Double Integrals 2 Figuring out the volume under z=xy^2
Double Integrals 2
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- You hopefully have a little intuition now on what a double
- integral is or how we go about figuring out the volume
- under a surface.
- So let's actually compute it and I think it'll all become
- a lot more concrete.
- So let's say I have the surface, z, and it's a
- function of x and y.
- And it equals xy squared.
- It's a surface in three-dimensional space.
- And I want to know the volume between this
- surface and the xy-plane.
- And the domain in the xy-plane that I care about is x is
- greater than or equal to 0, and less than or equal to 2.
- And y is greater than or equal to 0, and less
- than or equal to 1.
- Let's see what that looks like just so we have a
- good visualization of it.
- So I graphed it here and we can rotate it around.
- This is z equals xy squared.
- This is the bounding box, right? x goes from 0 to
- 2; y goes from 0 to 1.
- We literally want this-- you could almost view it the
- volume-- well, not almost.
- Exactly view it as the volume under this surface.
- Between this surface, the top surface, and the xy-plane.
- And I'll rotate it around so you can get a little bit better
- sense of the actual volume.
- Let me rotate.
- Now I should use the mouse for this.
- So it's this space, underneath here.
- It's like a makeshift shelter or something.
- I could rotate it a little bit.
- Whatever's under this, between the two surfaces--
- that's the volume.
- Whoops, I've flipped it.
- There you go.
- So that's the volume that we care about.
- Let's figure out how to do and we'll try to gather a little
- bit of the intuition as we go along.
- So I'm going to draw a not as impressive version of that
- graph, but I think it'll do the job for now.
- Let me draw my axes.
- That's my x-axis, that's my y-axis, and that's my z-axis.
- x, y, z.
- x is going from 0 to 2.
- Let's say that's 2.
- y is going from 0 to 1.
- So we're taking the volume above this rectangle
- in the xy-plane.
- And then the surface, I'm going to try my best to draw it.
- I'll draw it in a different color.
- I'm looking at the picture.
- At this end it looks something like this.
- And then it has a straight line.
- Let's see if I can draw this surface going down like that.
- And then if I was really good I could shade it.
- It looks something like this.
- If I were to shade it, the surface looks
- something like that.
- And this right here is above this.
- This is the bottom left corner, you can almost view it.
- So let me just say the yellow is the top of the surface.
- That's the top of the surface.
- And then this is under the surface.
- So we care about this volume underneath here.
- Let me show you the actual surface.
- So this volume underneath here.
- I think you get the idea.
- So how do we do that?
- Well, in the last example we said, well, let's pick an
- arbitrary y and for that y, let's figure out the
- area under the curve.
- So if we fix some y-- when you actually do the problem, you
- don't have to think about it in this much detail, but I want
- to give you the intuition.
- So if we pick just an arbitrary y here.
- So on that y, we could think of it-- if we have a fixed y, then
- the function of x and y you can almost view it as a function
- of just x for that given y.
- And so, we're kind of figuring out the value of this, of
- the area under this curve.
- You should view this as kind of an up down curve for a given y.
- So if we know a y we can figure out then-- for example, if y
- was 5, this function would become z equals 25x.
- And then that's easy to figure out the value
- of the curve under.
- So we'll make the value under the curve as a function of y.
- We'll pretend like it's just a constant.
- So let's do that.
- So if we have a dx that's our change in x.
- And then our height of each of our rectangles is going to be a
- function-- it's going to be z.
- The height is z, which is a function of x and y.
- So we can take the integral.
- So the area of each of these is going to be our function, xy
- squared-- I'll do it here because I'll run out of space.
- xy squared times the width, which is dx.
- And if we want the area of this slice for a given y, we just
- integrate along the x-axis.
- We're going to integrate from x is equal to 0
- to x is equal to 2.
- From x is equal to 0 to 2.
- Fair enough.
- Now, but we just don't want to figure out the area under the
- curve at one slice, for one particular y, we want to
- figure out the entire area of the curve.
- So what we do is we say, OK, fine.
- The area under the curve, not the surface-- under this curve
- for a particular y, is this expression.
- Well, what if I gave it a little bit of depth?
- If I multiplied this area times dy then it would give me a
- little bit of depth, right?
- We'd kind of have a three-dimensional slice of the
- volume that we care about.
- I know it's hard to imagine.
- Let me bring that here.
- So if I had a slice here, we just figured out the area of a
- slice and then I'm multiplying it by dy to give it a
- little bit of depth.
- So you multiply it by dy to give it a little bit of depth,
- and then if we want the entire volume under the curve we add
- all the dy's together, take the infinite sum of these
- infinitely small volumes really now.
- And so we will integrate from y is equal to 0
- to y is equal to 1.
- I know this graph is a little hard to understand, but you
- might want to re-watch the first video.
- I had a slightly easier to understand surface.
- So now, how do we evaluate this?
- Well, like we said, you evaluate from the
- inside and go outward.
- It's taking a partial derivative in reverse.
- So we're integrating here with respect to x, so we can treat
- y just like a constant.
- Like it's like the number 5 or something like that.
- So it really doesn't change the integral.
- So what's the antiderivative of xy squared?
- Well, the antiderivative of xy squared-- I want to make
- sure I'm color consistent.
- Well, the antiderivative of x is x to the 1/2--
- sorry. x squared over 2.
- And then y squared is just a constant, right?
- And then we don't have to worry about plus c since
- this is a definite integral.
- And we're going to evaluate that at 2 and 0.
- And then we still have the outside integral
- with respect to y.
- So once we figure that out we're going to integrate it
- from 0 to 1 with respect to dy.
- Now what does this evaluate?
- We put a 2 in here.
- If you put a 2 in there you get 2 squared over 2.
- That's just 4 over 2.
- So it's 2 y squared.
- Minus 0 squared over 2 times y squared.
- Well, that's just going to be 0.
- So it's minus 0.
- I won't write that down because hopefully that's a little
- bit of second nature to you.
- We just evaluated this at the 2 endpoints and
- I'm short for space.
- So this evaluated at 2y squared and now we evaluate
- the outside integral.
- 0, 1 dy.
- And this is an important thing to realize.
- When we evaluated this inside integral, remember
- what we were doing?
- We were trying to figure out for a given y, what the
- area of this surface was.
- Well, not this surface, the area under the surface
- for a given y.
- For a given y that surface kind of turns into a curve.
- And we tried to figure out the area under that curve
- in the traditional sense.
- This ended up being a function of y.
- And that makes sense because depending on which y we pick
- we're going to get a different area here.
- Obviously, depending on which y we pick, the area-- kind of a
- wall dropped straight down-- that area's going to change.
- So we got a function of y when we evaluated this and now we
- just integrate with respect to y and this is just plain old
- vanilla definite integration.
- What's the antiderivative of 2y squared?
- Well, that equals 2 times y to the third over 3,
- or 2/3 y to the third.
- We're going to evaluate that at 1 and 0, which
- is equal to-- let's see.
- 1 to the third times 2/3.
- That's 2/3.
- Minus 0 to the third times 2/3.
- Well, that's just 0.
- So it equals 2/3.
- If our units were meters these would be 2/3 meters
- cubed or cubic meters.
- But there you go.
- That's how you evaluate a double integral.
- There really isn't a new skill here.
- You just have to make sure to keep track of the variables.
- Treat them constant.
- They need to be treated constant, and then treat them
- as a variable of integration when it's appropriate.
- Anyway, I will see you in the next video.
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