AP®︎ Calculus AB (2017 edition)
- Shell method for rotating around vertical line
- Evaluating integral for shell method example
- Shell method for rotating around horizontal line
- Shell method with two functions of x
- Calculating integral with shell method
- Shell method with two functions of y
- Part 2 of shell method with 2 functions of y
- Shell method worksheet
- Shell method
Shell method with two functions of y
Stepping it up a notch, our solid is now defined in terms of two separate functions. Created by Sal Khan.
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- 5:40ishh why is the lenghth? of the the slice bottom function minus top function?? (y+1-(y-1)^2. Everything up to this point makes sense but now Im lost. . . I thought you do top minus bottom/ right most - left most(21 votes)
- Tilt your head 90 degrees to the right. Which is upper function now?
If we calculate in terms of y then the function output is x, and x=y+1 returns higher x'es then x=(y-1)^2, thus x=y+1 is the top function.(24 votes)
- at3:05, how is it y+2 for the radius?(11 votes)
- In other words, it is because the shape is not rotated around the x axis but rather around a line that is two units below the x-axis. Therefore to get the radius, you have to add those two units to the value of y, so, (y + 2) is the radius(7 votes)
- After the5:00mark, what does he do (y+1 - (y-1)^2) instead of the other way around? is x=(y-1)^2 not the larger radius, and he should therefore subract the smaller y+1 function to get the area between the two?(7 votes)
- We are integrating with respect to y, so the functions are in terms of y. To visualize this, tilt your head to the right (as Sal said) and look at the graph. Then you'll see y+1 is on top while (y-1)^2 is on the bottom. So in this case, y+1 is the larger radius.(6 votes)
- Is the radius always positive? Like, what if the axis of rotation was y=2 instead of y=-2? Would the radius still be y+2?(5 votes)
- A radius is a physical length, so it can't really have a negative value. No matter what the axis of rotation is, the radius will always be positive. :)(1 vote)
- So disk method is using area and shell method is using circumference ?(4 votes)
- 2:52, he denotes the distance from the x axis to the yellow function as y, but why does he use the upper function? Wouldn't there technically be two y's, one to the yellow and one to the blue function (and I realize that the yellow curve isn't an actual function, but I'm just calling it that for the sake of simplicity).(3 votes)
- Good question! It does look like he is defining y as the distance from the x-axis to the yellow function because of where he draws the vertical purple segment, but really it is giving the distance to the entire thin horizontal pink rectangle, not just to the point on the yellow curve. He could (and for clarity, he probably should) have drawn the purple segments a bit further to the right, to the middle of the thin rectangle.
Because the thin slices run horizontal, the y-values of the two curves are not what get subtracted, but instead we need to subtract the x-values of the curves. That's what is happening around4:52, when he talks about the "upper function" and the "lower function". Like you said, the yellow curve is NOT a function when we think of y as a function of x, but it IS a function if we consider its x-value as a function of y. I don't think he should refer to them as "upper" and "lower". Instead, I think he should call the blue curve the "right" curve and the yellow curve the "left" curve. So, to get the length of the thin rectangular slices, he does the x-value of the RIGHT curve (x = y + 1) minus the x-value of the LEFT curve (x = (y-1)²).(1 vote)
- At4:15- Out of curiosity, if I swapped the upper and lower functions, would it just result in a negative volume?(3 votes)
- I'm guessing it would be like putting an upper bound on the bottom of the integral which results in putting a negative sign infront of the integral to make the upper bound go on top; so I think basically yes(1 vote)
- Why is the thickness dy?
Shouldn't the thickness of shell be dx?(3 votes)
- dy is infinitely small
this is the shell method for functions of y, so we use dy here(1 vote)
- How could I solve this using disk method? Thanks:)
Here is what I written ,but I typed that into a calculator the answer wasn't correct
sum(0,4) pi (sqrt(x)+3)^2 dx
sum(0,4) pi (x+1)^2 dx
sum(0,1) pi 2^2 dx
sum(0,1) pi (x+1)^2 dx(1 vote)
- I'm not entirely sure how you built your integrals, but it looks like there might be one major problem keeping you from the right answer. There's a change at x=1 where the lower function changes, so you'll have to adjust your integrands and limits of integration to reflect that. This should work:
For x on the interval (0,1), the upper function is
y=sqrt(x)+1and the lower function is
y=-sqrt(x)+1, so the volume for this section (rotated about y=-2) will be given by
int(0,1) [ pi*(sqrt(x)+3)^2 - pi*(-sqrt(x)+3)^2 ] dx.
For x on the interval (1,4), the upper function is still
y=sqrt(x)+1, but the lower function is now
y=x-1, so the volume for this section will be given by
int(1,4) [ pi*(sqrt(x)+3)^2 - pi*(x+1)^2 ] dx.
The total volume will be the sum of those two integrals. Hope that helps.(3 votes)
- what is the answer to this integral? is it 63(pi)/2 ? thanks!(1 vote)
I'm going to take the region in between the two curves here, between the yellow curve-- defined as a function of y as x is equal to y minus 1 squared-- and this bluish-green looking line-- where y is equal to x minus 1. So I'm going to take this region right over here, and I'm going to rotate it around the line, y equals negative 2, to get this shape that looks like the front of a jet engine or something like that. And we're going to want to figure out what its volume is. And you can actually approach this with either the disk method or the shell method. In the disk method, you would create disks that look like this. And you would be doing integrating with respect to x. You will have to break up the problem appropriately, because you have a different lower boundary. You would have to break this up into two functions, an upper function and a lower boundary for this interval in x. And then a different one for that interval in x. But you could use the disk method. But instead of that, we don't feel like breaking up the functions and doing all of that. So we're going to do it in the shell method, especially because we've already expressed one of our functions as a function of y. And this one won't be too hard to do. So what we're going to do, once again, is imagine constructing these little rectangles that have height dy. And what we're going to do is rotate those rectangles around the line y equals negative 2. So let me draw that same rectangle over here. And when you do that, you construct a shell. So let me do that. So, go between these two points. And then this is what it would look like when it's down here. And then, let me make it clear that this constructs a shell of thickness dy or of depth dy. So let me do it like that. So that's my shell, and it has some thickness to it. And that thickness is dy. So that's the thickness of my shell. And let me shade it in a little bit. Make sure you can see the 3-dimensionality of my shell. So like we've done with all of these problems, our goal is to really just figure out what the volume of each shell is. And then we can enter it for a specific y in our interval. And then we integrate along all the y's of our interval. And we've done this many times before. The first thing we think about is what's the radius one of these shells? So what I'm doing right here in magenta, what is the radius of something like that? Well, it's essentially going to be the distance between y is equal to negative 2 and our y value for that specific y. So this distance right over here is y. And then we're going to have another 2. So the whole distance is going to be y plus 2. Another way of thinking about it is this is essentially y minus negative 2 to get the distance, which is going to be y plus 2. So the radius of one of our little shells is going to be y plus 2. If the radius is y plus 2, then we know that the circumference of this circle right over here is going to be 2 pi times y plus 2. And then the surface area, the outside surface area, of the shell, the stuff out here, is just going to be that circumference times I guess you could say the width of this shell times this distance right over here. Or we could say times this distance right over here. And what is that distance? Remember, we want everything expressed as a function of y. Well, it's going to be the upper function as a function of y minus the lower function. And we think about the upper function, it's the function that's giving us higher x values in that interval. And so this blue function is the upper function when we think in terms of y. But we have to express it as a function of y. So let me do that. So we can rewrite this as add 1 to both sides, you get x is equal to y plus 1. So that is our upper function. And then this is our lower function. If you were to tilt your head to the right and look at it that way, you'll see that this will be the upper function, and this will be the lower function for the same value of y. This gives a higher value of x than this one does for the same value of y. It gives the upper x values. So, the area is going to be the circumference times this dimension. Let me write this. The area of one of those shells is going to be 2 pi times y plus 2 times the distance between the upper function. So the distance between the upper function y plus 1, x is equal to y plus 1, and the lower function, x is equal to y minus 1 squared. I'll put the parentheses in that same color. And then if we want the volume of that shell, so we've got the outside surface area of the shell right now. We just multiply it by its depth, which is just dy. So that sets up our integral. The volume of one shell-- I'll do it all in one color now-- is 2 pi times y plus 2 times y plus 1 minus y minus 1 squared. Then we multiply that times the depth of each shell, dy. And then we integrate over the interval. So the volume is going to be this. And what's the interval? You might be able to eyeball it. But we can actually solve that explicitly. When do these two things equal each other? Well, you could just set y plus 1 to be equal to y minus 1 squared, so let's do that. So if we set y plus 1, this, to be equal to that, it's going to be equal to y minus 1 squared. So let me expand that out. That's going to be y squared minus 2y plus 1. And let's see. I could subtract y from both sides and I could subtract 1 from both sides. Minus y, minus 1. And then I am left with, on the left hand side, 0. And on the right hand side, I have y squared minus 3 y. And that's it, plus 0. So 0 is equal to y times y minus 3. So the zeros of this, when these are equal are when y is equal to 0 or y is equal to 3. And we see that right over here. When y is equal to 0, these two functions intersect. And when y is equal to 3, these two functions intersect. So our interval is going to be from y is equal to 0 to y is equal to 3. So using the shell method, we have been able to set up our definite integral. And now we can think about how we can evaluate this thing.