x minus 1. And then, of course,
we have our dx. So there you have it. Using the shell method, we have
set up our definite integral for the volume of this
strange-looking shape right over there.