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### Course: Multivariable calculus>Unit 4

Lesson 5: Double integrals

# Double integrals 6

Let's evaluate the double integrals with y=x^2 as one of the boundaries. Created by Sal Khan.

## Want to join the conversation?

• Aren't you still bumping into the curve when you go in the x direction…?
• 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.
• At shouldn't it be (y^3)/2 (instead of (y^3)/3)?
• Do double integrals work if one of the bounds is infinity?
• 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
• 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.
• 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.
http://bajasound.com/khan/Khan_IntegrationLimitsEx1.pdf
• what if we didn't have double integrals? What IF we didn't have integrals at all?
• 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.
• Are there any videos that explain why we use a Jacobian when changing coordinates?
• is there videos related to polar coordinates?
• 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...
• 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.

http://www.wolframalpha.com/input/?i=graphing&a=*C.graphing-*Calculator.dflt-&a=FSelect**Plot-.dflt-&f3=xy%5E2&f=Plot.plotfunction%5Cu005fxy%5E2&f4=x&f=Plot.plotvariable%5Cu005fx&f5=0&f=Plot.plotlowerrange_0&f6=1&f=Plot.plotupperrange_1&f7=0&f=Plot.ymin%5Cu005f0&f8=1&f=Plot.ymax%5Cu005f1
• In the previous video, Sal starts of by saying that we're finding the volume under `z = xy²` that's bound by `y = x²` and the x-axis. So he figure it out from anywhere, it's part of the initial problem.