Double integral 1
Introduction to the double integral. Created by Sal Khan.
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- I might be getting ahead of myself, but with this new definition do we use the same techniques (Integration by Parts, Trig. Substitution, etc) on Double Integrals as we did with single Definite Integrals? or do we need to redefine them?(13 votes)
- The beautiful thing is that all of the old techniques apply. A double integral is just one regular integral inside of another. Thus, you can use integration techniques on the inside integral, then use integration techniques for the outside integral.(25 votes)
- Would i have to do this i college if i did calculus.(3 votes)
- That depends on the school. Often the first semester is differential calculus, the second is single integral calculus. After that, multivariable and vector calculus.
If you need the calculus for a non math degree, often the first two semesters are enough and often, the college will have a special calculus program for non-science students.
The question is: How far down the rabbit hole do you want to go?
The deeper you go, the more fun and amazing . . . . jump in!(12 votes)
- when in life would i use this?(1 vote)
- When you ask that question, you sort of prevent yourself from ever getting an answer. Learn it, see what it means, see how it works, and where it can be applied will reveal itself to you. You'll begin to see where others use it, and how such things can be used to describe our world, and how they can be used to try and make the world a better place.(12 votes)
- So this is a small question, probably irrelevant to the whole solving process...but when we labeled the surface f(x,y)dx as f(y),
how is f(y) the surface?...like how does f(y) represent the surface we integrate across y to find the volume?(4 votes)
- f(x,y)dx isn't a surface, neither is f(y).
f(x,y)dx is the area of a small vertical planar element of the cross section of the volume under the surface f(x,y).
f(y) is the integral of f(x,y)dx so it's the sum of these areas and so gives the total area under the cross-section of the volume underneath the surface f(x,y) for a given value of y.
So, f(x,y) is a surface, but f(x,y)dx and f(y) refer to areas not on the surface, but of a section of the volume underneath it.
Following on further:
f(y)dy is the volume of a small section of the total volume under the surface f(x,y) whilst the integral of f(y)dy is the total volume of the volume under the surface f(x,y)(5 votes)
- Is this section just videos or are there problem sets attached to them? I'm using Khan Academy as a refresher for the GRE subject test but I can't seem to find problems after their Integral Calculus Section.(4 votes)
- At the moment there are no problems.
I can send you problems with solutions if you like or search the net for them - they are out there.(6 votes)
- at8:20Sal says "0 to b". why wouldn't it be a to b?(4 votes)
- No because "a" is some constant y value and he is only evaluating the function with respect to changes in X from 0 - b (or some number more than 0 on the X axis). The inner integral, integrated from 0 to b provides the 2D value of area. Notice, however, that when a change of y is introduced, volume (a 3D value) can be calculated. In order to do that, there needs to be some change in y (0-a) by which the integral of 0 - b can be evaluated. Hope this mades sense... explaining calc with text is a bit tricky ;)(2 votes)
My questions concerns solving double integrals through conversion to polar coordinates. I was wondering why dxdy is sometimes replaced by dthetadr and other times by drdtheta? Does the order matter?(3 votes)
- When integrating multiple variables, the order in which the integrals are evaluated does not matter. Integrating by the radius then theta will give the same answer as integrating theta then the radius.(2 votes)
- What?! No problems/questions?! I don't just want to watch a video. What fun is that?!(3 votes)
- I suspect the problems for these videos are forthcoming, but who knows when.
I am happy to send you problems at the level of each video, with the answer, if you want.
When you are done I can send you the outline the solutions as well, if any of the problems I sent gave you trouble.(2 votes)
- Whats the difference between double integrals and triple integrals?(2 votes)
- The main difference is a double integral is integrating a function over an area and a triple integral integrates over a volume.
This may seem strange, but for all of the double integrals the integrand was a function in R3 (3D) (e.g. f(x,y) you have a length, width, and height). The answer was a volume, as you have three dimensions. Another way to depict a double integral that illustrates this difference more effectively is ∫∫_a f(x,y) da where a is the area you are integrating over.
A triple integral is integrating a function in R4 over a volume. The density function example shows this well. p(x,y,z) assigns a density to each point in space (for every point in space you pick the functions returns a density). This means the function has length, width, height, and another height which is the density (4D). The triple integral integrates over a volume and give a hyper volume (units^4). In the case of the density function, the integral represents a mass because pv = m (basic density equation). Each small volume is being multiplied by a height to yield a mass. A triple integral can be generalized for a volume v as follows: ∫∫∫_v f(x,y,z) dv.(3 votes)
- How we will calculate the volume when x and y are not bounded ??(2 votes)
- There has to be some limits in order to calculate a volume, otherwise you wouldn't know what figure you have to calculate the volume to. Many times you have to obtain the integration limits from the geometry of the object you are trying to get the volume.(3 votes)
So far, we've used integrals to figure out the area under a curve. And let's just review a little bit of the intuition, although this should hopefully be second nature to you at this point. If it's not, you might want to review the definite integration videos. But if I have some function-- this is the xy plane, that's the x-axis, that's the y-axis-- and I have some function. Let's call that, you know, this is y is equal to some function of x. Give me an x and I'll give you a y. If I wanted to figure out the area under this curve, between, let's say, x is equal to a and x is equal to b. So this is the area I want to figure out. What I do is, I split it up into a bunch of columns or a bunch of rectangles. Where-- let me draw one of those rectangles-- where you could view-- and there's different ways to do this, but this is just a review. Where you could review-- that's maybe 1 of the rectangles. Well, the area of the rectangle is just base times height, right? Well, we're going to make these rectangles really skinny and just sum up an infinite number of them. So we want to make them infinitely small. But let's just call the base of this rectangle dx. And then the height of this rectangle is going to be f of x, at that point. It's going to be f of-- if this is x0, or whatever, you can just call it f of x, right? That's the height of that rectangle. And if we wanted to take the sum of all of these rectangles-- right? There's just going to be a bunch of them. One there, one there. Then we'll get the area, and if we have infinite number of these rectangles, and they're infinitely skinny, we have exactly the area under that curve. That's the intuition behind the definite integral. And the way we write that-- it's the definite integral. We're going to take the sums of these rectangles, from x is equal to a, to x is equal to b. And the sum, or the areas that we're summing up, are going to be-- the height is f of x, and the width is d of x. It's going to be f of x times d of x. This is equal to the area under the curve. f of x, y is equal to f of x, from x is equal to a to x is equal to b. And that's just a little bit of review. But hopefully, you'll now see the parallel of how we extend this to taking the volume under a surface. So first of all, what is a surface? Well, if we're thinking in three dimensions, a surface is going to be a function of x and y. So we can write a surface as, instead of y is a function of f and x-- I'm sorry. Instead of saying that y is a function of x, we can write a surface as z is equal to a function of x and y. So you can kind of view it as the domain. Right? The domain is all of the set of valid things that you can input into a function. So now, before, our domain was just-- at least, you know, for most of what we dealt with-- was just the x-axis, or kind of the real number line in the x direction. Now our domain is the xy plane. We can give any x and any y-- and we'll just deal with the reals right now, I don't want to get too technical. And then it'll pop out another number, and if we wanted to graph it, it'll be our height. And so that could be the height of a surface. So let me just show you what a surface looks like, in case you don't remember. And we'll actually figure out the volume under this surface. So this is a surface. I'll tell you its function in a second, but it's pretty neat to look at. But as you can see, it's a server. It's like a piece of paper that's bent. Let's see, let me rotate it to its traditional form. So this is the x direction, this is the y direction. And the height is a function of where we are in the xy plane. So how do we figure out the volume under a surface like this? How do we figure out the volume? It seems like a bit of a stretch, given what we've learned from this. But what if-- and I'm just going to draw an abstract surface here-- let me draw some axis. Let's see, that's my x-axis. That's my y-axis. That's my z-axis. I don't practice these videos ahead of time, so I'm often wondering what I'm about to draw. OK. So that's x, that's y, and that's z. And let's say I have some surface. I'll just draw something. I don't know what it is. Some surface. This is our surface. z is a function of x and y. So, for example, you give me a coordinate in the xy plane. Say, here, I'll put it into the function and it'll give us a z value now. And I'll plot it there and it'll be a point on the surface. So what we want to figure out is the volume under the surface. And we have to specify bounds, right? From here, we said x is equal to a, to x is equal to b. So let's make a square bound first, because this keeps it a lot simpler. So let's say that the domain or the region-- not the domain-- the region of-- the x and y region of this part of the surface under which we want to calculate the volume. Let's say, the shadow-- if the sun was right above the surface, the shadow would be right there. Let me try my best to draw this neatly. So this is what we're going to try to figure out the volume of. And let's say-- so, if we wanted to draw it in the xy plane, like you can kind of view the projection of the surface of the xy plane, or the shadow of the surface of the xy plane. What are the bounds? You can almost view-- what are the bounds of the domain? Well, let's say that this point-- let's say that this right here, that's 0, 0 in the xy plane. Let's say that this is y is equal to-- I don't know, that's y is equal to a. That's this line right here. Y is equal to a. And let's say that this line right here is x is equal to b. Hope you get that, right? This is the xy plane. If we have a constant x, it would be a line like that. A constant y, a line like that. And then we have the area in between it. So how do we figure out the volume under this? Well, if I just wanted to figure out the area of-- let's just say, this sliver. Let's say we had a-- well, actually let me go the other way. Let's say we had a constant y. So let's say I had some sliver. I don't want to confuse you. Let's say that I had some constant y. I just want to give you the intuition. You know, let's say. I don't know what that is. It's an arbitrary y. But for some constant y, what if I could just figure out the area under the curve there? How would I figure out just the area under that curve? It'll be a function of which y I pick, right? Because if I pick a y here, it'll be a different area. If I pick a y there, it'll be a different area. But I could view this now as a very similar problem to this one up here. I could have my dx's-- let me pick a vibrant color so you can see it. Let's say that's dx, right? That's a change in x. And then the height is going to be a function of the x I have and the y I picked. Although I'm assuming, to some degree, that that's a constant y. So what would be the area of this sheet of paper? It's kind of a constant y. It's part of-- it's a sheet of paper within this volume, you can kind of view it. Well, it would be-- we said the height of each of these rectangles is f of xy, right? That's the height. It depends which x and y we pick down here. And then its width is going to be d of x. Not d of x, dx. And then if we integrated it, from x is equal to 0, which was back here, all the way to x is equal to b, what would it look like? It would look like that. x is going from 0 to b. Fair enough. And this would actually give us a function of y. This would give us an expression so that I would know the area of this kind of sliver of the volume, for any given value of y. If you give me a y, I can tell you the area of the sliver that corresponds to that y. Now what could I do? If I know the area of any given sliver, what if I multiply the area of that sliver times dy? This is a dy. Let me do it in a vibrant color. So dy, a very small change in y. If I multiplied this area times a small dy, then all of a sudden I have a sliver of volume. Hopefully that makes some sense. I'm making that-- that little cut that I took the area of-- by making it three dimensional. So what would be the volume of that sliver? The volume of that sliver will be this function of y times dy, or this whole thing times dy. So it would be the integral from 0 to b of f of xy dx. That gives us the area of this blue sheet. Now if I multiply this whole thing times dy, I get this volume. It gets some depth. This little area that I'm shading right here gets depth of that sheet. Now if I added all of those sheets that now have depth, if I took the infinite sum-- so if I took the integral of this from my lower y bound-- from 0 to my upper y bound, a, then-- at least based on our intuition here-- maybe I will have figured out the volume under this surface. But anyway, I didn't want to confuse you. But that's the intuition of what we're going to do. And I think you're going to find out that actually calculating the volumes are pretty straightforward, especially when you have fixed x and y bounds. And that's what we're going to do in the next video. See you soon.