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Current time:0:00Total duration:9:51

Video transcript

in all of the double integrals we've done so far the boundaries on X&Y were fixed now we'll see what happens when the boundaries on x and y are variable so let's say I have the same surface and I'm just going to I'm not going to draw it the way it looks I'll just kind of draw it figuratively but the problem we're actually going to do is Z and this is the exact same one we've been doing all along but the point of here isn't to show you how to integrate the point of view is to show you how to visualize and think about these problems and frankly in double integral problems the hardest part is figuring out the boundaries once you do that the integration is pretty straightforward it's really not any harder than single variable integration so let's say that's our surface Z is equal to XY squared and let me draw the axes again so that's my x-axis that's my z-axis that's my y-axis x y and z and you saw what this graph look like several videos ago I took out the whole graph ER and we rotated in things I'm not going to draw the graph the way it looks I'm just going to draw it fairly abstractly it's just an abstract surface because the point here is really to figure out the boundaries of integration so before I actually didn't draw the surface I'm going to draw the boundary so every every the first time we did this problem we said okay X goes from 0 to 2 Y goes from 0 to 1 and then we figured out the volume above that bounded domain now let's do something else let's figure out let's say that X goes from 0 to 1 so X goes from 0 to 1 and let's say that the domain the the volume that we want to figure out under the surface it's not from a fixed Y to an upper bound Y although it kind of well I'll show you it's actually a curve so this is this is all on the XY plane everything I'm drawing here and this curve is we could view it two ways we could say Y is a function of X Y is equal to x squared or we could write X is equal to square root of Y we don't have to write plus or minus or anything like that because we're in the first quadrant so this is the area above which we want to figure out the volume let me and doesn't hurt to color it in just so we can really hone in on what we care about so that's the area above which we want to figure out the volume or kind of you could kind of say that's our bounded domain and so X goes from 0 to 1 and then this point is going to be what that's that point is going to be 1 comma 1 right 1 is equal to 1 squared 1 is equal to square root of 1 so this point is y is equal to 1 all right that's y is equal to 1 and then I'm not going to draw this surface exactly I'm just trying to give you a sense of what the volume of the figure we're trying to calculate is so the top of if this is just some arbitrary surface let me do it in a different color so this is the top this is what this line is going vertical in the Z direction all right actually I can draw it like this like it's a curve and then this curve back here is going to be like a wall right that's going to be like a wall and maybe I'll paint this side of the wall just so you can see what it kind of looks like trying my best think you get an idea let me make it a little darker this is actually more of an exercise in art than in then in math in many ways if you get the idea and then the boundary here is like this and this top isn't flat you know it could be a curved surface I do it a little like that but it's a curved surface and we know and the example we're about to do that this surface right here is Z is equal to x squared so we want to figure out the volume under this so how do we do it well let's think about it we could actually use the intuition that I just gave you we're essentially just going to take a DA right which is a little small square down here right and that little area that's the same thing as DX let me use a darker color as the DX times the dy and then we just have to multiply it times f of X Y or which is this for each area and then sum them all up and then we could take a sum in the X direction first or the Y direction first now before doing that just to make sure you that you have the intuition because the boundaries are the hard part let me just draw our XY plane so let me rotate it up like that I'm just going to draw our XY plane because that's what matters because the hard part here is just figuring out our bounds of integration so the curve is just y is equal to x squared so it'll look something like that y is equal to x squared this is the point y is equal to 1 this is y axis this is the x axis this is the point X is equal to 1 X is equal to 1 there you go that's not an X that's a 1 this is the x axis anyway so we want to figure out how do we sum up these d this DX times dy or this da along this domain along this domain so let's draw it let's visually draw it it doesn't hurt to do this when you actually do the problem because this frankly is the hard part a lot of calculus teachers will just have you set up the integral and then say okay well the rest is easy or the rest is calc 1 ok so this area this area here is the same thing as this area here so it's base is DX and its height is dy right and then we're going to and then you can imagine that we're looking at this thing from above so the surface is up here someplace and we're looking straight down on it and so this is just this area right so let's say we wanted to take the integral with respect to X first so we want to sum up we want to sum up the so if we want the volume above this column first of all is this area times DX dy right so let's write the volume above that column it's going to be the value of the function right the height at that point which is XY squared so it's going to be XY squared times DX dy DX dy that this expression gives us the volume above this area or this column right here and let's say we want to sum in the x-direction first so we want to sum that DX some one here some here etc etc right so we're going to sum in the x-direction so my question to you is what is our lower bound of integration what is our lower bound of integration well for any we're kind of holding our Y constant right and so if we go to the left if we go lower and lower X's we kind of bump into the curve here right so the lower bound of integration is actually the curve and what is this curve if we were to write X as a function of Y right this curve is y is equal to x squared or X is equal to the square root of Y so if we're integrating with respect to X for any for a fixed Y right here right where we're integrating in the horizontal direction first our lower bound is X is equal to the square root of Y X is equal to the square root of Y that's interesting I think this is the first time you've probably seen a variable bound and integral but it makes sense because for this row that we're adding up right here the upper bound is easy the upper bound is X is equal to 1 the upper bound is X is equal to 1 but the lower bound is X is equal to the square root of Y right because as you go back like oh I bump into the curve and what's the curve well the curve is X is equal to square root of Y because we don't know which Y we picked fair enough so once we figured out the volume so that'll give us the volume above this rectangle right here and then we want to add up the D wise right and remember there's a whole volume above what I'm drawing right here right I'm just drawing this part in the XY plane all right so now so what we've done just now this expression as it's written night now figures out the volume above that rectangle now if we want to figure out the entire volume of the of the solid we integrate along the y axis or we add up all the D Y's this was a dy right here not a DX my DX is and dy is a similar so now what is the lower bound what is the lower bound on the y-axis if I'm summing up these rectangles well the lower bound is y is equal to zero right so we're going to go from Y is equal to zero to what what is the upper bound to Y is equal to one and there you have it let me rewrite that integral so the double integral is going to be from X is equal to square root of Y to X is equal to 1 X Y squared DX dy and then the Y bounds Y goes from 0 to Y to 1 I've just realized I've run out of time in the next video we'll evaluate this and then we'll do it in the other order see you soon