Shell method around a non-axis line Taking the revolution around something other than one of the axes.
Shell method around a non-axis line
- Let's do a couple more rotational volume problems, and
- I'm going to make these a little bit more difficult.
- And hopefully after these if you've understood everything
- we've done up to now and the ones I'm about to do, I think
- you're pretty set for most of what you should face
- in most math classes.
- And definitely I think you'll be set for the AP exam,
- and either ab or bc on this concept.
- So let's do another example.
- So let's say that I want to-- actually, let me
- do something different.
- Let me draw here.
- So that's my y-axis.
- This is my x-axis.
- Now let me draw the function y is equal to x squared.
- And we know that that could be written as y is equal to x
- squared, or we could write that as x is equal to square root of
- y, depending on what we want to be a function of what.
- This is the y-axis.
- This is the x-axis.
- Let's say I also have the line y equals 2.
- Goes over what I just wrote.
- y equals 2.
- Now this problem is going to be slightly different then
- what we've done so far.
- I'm going to take a rotation, but instead of taking a
- rotation around the y or the x-axis, I'm going to take a
- rotation around another line.
- So let's say I want to take a rotation between
- x is equal to 0.
- Actually let me do something arbitrary.
- Let me say between x is equal to 1, so that's that point,
- and x is equal to 2.
- That's where they intersect.
- This is the point right here, this is 2,2.
- I'm sorry, no, 2,4.
- Because y is equal to x squared.
- So this is 2,4.
- This is the point 4.
- So what point is this.
- This is the point 1, right?
- So our y values go from 4 to 1, our x values go from 1 to 2.
- And that makes sense, because y is x squared.
- And so if we were to kind of take the area that we're going
- to rotate, and I haven't told you what we're going to rotate
- it around yet, and this might prove to be shocking to you.
- So this is the area we're going to rotate.
- Instead of rotating it around the y-axis, I want to rotate
- it around the line y is equal to minus 2.
- So if that's 2, y equals minus 2 should be roughly here.
- So I'm going to rotate this area around this line.
- So what's it going to look like?
- It's going to be a fairly big ring.
- Like if I were to try to draw it-- let me see if I can
- even make an attempt.
- Once again this is always the hardest part, just drawing
- what I'm trying to rotate.
- I'll try to do it from an upward perspective.
- So that's kind of the inner loop, and then there
- will be an outer loop.
- The top is flat right, because it's defined by y is equal
- to 4, so that's the top.
- And the inside is also going to be a hard edge.
- But then the outside is going to curve inward.
- I don't know if you see what I'm saying, because this is the
- outside, it's curving inward.
- So it's going to be a big ring.
- So if I were to draw the axis, this would be the y-axis access
- coming in-- no, no, sorry.
- The y-axis is actually going to be closer to this hand side.
- The y-axis is going to be in the middle of kind of-- so
- this is going to be the y-axis coming up here.
- And then the x-axis is going to come below that.
- I'm drawing everything at an angle as best as I can.
- The x-axis is going to come a below that.
- And then this line we're rotating it around,
- that's going to be someplace over here.
- That's going to be something like that.
- It's going to go behind there and come back over there.
- Hopefully that makes sense.
- We're just getting a big ring.
- So how are we going to do this?
- Well actually there's a couple of things we can do.
- First we could just use the shell method,
- using the x value.
- So how do we do that?
- The important thing is to always visualize
- the shell or the disk.
- So the shell method we're going to take slivers like this,
- where the width of that sliver is dx.
- I could draw it really big.
- So that's our direct angle, is going to be dx.
- What's the height going to be of the sliver?
- Well it's going to be the top function minus
- the bottom function.
- It's going to be y equals 4 minus x squared.
- So this is going to be 4 minus x squared, the height at
- any point right here.
- And then if I were to do the shell just like we did
- before-- let me see if I can draw a decent shell.
- I think I'm getting better at this.
- This is one edge of the shell, that's the other shell.
- We already figured out that the width of the shell is dx.
- The height is 4 minus x squared, the top function minus
- the bottom function because of the distance between the two.
- And then what's the radius going to be?
- What's the radius of that shell going to be?
- Well, is it going to be just x?
- Is it just going to be x value?
- No, the x value will tell you the distance from the y-axis to
- that shell, and it's going to be from minus 2 to the e value.
- So it's going to be essentially 2 plus x.
- That's going to be the radius at any point.
- And this is where we diverge from what we've done before.
- Before the radius was just x, but now it's 2 plus x.
- So what's the circumference of each shell going to be?
- Well circumference is equal to 2 pi r, and
- our radius is 2 plus x.
- So it's 2 pi times 2 plus x, which equals 4 pi
- plus 4 pi plus 2 pi x.
- That's the circumference.
- And then what's the surface area of this?
- Well it's going to be the circumference times the height.
- So surface area is equal to that.
- The circumference 4 pi plus 2 pi x, all of that times the
- height-- times 4 minus x squared.
- And let's see if we can foil this out or
- distribute this out.
- So 4 pi times 4 is 16 pi.
- 4 pi minus x squared minus 4 pi x squared.
- 2 pi x times 4 plus 8 pi x, and then 2 pi x times minus x
- squared, so that's minus 2 pi x to the third.
- So that's the surface area of each ring.
- And then if we want the volume of each shell, essentially
- we multiply it times the width, the dx.
- So that's the volume of each shell, and if we want
- the volume of all the shells, we sum them up.
- So we take the integral, that's an integral sign, and I'm
- running out of space like I always do.
- And where did I take the integral from?
- I take the integral from x is equal to 1 to x is equal to 2.
- From 1 to 2.
- That's probably too small for you to read.
- So let's see if we can take the antiderivative of this.
- Let me make some space free just so I don't
- have to write so small.
- So I'll keep this down here, because that's the set
- up of the problem.
- I think I can get rid of a lot of this.
- I think that is pretty good.
- And let me switch to another color.
- And we're going to take the antiderivative.
- So what's the antiderivative of this?
- So the antiderivative of 16 pi is 16 pi x.
- And then what's the antiderivative of
- 4 pi x squared?
- It's going to be x to the third over 3, so it'll be minus 4
- pi over 3-- sorry-- 4 pi over 3 x to the third.
- Well this will be x squared over 2, so it'll be
- plus 4 pi x squared.
- And then minus, this will be x to the fourth over
- 4, so minus pi over 2.
- Just divided by 4.
- x to the fourth.
- I'm going to evaluate that.
- This is a much hairrier problem then what we've been doing.
- 2 and at 1.
- So what is it evaluated at 2?
- It is 32 pi minus 4 pi over 3 times 8 plus 4 pi times 4 plus
- 16 pi minus 2 to the fourth is 16 divided by 2 minus 8 pi.
- And then I just realized I'm running out time, so I will
- continue this in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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