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## Integral Calculus

### Course: Integral Calculus > Unit 3

Lesson 10: Volume: disc method (revolving around other axes)# Disc method rotation around horizontal line

AP.CALC:

CHA‑5 (EU)

, CHA‑5.C (LO)

, CHA‑5.C.2 (EK)

Solid of revolution constructing by rotating around line that is not an axis. Created by Sal Khan.

## Want to join the conversation?

- if f(x) is the outermost function and g(x) is the innermost function, shouldn't it be (f(x))^2 - (g(x))^2 instead of [(f(x)) - (g(x))]^2 like in this video?(14 votes)
- Good question.

However, he is not subtracting the volume of the solid created if you rotate y=1 around the x-axis from the volume of the solid created if you rotate y=√x around the x-axis, but rather rotating y=√x around y=1. Thus, this is not about outermost and innermost functions, but rather about shifting the function so that we are rotating it around the x-axis but getting the same volume as if we were rotating it around y=1.

I hope this clarifies the problem!(26 votes)

- Isn't the washer method supposed to work here too because we have two functions?

I keep trying to use the washer method and it gives a difference answer.(6 votes)- No the washer method wouldn't work because there is no 'opening' in the middle of the revolution. Had it been revolved around the y-axis, then you would have to use the washer method.(8 votes)

- If the equation was
`y = lnx`

instead of`y = √x`

, and I was rotating around of`y = -1`

instead of`y = 1`

, I presume that each washer would equate to`π(lnx + 1)`

instead of - 1. Is that presumption accurate? See2:25.(3 votes)- That's very close. Remember, we are finding the volume, so we will need to find the area of the circle. Since the radius is ln(x)+1, the area is π(ln(x) + 1)^2. Also, to find the volume, we will need to multiply the area of the circle by the width of each washer, which is dx. The dx is important. With dx, you will summing up an infinite number of finite values, which is infinite, but with dx, you will be summing up an infinite number of infinitesimal values, which is finite.

I hope this helps!(8 votes)

- What would the radius be if the function was being rotated around y = 5?(3 votes)
- The expression √x -5 gives your new radius, because the radius is still just the distance between the axis of rotation and the function

The full integral would be v = ∫π(√x - 5)^2dx, on whatever definite interval you want to take the volume(4 votes)

- how do you know whether you're supposed to subract/add the line rotated around to the radius?(2 votes)
- You would subtract the line if it is below the function, or subtract the rotated function if it is below the line. This is because the radius of the solid at any given x value is the
*distance*between the rotated function curve and the line it is being rotated around. Distance technically is found by taking the absolute value of their difference, however assuming the the two never cross inside the section we are taking the volume of, we can just pick the one that is lowest and subtract it from the other.(4 votes)

- At3:45, how do we know that the interval is from 1 to 4?(1 vote)
- The interval comes from the setup of the problem in the first 45 seconds of the video. We're working with a figure that begins at the point where the curve y = sqrt x crosses the line y = 1, and that occurs when x = 1. The figure ends at x = 4, which is just an arbitrary ending point that Sal states early in the video.(3 votes)

- Hello, Calculus II student here. Some of these easier integration problems take me 5-10 minutes especially when I write out every step. My question is: Is this normal? As in, do most average-above average students write EVERYTHING out? And by everything, I mean how Sal does it.(2 votes)
- Why don't you use U-substitution for the integral(1 vote)
- If we let u = sqrt(x) - 1, then the derivative of u would be 1/2*sqrt(x), which does not appear in the expression that we're integrating; therefore u substitution is not the best method when we could take a much more feasible option and multiply out the expression that we're integrating.(3 votes)

- What about the x interval [0,1]? The problem asks us to rotate the (whole?) function around the given axis, so i thought there should be something happening in that interval.

Isn't there a sort of a cone with the head pointing to the right? Shouldn't we add the volume of this smaller cone to the one Sal found?(1 vote)- Sal didn't say what volume we were computing too explicitly, so we can just assume that the volume he cared about didn't include the part above [0, 1]. But you're correct that there would be another figure there, shaped sort of like the mouth of a trumpet and coming to a point at (1,1).(2 votes)

- For the radius of the first circle discussed, would it make sense to think of the radius as [(x)^(1/2)]/2? Since the radius of a regular disc seems to be equivalent to the diameter of this disc. I want to know why that is wrong(since I got a different answer) considering my line of thought is logical. And another thing that perplexed me is how when I just manipulated the original function with phase shifts to form [(x-1)^(1/2)]+1 and used as as the radius to integrate in the interval of (1,4), I still didn't get similar results as Sal. I will really appreciate it if someone out there answers my questions. I hope you understand me well. If I seem to be incomprehensible please let me know what I should elucidate. Best(1 vote)
- These are much more difficult to answer with words than pictures but I'll do my best. To put simply though, the bit of the function that Sal integrated and the 2 that you did in your 2 questions are 3 different shapes.

1: Perhaps it seems that way because of the place he chose to draw his disc. I assume by "regular disc," you mean one that has been rotated around y=0 (the x-axis.) The diameter of the disc whose "circle part" is centered on y=1 is only the same as the radius of a disc centered on y=0 at x=4 (in this example.) Think about moving to a different spot along the function. I'll arbitrarily choose x=25. The function y=√(x) will have a y value of 5. So the radius when it's rotated around y=1 will be 4 (5-1) making the diameter 8. The radius if the function is rotated around the x-axis is 5 (5-0). Of course, 4 ≠ 5.

2: You shifted the whole function up 1 and to the right 1. We already went over going up 1 in the first answer. Unless I'm drunk, with just the shift up 1, rotating around y=1 should get you the same answer as rotating the un-shifted function around y=0 so the reasoning from before still stands. With the shift to the right, think of it as adding on the little bit at the beginning ( x=0 to x=1) that Sal didn't include as part of his problem but removing the (larger) bit at the other end (x=3 to x=4.) The shape moved but the boundaries of integration didn't so you're integrating a different part of the overall shape. The problems caused by each shift are unrelated and so they don't cancel each other out if that's what you were thinking.(1 vote)

## 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.