Let's evaluate the double integrals with y=x^2 as one of the boundaries. Created by Sal Khan.
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- Aren't you still bumping into the curve when you go in the x direction…?(21 votes)
- Assuming you're integrating over dy first, the bumping is taken into account by using x*2 as upper bound in the first definite integration. You now have an infinitesimal rectangle bounded by y = 0 on the lower side and y = x*2 on the upper. When you do the 2nd integration (over dx, from x = 0 to x = 1), you are summing up infinitesimal rectangles that vary in height, as per the function y = x**2. This variation in height accounts for "bumping into the curve" (in the y direction). In the x direction you start with x = 0 (lower bound) and "bump" into the line x = 1 (upper bound).
Hope this helps. You could check the answers to a similar question for the previous video.(44 votes)
- Do double integrals work if one of the bounds is infinity?(6 votes)
- The original definition of the Riemann integral does not apply to such a function because in this case the domain of integration is unbounded ("bounded" by infinity).
However, the Riemann integral can often be extended by continuity, by defining the improper integral instead as a limit. In that case indeed the double integral can be evaluated, but one has to regard the "bound" which goes to infinity as a limit.
Also, the easiest way to do this is by taking the unbounded integrand last. That is(7 votes)
- Would be interesting to know how to actually arrive at the functions of the curves in the 2d plane (i.e. how you arrived at y=x^2 for example). And perhaps some abstract example of how you can have variable bounds in both integrals.. That would be perfect.(3 votes)
- Hi. Here is the first example of the set I am developing. This example may be on the easy side, but is more about explaining a process to finding the limits of integration. I am interested in your opinion if it will be useful. Thanks.
- what if we didn't have double integrals? What IF we didn't have integrals at all?(2 votes)
- I don't know if you are asking "If integrals didn't existed how would we solve this kind of problems?", or "What if we are given a problem without integrals?".
The answer to the first questions is that there are much more complicated methods to calculating areas and volumes, geometric methods, that were used before integrals were invented.
The answer to the second question is that if a problem does not need integrals, then you are not require to use them.(5 votes)
- is there videos related to polar coordinates?(2 votes)
- There is a whole section of the site dedicated to them: https://www.khanacademy.org/math/precalculus/parametric_equations/polar_coor(3 votes)
- Very helpful video and explanation of how to find volumes! I also like how you stepped through the problem both directions. BUT... it seems counter-intuitive that the volume you calculated is in fact 1/24. When I look at the image, it does NOT appear to have a volume of 1/24.... it looks far more like it would have a volume of a little bit less than 1/2!? Is this just an optical illusion? Is the curve you drew not really scaled due to it being only a sketch? I'm confused on that...(2 votes)
- Good eye!
The x and y limits are pretty well to scale, though to me his graph of x=sqrt(y) is a little too fat. The part that is not correct is the way Sal drew the "roof" of the function z=xy². It is NOT a flat surface as appears in his graph, it is a curved surface that slopes down toward the origin in the 1st quadrant - and obviously, when x=0 and or y=0, z=0 as well. The ONLY time z=1 is when both x=1 and y=1.
Here is a graph of the function - I hope it helps you see the more diminutive result.
PS - you may need to cut and paste the entire URL below into your browser as Khan may not render this link correctly.
- At4:56, where is Sal getting that y=x^2 (or x=sqrt(y)). Is this just intuition based upon looking at the 3-D graph from the z-axis, or is there a specific method used to solve this and set it as a bound?(1 vote)
- In the previous video, Sal starts of by saying that we're finding the volume under
z = xy²that's bound by
y = x²and the x-axis. So he figure it out from anywhere, it's part of the initial problem.(3 votes)
- how did sal know that the xy plane will have a y=x^2??(1 vote)
- Because that is one of the functions that defines the solid. In other words, projecting the solid into the xy plane would yield y=x^2.(3 votes)
Welcome back. In the last video we were just figuring out the volume under the surface, and we had set up these integral bounds. So let's see how to evaluate it now. And look at this. I actually realized that I can scroll things, which is quite useful because now I have a lot more board space. So how do we evaluate this integral? Well, the first integral I'm integrating with respect to x. Right? I'm adding up the little x sums. So I'm forming this rectangle right here. Right? Or you could kind of view it I'm holding y constant and integrating along the x-axis. I should switch colors. So what's the antiderivative of x y squared with respect to x? Well it's just x squared over 2. And then I have the y squared-- that's just a constant-- all over 2. And I'm going to evaluate that from x is equal to 1 to x is equal to the square root of y, which you might be daunted by. But you'll see that it's actually not that bad once you evaluate them. And then let me draw the outside of the integral. This is y is equal to 0 to y is equal 1. dy. Now, if x is equal to 1 this expression becomes y squared over 2. Right? y squared over 2, minus-- now if x is equal to square root of y, what does this expression become? If x is equal to the square root of y, then x squared is just y. And then y times y squared is y to the third. Right? So it's y to the third over 3. Fair enough. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction. 0, 1. This is with respect to y. And that's cool, right? Because when you take the first integral with respect to x you end up with a function of y anyway, so you might as well have your bounds as functions of y's. It really doesn't make it any more difficult. But anyway, back to the problem. What is the antiderivative of y squared over 2 minus y to the third over 3? Well the antiderivative of y squared-- and you have to divide by 3, so it's y cubed over 6. Minus y to the fourth-- you have to divide by 4. Minus y to the fourth over-- did I mess up some place? No, I think this is correct. y to the fourth over 12. Oh wait. How did I get a 3 here? That's where I messed up. This is a 2, right? Let's see. x is square root of y. Yeah, this is a 2. I don't know how I ended up. Right? Square root of y squared is y, times y squared y to the third over 2. Right. And then when I take the integral of this it's 4 times 2. 8. Got to make sure I don't make those careless mistakes. That's the tough part. I just want to make sure that you got that too. I hate it when I do that. But I don't want to re-record the whole video. So when I evaluated this-- right. This is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8. And now I evaluate this at 1 and 0. And that give us what? 1/6 minus 1/8 minus-- well both of these when you evaluate them, are going to be 0's. So this is going to be another 0 minus 0. So you don't have to worry about that. So what's 1/6 minus 1/8? Let's see. 24. That's 4 minus 3 over 24, which is equal to 1/24-- is the volume of our figure. So this time, the way we just did it, we took the integral with respect to x first, and then we did it with respect to y. Let's do it the other way around. So let me erase some things. And hopefully I won't make these careless mistakes again. I'll keep this figure, but I'll even erase this one. Let me erase all of this stuff. We have room to work with. So I kept that figure. But let me redraw just the xy plane, just so we get the visualization right. It's more important to visualize the xy plane in these problems than it is to visualize the whole thing in 3-d. That's the y-axis, that's the x-axis. OK. Our upper bound, you can say, is the graph y is equal to x squared. Or, you could view it as the bound x is equal to the square root of y. That is x is equal to 1, that's y is equal to 1. And we care about the volume above the shaded region. Right? That shaded region is this yellow region right here. And let's draw our da. I'll draw a little square actually. And I'll do it in magenta. So that's our little da. And the height is dy. That's a y. dy. And it's with this dx. Right? So the volume above this little square-- that's the same thing as this little square. Just like we said before, the volume above it is equal to the value of the function. Right? The height is the value of the function, which is x y squared. And then we multiply it times the area of the base. Well the area of the base, you could say it's da, but we know it's really dy times dx. I didn't have to write that times there. You can ignore it. And I wrote the y first just because we're going to integrate with respect to y first. We're going to sum in the y direction. So what does summing in the y direction mean? It means we're going to add that square to that square to that square-- [COUGHING] Excuse me. So we're adding all the dy's together, right? So my question to you is, what is the upper bound on the y? Well once again, we bump into the curve. Right? So the curve is the upper bound when we go upwards. And what is the upper bound on the curve? Well we're holding x fixed, so for any given x what is this point? Well that's going to be x squared. Right? Because this is the graph of y equals x squared. So our upper bound is y is equal to x squared. And what's our lower bound? We can keep adding these squares down here. We're adding all the little changes in y. So what's our lower bound? Well our lower bound is just 0. That was pretty straightforward. So this expression, as it is written right now, is the volume above this rectangle. Right? Let me draw it. Is the volume above this rectangle. These are the same rectangles. The volume above that rectangle. Now what we want to do is add up all the dx's together, and we'll get the volume above the entire surface. So that rectangle, now we'll add it to another dx one, dx one like that. So what's the upper and lower bounds on the x's? We're going from x is equal to 0, right? We don't bump into the graph when we go all the way down. So we go from x is equal to 0. And then our upper bound is x is equal to 1. So x is equal to 0, x is equal to 1. And in general, one way to think about it is when you're doing kind of the last sum or the last integral, you really shouldn't have variable boundaries at this point. Right? Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract. But our final answer is going to have a number. So if you had some variables here you probably did something wrong. And it's always useful, I think, to draw that little da and then figure out-- OK, I'm summing to dy first. When I go upwards I bump into the curve. What is that upper bound if x is constant? Oh, it's x squared. y is equal to x squared. If I go down I bump into the x-axis, or y is equal to 0. And so forth and so on. So now let's just evaluate this and confirm that we get the same answer. So we're integrating with respect to y first. So that's x y to the third, over three. At x squared and 0. And then we have our outer integral. x goes from 0 to 1. dx. If we substitute x squared in for y. x squared to the third power is x to the sixth. x to the sixth times x-- let me write that. So we have x times x squared to the third power, over 3. That equals x to the seventh. Right? x squared to the third-- you multiply the exponents, and then you add these. x to the seventh over 3, minus this evaluated with y is 0. But that's just going to be 0. Right? And then we evaluate that from 0 to 1, dx. We're almost there. Increment the exponent. You get x to the eighth over 8. And we already have a 3 down there, so it's over 24. And you evaluate that from 0 to 1. And I think we get the same answer. When you evaluate it at 1 you get 1/24 minus 0. So once again, when we integrate in the other order you still get the volume of the figure, being 1/4, whatever, cubic units. Anyway, see you in the next video.