Disk method: rotating x=f(y) around the y-axis Using the disk method around the y-axis.
Disk method: rotating x=f(y) around the y-axis
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- In the last video we took essentially the length of y
- equals x squared, not the length, but we went from zero
- to 1 on the x-axis and you can view it as this area.
- And we rotated around the y-axis to get this figure here
- and we figured out the volume, I think our answer if I
- remember was pi over 2.
- And we use what I called and what everyone calls
- the Shell method.
- I want to show you that you could actually use
- the disk method here.
- But then we'll just have to switch the ys and xs.
- Instead of writing this the function as a function of x,
- we'll just take the inverse of it and write it
- as a function of y.
- so this curve y equals x squared, what it can also be
- written as, we can just take the squared root of both
- sides as x is equal to the square root of y.
- Just take the square root of both sides.
- And now we can use this information to do the disk
- method, but everywhere where we had an x in the past
- we'll now have a y.
- So let's think about how we would do it.
- Once again it's this area-- really the hardest thing about
- all of these problems is just the visualization.
- I think that's why they do it.
- Just to make sure that you know how to visualize things and
- maybe understand the calculus.
- It's more visualization then calculus really.
- So we're still dealing with-- if we take a cross section, if
- we were to cut this figure this area would be the
- cross section.
- And we're still rotating around the y-axis just like we
- did in the last video.
- We're still rotation around the y-axis, we get this figure.
- So how would we deal with disks.
- Well a disk would look like this at some point-- let
- me pick another color.
- At some point we'd have a disk like that.
- That would be the top of the disk, and it would have some
- depth like we did before.
- And its height, or the radius of that disk,
- would be equal to x.
- It's equal to x.
- I know you're thinking it looks like the shell method.
- But what is x equal to?
- It's equal to the square root of y.
- It equals the square root of y.
- And what would be the width of that disk?
- Now we are making everything as a function of y, so the width
- would be just a very small distance d y, the
- differential y.
- That's essentially all we need to know.
- So the volume of that disk would be the radius squared
- times pi times d y.
- Hopefully that makes a little bit sense, but there's another
- hitch on this problem.
- This is actually similar to what we did two videos ago.
- Because when you view it in the y-axis, from this y frame of
- reference, what we're going to do is we're going to take the
- volume, we take the volume of x is equal to 1.
- So what would be the volume of x is equal to 1 rotated
- around the x-axis?
- It would just be the entire cylinder.
- It's really important that you visualize this
- right, what we're doing.
- We're going to figure out this volume, volume
- of x is equal to 1.
- Let me draw the axis just so you know what we're doing.
- So this would be the y-axis, that would be the x-axis
- as best as I can draw.
- Originally, we can figure out the volume of
- the entire cylinder.
- This is x is equal to 1, x is equal to 1 from what y points?
- From the point, well what is this?
- This is y equals 1.
- y equal 1 to zero.
- So we would figure out this volume from y equals 1 to zero.
- And how would we do that?
- What would be the integral?
- Remember everything is in y now, so it might seem
- a little confusing.
- For each of these disks, this is going to be made up of
- a bunch of disks, let me draw one of them.
- Let me draw the top one.
- The top disk, and it has a width, it's not going to be
- this entire cylinder, its width is d y, its width
- is just going to be this.
- d y, a very small sliver.
- d y, and its radius is x or you could say f of y.
- What would be the volume of that disk?
- What would be the surface area of the top?
- It would be f of y squared.
- Remember we're dealing everything with y. f of y
- squared times pi, that gives the area of the top.
- Put the pi outside of the integral times d y.
- d y, and we said y is going from zero to 1.
- y is going from zero to 1.
- So that's that entire cylinder.
- And what we'll want to do is subtract out, essentially cut
- out the volume of the inner bowl.
- So minus.
- How would we figure out the inner bowl?
- That's where we will deal with the x is equal to the
- square root of y function.
- Because here, what's each disk?
- Once again it's f of y.
- I wrote this generally, this is f of y of the outside.
- I'm going to now do it for the inside.
- I'm going to cut out the inside volume and because I kept
- it general I could do it.
- We're still going from zero to 1, from y equals
- zero to y equals 1.
- Remember we're doing with y now.
- I will take f of the inside f of y squared d of y.
- You can imagine in this case the inside function, we're
- going to take the volume.
- That original disk I drew is actually one of the disks
- for the inside function.
- Where the height of that disk is d y, that should be d y.
- The radius of the disk is f of y.
- And of course the area of the top of the desk is pi
- times the radius squared.
- How do we apply it to this particular situation?
- What is the outside function?
- f of y, x is equal to f of y.
- We are just switching the variables.
- We say that x is equal to 1.
- This is just a big cylinder, right?
- So f of i is equal go to 1 in this function, so we get pi--
- let me switch colors-- times from zero to 1.
- 1 squared d y minus pi, and we're still going
- from zero to 1.
- Remember, that's the y boundaries.
- What's f of y?
- Here f of y is square root of y squared d y.
- Let's take the pi out.
- So pi times, well 1 squared is 1.
- What's the anti-derivative of 1?
- What function's derivative is 1?
- It's x, right?
- x, that's the anti-derivative there.
- We can merge these.
- We took the pi out.
- This square root of y squared, this is just y.
- Sorry, we have that derivative 1.
- See, even I get quite confused here.
- We're taking this with respect to y, right?
- The anti-derivative of 1 now is y.
- We are setting up a bunch of ys.
- I'm sorry.
- I find this even a little perplexing when I start
- switching x and y.
- The second function.
- Remember, we can merge these because it's the same
- boundaries and they're both in respect to y.
- Square root of y squared is y.
- What's the anti-derivative of y, the function y or y d y.
- Was y squared over 2, minus y squared over 2.
- We are going to evaluate that at 1 and zero.
- So what does that get us?
- We get pi-- we get 1 minus 1/2 minus zero minus zero
- minus zero plus 0/2.
- Whatever these are zero.
- 1 minus 1/2 that's 1/2 and times pi is
- equal to pie over 2.
- Which is the exact same result.
- And I was worried, because you never know when you might make
- a careless mistake as I often do.
- We got the exact same result as I got in the previous video
- when we used the shell method.
- The hardest thing here is just remembering that you're doing
- everything in terms of y.
- When we did the disk method traditionally, the disks
- were vertical disks.
- Now they're horizontal disks but it's the exact same thing.
- You just have to get your brain around the idea that we're
- dealing with the ys, that the boundaries on the integrals
- are now y values.
- We're taking the width of the disks, or the height of the
- disks are d y and the radius is now the function of y.
- Hopefully I didn't confuse you too much.
- My own brain malfunctioned a bit when I took the
- anti-derivative of 1 d y, it should have been y.
- I will see you in the next video where we will do even
- more complicated problems, where I'm sure I'll make
- even more mistakes.
- See you soon.
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