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# Double integrals 5

Finding the volume when we have variable boundaries. Created by Sal Khan.

## Want to join the conversation?

• Dont you bump into the y=x^(2) also on the y axis? (i am questioning y=0 and y=1 as bounderies)
• When you integrate x from sqrt(y) to 1 it eliminates the need to consider the curve when integrating y. This can be seen at by the shorter rectangle on top of the longer one. The dx determines the length of the rectangle so the dy only needs to consider how high these rectangles are stacked. Hope that makes sense.
• how would you know that y=x^2 if all you are given is that z=xy^2 ?
• They are not dependent on each other.

In the last video, y was a constant or a straight line. In this video, y is a function and y=x^2 was just chosen randomly because it's a familiar graph. Pretty much any function could have been chosen as an example as the y boundary.

z=xy^2 defines the surface at the top of the solid. Sal drew the top as a flat solid, but he does mention the fact that the top actually has the shape of the surface z=xy^2. Visualize the wedge as having a top that is not flat.
• If z=xy^2, shouldn't the wedge thingy shrink on the z-axis as both x and y approach 0? For example, at (x,y) = (0,0) shouldn't the z (or height) be 0 as well? Thanks.
• You're right that z should go to 0 near the origin, but Sal wasn't drawing an exact representation of xy^2 in his diagram. He was just drawing a generic surface to show the process.
• If you integrated with respect to y first (dy*dx, not dx*dy) - would the lower boundary then be y=x^2?
• The lower would be 0, the upper y=x^2
• Thank you so much, it's a really interesting website,
but i have a question, and it seems I did not get the idea,
how do you determine y=x² ? I think it will help me in the determination of the bounds... I always have some difficulties with that. Thanks a lot for your help.
• I think it is just an abitrary fuction given in maths problems to bound a region so that you have a bounded area to find the volume above it. I don't think you need to find it yourself.
• in question z=xy^2 why did you draw y=x^2
• In this problem you are given 3 pieces of information in order to solve for the volume under the curve z=xy^2.
1) The limits of integration over the z axis, those being z=0 and the curve itself, z=xy^2, which represents the ceiling or upper boundary or limit of integration of the volume on the z axis.
2) The limits of integration over the x axis, those being x=0 to x=1.
3) The limits of integration over the y axis, those being y=0 and y=x^2.
So Sal drew y=x^2 because it was a limit of integration for the problem, NOT because it has any relation with the curve z=xy^2.
This example is to show how to proceed when your limits of integration are themselves functions, or a Sal says, "variable boundaries".
This is a SUPER IMPORTANT skill in multivariable calculus - often figuring out the boundaries of integration is harder than doing the integration itself!
Keep Studying and Keep Asking Questions!
• I don't get intuitively why y does not have variable bounds. Doesn't y "bump into" the curve on the top?
• I don't know if this helps... but.. The first integral we must take into account for the curve, that is why we integrate from the edge of the curve to the x-value "1". Then, when we integrate along the y-axis, we have already taken account of the curve, so we don't need to re-integrate it again. So, we just integrate along it's minimum and maximum limits, from 0 to 1. I hope this makes sense. If not, Sal explains it in Double Integral video number 6.
• at I do not understand why it is our lower bound sqrt of y, from were we concluded that that's the funtion of our curved? thanks
• The curved shape that we are integrating at that moment of the video is bounded by x = 1 on the upper end and by the curve formed by y = x² at the lower end. As Sal says, we don't know what point we would chose, so we need a value of x . We need to put the bounds in terms of x so the next step is to find the inverse of y = x² . We can do that by taking the square root of both side of y = x² which gives √y = x
• I don't get this at all !! at why can't we take the limits for x between 0 and 1 itself?? When we know that x varies between 0 and 1 why are we taking variable limits?? and why have we taken the function y=x^2? why not y=x?? (isn't that way easier to integrate??)
Finally, is a double integral and a surface integral the same???
thanx!
• We might want to answer your questions in a slightly different order that you asked them in, just to get some foundation.
Firstly, why are we integrating over y=x^2. This is a good question, and for this video, it is totally arbitrary. What Sal is doing is finding the volume under the surface given by z=f(x,y)=xy^2, but only finding that volume under the domain (x,y plane) of y=x^2. He could just as easily have chosen y=x as one of the domain bounds, or could have chosen the domain to be x^2+y^2 = 1 (then he would be finding the volume contained under a circle of radius 1 on the x-y plane. Choosing y=x is slightly easier, but you have to remember that this is not the function that is being integrated. They only determine the bounds of where we want our integration to start and stop within our domain.

Secondly, why can't we just say x varies from 0-1. Well... This has to do with the bounds given again. if we did that, then we would be finding the volume of the function under the rectangle from x=0 to x=1 and y=0 to y=1. We don't want that. we want to find the volume under the specific curve y=x^2. Now... lets look when y=1/2. At this fixed value of y, we do not start integrating x (the yellow) at 0, but rather at the start of our "curve" on the domain, or y=x^2. This is why this must be used as the lower bound of integration. It has to do with the domain that we are integrating over.

Finally, No this is not the same as a surface integral. Sal covers this in future videos.