Double Integrals 5 Finding the volume when we have variable boundaries.
Double Integrals 5
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- In all of the double integrals we've done so
- far, the boundaries on x and y were fixed.
- Now we'll see what happens when the boundaries on
- x and y are variables.
- So let's say I have the same surface, and I'm not going to
- draw it the way it looks, I'll just kind of draw
- it figuratively.
- But the problem we're actually going to do is z, and this is
- the exact same one we've been doing all along.
- The point of here isn't to show you how to integrate, the point
- of here is to show you how to visualize and think
- about these problems.
- And frankly, in double integral problems the hardest part is
- figuring out the boundaries.
- Once you do that, the integration is pretty
- It's really not any harder then single variable integration.
- So let's say that's our surface: z is equal
- to xy squared.
- Let me draw the axes again.
- So that's my x-axis.
- That's my z-axis.
- That's my y-axis.
- x, y, and z.
- And you saw what this graph looked like several videos ago.
- I took out the whole grapher and we rotated and things.
- I'm not going to draw the graph the way it looks; I'm just
- going to brought fairly abstractly as just an
- abstract surface.
- Because the point here it's really to figure out the
- boundaries of integration.
- Before I actually even draw the surface, I'm going
- to draw the boundary.
- The first time we did this problem we said, OK, x goes
- from 0 to 2, y goes from 0 to 1, and then we figured out the
- volume above that bounded domain.
- Now let's do something else.
- Let's say that x goes from 0 to 1.
- And let's say that the volume that we want to figure out
- under the surface, it's not from a fixed y to
- an upper-bound y.
- I'll show you: it's actually a curve.
- So this is all on the xy plane, everything I'm drawing here.
- And this curve, we could view it two ways: we could say y is
- a function of x, y is equal to x squared.
- Or we could write is equal to square root of y.
- We don't have to write plus or minus or anything like that
- because we're in the first quadrant.
- So this is the area above which we want to
- figure out the volume.
- Let me, yeah, it doesn't hurt to color it in just so we
- can really hone in on what we care about.
- So that's the area above which we want to
- figure out the volume.
- You could kind of say, that's our bounded domain.
- And so x goes from 0 to 1, and then this point
- is going to be what?
- That point's going to be 1 comma 1, right?
- 1 is equal to 1 squared, 1 is equal to the square root of 1.
- So this point is y is equal to 1.
- And then I'm not going to draw this surface exactly.
- I'm just trying to give you a sense of what the volume of
- the figure we're trying to calculate is.
- If this is just some arbitrary surface-- let me do it in a
- different color --so this is the top.
- This line is going vertical in the z-direction.
- Actually, I could draw it like this, like it's a curve.
- And then this curve back here is going to be like a wall.
- And maybe I'll paint this side of the wall just so you can see
- what it kind of looks like.
- Trying my best.
- Think you get an idea.
- Let me make it a little darker; this is actually more of an
- exercise in art than in math, in many ways.
- You get the idea.
- And then the boundary here is like this.
- And this top isn't flat, you know, it could
- be curved surface.
- I do a little like that, but it's a curved surface.
- And we know in the example we're about to do that the
- surface right here is z is equal to x squared.
- So we want to figure out the volume under this.
- So how do we do it?
- Well, let's think about it.
- We could actually use the intuition that I just gave you.
- We're essentially just going to take a da, which is a little
- small square down here, and that little area, that's the
- same thing as the dx-- let me use a darker color --as a dx
- times a dy, and then we just have to multiply it times f of
- xy, which is this, for each area, and then
- some them all up.
- And then we could take a sum in the x-direction first
- or the y-direction first.
- Now before doing that, just to make sure that you have
- the intuition because the boundaries are the hard part,
- let me just draw our xy plane.
- So let me rotate it up like that.
- I'm just going to draw our xy plane.
- Because that's what matters.
- Because the hard part here is just figuring out our
- bounds of integration.
- So the curve is just y is equal to x squared, look
- something like that.
- This is the point y is equal to 1.
- This is y-axis, this is the x-axis, this is the
- point x is equal to 1.
- That's not an x, that's a 1.
- This is the x.
- Anyway, so we want to figure out, how do we sum up this dx
- times dy, or this da, along this domain?
- So let's draw it.
- Let's visually draw it and it doesn't hurt to do this when
- you actually have to do the problem because this
- frankly is the hard part.
- A lot of calculus teachers will just have you set up the
- integral and then say, OK, well the rest is easy.
- Or the rest is Calc 1.
- OK, so this area, this area here is the same thing
- as this area here.
- So its base is dx and its height is dy.
- And then you could imagine that we're looking at
- this thing from above.
- So the surface is up here some place and we're looking
- straight down on it, and so this is just this area.
- So let's say we wanted to take the integral with
- respect to x first.
- So we want to sum up, so if we want the volume above this
- column, first of all, is this area times dx, dy, right?
- So let's write the volume above that column.
- It's going to be the value of the function, the height at
- that point, which is xy squared times dx, dy.
- This expression gives us the volume above this area, or
- this column right here.
- And let's say we want the sum in the x direction first.
- So we want to sum that dx, sum one here, sum here,
- et cetera, et cetera.
- So we're going to sum in the x-direction.
- So my question to you is, what is our lower
- bound of integration?
- Well, we're kind of holding our y constant, right?
- And so if we go to the left, if we go lower and lower x's we
- kind of bump into the curve here.
- So the lower bound of integration is
- actually the curve.
- And what is this curve if we were to write x
- is a function of y?
- This curve is y is equal to x squared, or x is equal
- to the square root of y.
- So if we're integrating with respect to x for a fixed y
- right here-- we're integrating in the horizontal direction
- first --our lower bound is x is equal to the square root of y.
- That's interesting.
- I think it's the first time you've probably seen a
- variable bound integral.
- But it makes sense because for this row that we're adding up
- right here, the upper bound is easy.
- The upper bound is x is equal to 1.
- The upper bound is x is equal to 1, but the lower bound is
- x is equal to the square root of y.
- Because you go back like, oh, I bump into the curve.
- And what's the curve?
- Well the curve is x is equal to the square root of y because we
- don't know which y we picked.
- Fair enough.
- So once we've figured out the volume-- so that'll give us the
- volume above this rectangle right here --and then we
- want to add up the dy's.
- And remember, there's a whole volume above what
- I'm drawing right here.
- I'm just drawing this part in the xy plane.
- So what we've done just now, this expression, as it's
- written right now, figures out the volume above
- that rectangle.
- Now if we want to figure out the entire volume of the solid,
- we integrate along the y-axis.
- Or we add up all the dy's.
- This was a dy right here, not a dx.
- My dx's and dy's look too similar.
- So now what is the lower bound on the y-axis if I'm summing
- up these rectangles?
- Well, the lower bound is y is equal to 0.
- So we're going to go from y is equal to 0 to what--
- what is the upper bound?
- --to y is equal to 1.
- And there you have it.
- Let me rewrite that integral.
- So the double integral is going to be from x is equal to square
- root of y to x is equal to 1, xy squared, dx, dy.
- And then the y bound, y goes from 0 to y to 1.
- I've just realized I've run out of time.
- In the next video we'll evaluate this, and then we'll
- do it in the other order.
- See you soon.
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