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## Volume with disc method: revolving around other axes

Current time:0:00Total duration:9:54

# Disc method rotation around horizontal line

AP Calc: CHA‑5 (EU), CHA‑5.C (LO), CHA‑5.C.2 (EK)

## Video transcript

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.

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