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## Integral Calculus

### Course: Integral Calculus > Unit 3

Lesson 12: Volume: washer method (revolving around other axes)- Washer method rotating around horizontal line (not x-axis), part 1
- Washer method rotating around horizontal line (not x-axis), part 2
- Washer method rotating around vertical line (not y-axis), part 1
- Washer method rotating around vertical line (not y-axis), part 2
- Washer method: revolving around other axes

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# Washer method rotating around horizontal line (not x-axis), part 2

Doing some hairy algebra and arithmetic to evaluate the definite integral from the last video. Created by Sal Khan.

## Want to join the conversation?

- I might sound silly but why cant we just sink the figure into water and measure how much water fell of?(14 votes)
- That is not always practical. And, it is not, in real world practice, completely accurate. If you are making a part for a high performance engine, you may have tolerances of a few microns -- much too small to try to measure displaced water. Also, if the object in question is porous, it will absorb some of the water, thus invalidating the method.(81 votes)

- How would you find the area of an equation/equations that was rotated around a line that wasn't horizontal or vertical?(5 votes)
- If you were to revolve the functions about y=-4, would you find the same volume?(3 votes)
- It would NOT be for two reasons. since the general formula is (top curve- bottom curve) you would subtract a negative 4 instead of a positive one.Also, the curve causing the gap between the axis of rotation would be different. in the video, the curve causing the gap, or empty space would be y=x, and the washer's height would be dictated by the other function. in your question, it would be the other way around. the integral set up would be the integral (from 0 to 3) of [(x+4)^2-(x^2-2x+4)].(3 votes)

- Could you do U-substitution or something instead of multiplying the polynomial?(2 votes)
- He could use the identity a^2-b^2=(a+b)(a-b)(4 votes)

- When you raise x to the fourth at a little before6:00and the result is 81, wouldn't that negative sign become positive? ie Yielding positive 81 in lieu of negative 81??(3 votes)
- It would become positive if we had (-x)^4, because then we'd be multiplying -x by itself four times, But in this video we have -x^4 instead. The way to evaluate that expression is to multiply x by itself four times and then applying the minus sign. So when x = 3 we get -(3*3*3*3) which is -81.(2 votes)

- Is the same if you put the 4 at the end of each equation instead that at the beginning?

for example: x^2-2x-4 instead of 4-x^2+2x(1 vote)- Sal did put the polynomial into standard form before he squared it.

His order was -x² + 2x + 4

Yes, the result is exactly the same when you square x²- 2x- 4 and compare with the square of 4 - x²+ 2x and the square of -x² + 2x + 4. This is also true if you change the signs of 4 - x before squaring.

For many purposes, you can switch the signs of the coefficients as long as you switch all of them. You can solve for x, for example after switching all of the signs. You will get the same answer when you square the polynomials after switching the signs:

(-x² +2x +4)² is x⁴ - 4x³ - 4x² + 16x + 16

and

(x²- 2x- 4)² is x⁴ - 4x³ - 4x² + 16x + 16

likewise (4 - x)² = (x - 4)² = x² - 8x + 16

But when finding integrals, you should keep the signs the way they were.

∫ (-x² + 2x + 4) dx

= ∫ -x² dx + ∫ 2x dx + ∫ 4 dx

= - ⅓ x³ + x² + 4x

If you evaluate the integral between 0 and 4, the result is -64/3 + 32 = 32/3

Meanwhile:

∫ (x² - 2x - 4) dx

= ∫ x² dx - ∫ 2x dx - ∫ 4 dx

= ⅓ x³ - x² - 4x

If you evaluate the integral between 0 and 4, the result is 64/3 - 32= -32/3, the exact opposite of the correct answer.

Also, make sure you do not try to graph after switching the signs: the shape will be flipped.(5 votes)

- For this example why did we not have to account for the area under the x axis formed by the curve y= x^2 -2x in the definite integral. From my understanding that area is negative and we have to take the absolute value of that interval after finding the rest of the area. Does it not matter for the washer/ disk method?(2 votes)
- when subtracting the 4-x^2+2x..... shouldn't it be 4-(x^2+2x) and the negative sign is distributed? Can you please explain this- thank you!(2 votes)
- No because when you remove the negative from (-x^2) and (2x), you would get (x^2) and (-2x), so you would get 4-x^2+2x=4-(x^2-2x). You are distributing the negative sign to both values in the now separated polynomial by making the second value negative.(1 vote)

- What is being done when substracting the function from 4 isn't basically working with the figure on the y=0 axis? Please correct me if wrong, thank you.(1 vote)
- For the most part, just ignore the x-axis. The act of subtracting either function from 4 gives you the actual magnitude of the radius for both the inner and outer circle.

Take y = x for example as the radius of the inner circle. At x = 1, the y value is 1. The axis of rotation is y = 4 so the y-value is 4. The radius of the inner circle is the difference between these two values all the way around. 4 - 1 = 3 at that value. At x = 2, the y-value of the function is 2 and so the radius is 4 - 2 = 2.

The same thing applies to y = x^2 - 2x. Subtracting the value of this function from 4 gives you the magnitude of the radius of the outer circle. At x = 1, the y-value is -1. Since we don't care about negative values in this application because we want positive (real) volume of solids, subtracting from 4 gives us the magnitude of this distance. So 4 - -1 = 5. It's a distance of 4 from y = 4 to y = 0 plus the additional distance from y = 0 to y = -1. That total distance gives us a radius of 5.

If you want to think about the concept in general, just forget about the y and x axes and think of y = 4 as the only axis. Subtracting from that starting point will give you the difference from the starting point to the function. That difference is what we care about since that represents the radius that we are rotating around the origin ( y = 4).(3 votes)

- I'm curious, to get your inner and outer radius

I evaluated my outer radius as (f(x)-4)^2 as opposed to (4-f(x))^2

Because the way I understood it was, if you're above the x-axis, you are some value less than the x-axis (so subtract 4) and if you're below the x-axis you're some value greater than the x-axis (so add the value).

Is this thinking/logic incorrect?(1 vote)- You might wanna give the previous video another watch. That's where he explains why it is 4 - f(x) and not f(x) - 4.

However, I will say this. You can cheat at this a bit and use f(x) - 4. Why? It's because at the end of the day, you're gonna have to square it, and when you square, what order you subtract doesn't matter. So, f(x) - 4 and 4 - f(x), on squaring, return the same result.

Now, if you were to find the surface area, this would be an issue, as you'd use 2(pi)r and not (pi)r^(2). But, as long as it is volume, using either would give the same answer.

Also, there's another thing you could do. You could translate all functions down by 4 units. That way, your "axis of rotation isn't the x axis" becomes "the axis is now the x axis". That way, it becomes simpler, plus your functions are already adjusted accordingly.(2 votes)

## Video transcript

Where we left off
in the last video, we'd actually set up
our definite integral to figure out the
volume of this figure. So now we just have
to evaluate it. And really the
hardest part is going to be simplifying that
and that right over there. So let's get to it. So what is this thing squared? Well, looks like
we're going to have to do a little bit of
polynomial multiplication here. So I'll go into that
same color I had. So we're going to have 4 minus x
squared plus 2x times 4 minus-- actually, let me write that
in the order of the terms, or the degree of the terms. It's negative x squared
plus 2x plus 4-- I just switched the order of
these things-- times negative x squared plus 2x plus 4. So we're just going to
multiply these two things. I shouldn't even write
a multiplication symbol, it looks too much like an x. So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared
is negative 4x squared. 2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times
negative x squared is negative 2x to
the third power. And now we just have to multiply
negative x times all this. So negative-- or
negative x squared. Negative x squared times
4 is negative 4x squared. Negative x squared times 2x is
negative 2x to the third power. And then negative
x squared times negative x squared is
positive x to the fourth. So it's going to be
positive x to the fourth. And now we just have to
add up all of these terms. And we get, let's see,
we get x to the fourth-- add these two-- minus
4x to the third. And then this cancels
with this, but we still have that, so minus 4x squared. You add these two right
over here, you get plus 16x. And then you have plus 16. So that's this expanded out. And now if we want to-- 4
minus x times 4 minus x. So if we just have 4
minus x times 4 minus x, we could actually do
it this way as well. But that's just going
to be 4 times 4, which is 16, plus 4 times negative
x, which is negative 4x, negative x times 4,
another negative 4x, and then negative x times
negative x is equal to plus x squared. So if we were to
swap the order, we get x squared minus 8x plus 16. But we're going
to subtract this. So if you have the
negative sign out there, we're going to
subtract this business. So let's just do
it right over here since we already have it set up. So we have this minus this. We're going to
subtract this, or we could add the negative of it. So we'll put negative x
squared plus 8x minus 16. So I'm just going to add
the negative of this. And we get-- and I'll
do this in a new color-- let's see, we get x
to the fourth minus 4x to the third power minus
5x squared plus 24x, and then these cancel out. So that's what we are left with. And so that's the
inside of our integral. So we're going to take the
integral of this thing, just so I don't have to
keep rewriting this thing, from 0 until, if I
remember correctly, 3. Yup, from 0 to 3, of this, dx. And then we had a
pi out front here. So I'll just take that
out of the integral. So times pi. Well, now we just to
take the antiderivative. And this is going to be equal
to pi times antiderivative of x to the fourth is
x to the fifth over 5. Antiderivative of
4x to the third is actually x to the fourth. So this is going to be
minus x to the fourth-- you can verify that. Derivative is 4x, and
then you decrement the exponent, 4x to the third. So that works out. And then the antiderivative
of this is negative 5/3 x to the third. Just incremented the
exponent and divided by that. And then you have plus 24x
squared over 2, or 12x squared. And we're going
to evaluate that. I like to match colors
for my opening and closing parentheses. We're going to evaluate
that-- actually, let me do it as brackets. We're going to evaluate
that from 0 to 3. So this is going to be
equal to pi times-- let's evaluate all this business at 3. So we're going to get
3 to the fifth power. So let's see, 3 to the
third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be
equal to-- this times 3 is 243. So it's going to
be 243-- it's going to be some hairy math-- 243 over
5 minus-- well, 3 to the fourth is 81. Minus 81. Minus, let's see, we're going
to have 3 to the third times-- let's see, so it's going
to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. So let's just simplify that. So this is going
to be equal to 45. Did I do that right? Yeah. It's essentially going to
be like 3 squared times 5, because you're
dividing by 3 here. So it's going to have 45. And then finally,
3 squared is 9. 9 times 12 is 108. So plus 108. These problems that
involve hairy arithmetic are always the most
stressful for me, but I'll try not to
get too stressed. And then we're going to subtract
out this whole thing evaluated at 0, but lucky for us,
that's pretty simple. This evaluated is 0, 0, 0, 0, 0. So we're going to
subtract out 0, which simplifies
things a good bit. So now we just have to do some
hairy fraction arithmetic. So let's do it. So what I'm going to do first
is simplify all of this part. And then I'm going to
worry about putting it over a denominator of 5. So let's see what we have. We have negative 81 minus 45. So these two right over
here become negative 126. And then negative 126 plus 108. Well, that's just going to be
the same thing as negative 26 plus 8, which is going
to be negative 18. So this whole thing
simplifies to negative 18. Did I do that right? So this is negative 126. And then negative 126-- yes. It will be negative 18. So now we just have to
write negative 18 over 5. Negative 18 over 5 is the same
thing as negative-- let's see, 5 times 10 is 50, plus 40. So that's going to be
negative 90 over 5. So this whole thing
has simplified to-- is equal to pi
times 243 over 5 minus 90 over 5, which is
equal to pi-- guys, I deserve a drum roll now. This was some hairy mathematics. So 243 minus 90 is
going to be 153 over 5. Or we can write this
as 153 pi over 5. And after all that
math, you sometimes forget what we were
doing in the first place. We were figuring out
this right over here is the volume of
this figure that had this little inside
of it bored out.