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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.