Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rewrite the integral as an equivalent iterated integral in the order

  1. Aug 1, 2011 #1
    I'm writing this on my iPhone. So, forgive me if the formatting isn't just write.
    1) problem statement: here is the region of the integration of the integral: -1 to 1, x^2 to 1, 0 to 1-y dzdydx. Those bounds go from the outermost integral to the innermost. The problem asks to rewrite the integral as dxdzdy and dzdxdy. I have the 3d drawing of the region as well as the xy, xz, yz plane drawings. I know the drawings are correct, because my professor gave them to me. I'm struggling to see why some of the bounds are what they are. Like, dzdxdy has 0 to 1-y as the bounds for dz, but based on the plane drawings, I don't see why it would not be -sqrt(1-z) to sqrt(1-z). Also, with dxdzdy, the bounds for dx are: -sqrt(y) to sqrt(y), but again I had -sqrt(1-z) to sqrt(1-z). This unit has given me a lot of trouble, so any help is greatly appreciated!
  2. jcsd
  3. Aug 1, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Can you scan the 3D drawing & the coordinate plane drawing? -- & post them.

    As far as your following question goes:
    After doing the inner integration, including plugging in the integration limits, (in this case the variable is z) the inner variable (z) should no longer be present.
    Furthermore, in the initial order of integration, it's y that goes from x2 to 1, so in integrating over x first, I would expect the bounds to include the √(y) .​

    z is related to y linearly, so it makes sense that y is in turn related to z linearly.

    Try doing the integrations using the various orderings. If they agree (and they should) use a fairly simple function (linear) as the integrand and again integrate using the various orderings. If various orderings don't agree in these two cases, something's wrong.
  4. Aug 1, 2011 #3
    Firstly, if we're interested in the order dzdxdy, then

    [tex]\int_a^b \int_{f_1(y)}^{g_1(y)} \int_{f_2(x,y)}^{g_2(x,y)} \ dz \ dx \ dy. [/tex]

    That is to say, the limits of integration on y must be constant (for otherwise you'd end up with variables in your final answer!), the limits of integration on x must be functions of y only (note that this includes constants), and the limits on z can be functions of both x and y (this also includes constant functions).

    It's easy to see why this must be true. If the limits on x were functions of z, then that z would never be "integrated out" and you'd end up with z's in your final answer.

    Here's what I like to do: make sure you have plots of the region in the x-y, x-z, and y-z planes. These are your targets, your reference. These pictures are what you're going to try to recreate when you switch order. You said your professor already gave you these. Good.

    On your paper, draw your own x-y, x-z, and y-z planes. These are going to be what we are drawing on and erasing, drawing and erasing, drawing and erasing, etc. You get the idea.

    Now start with the inner most variable. In this case that would be z. We have to try to find functions [itex] f_2 (x,y),g_2(x,y)[/itex] that bound z, i.e., functions such that

    [tex] f_2 (x,y) \leq z \leq g_2 (x,y).[/tex]

    This means we don't care about the x-y plot right now. Take a look at your y-z plane. What functions of y is z bounded by? It looks like on the bottom 0 and on the top 1-y. What bout the x-z plane? What functions of x is z bounded by? It looks like 0 and 1. So we have

    [tex]0 \leq z \leq 1-y[/tex]


    [tex] 0 \leq z \leq 1[/tex]

    from the y-z and x-z planes, respectively. Well we know what [itex]f_2[/itex] and [itex]g_2[/itex] are now! It's pretty clear that [itex]f_2(x,y)=0[/itex], but what about [itex]g_2[/itex]? Remember that the x-z plane is when y=0, so if [itex] g_2(x,y)=1-y[/itex], then that function will satisfy both of those inequalities.

    EDIT: Or, rather, at this point you could take notice that z is bounded below and above by the PLANES z=0 and z=1-y.

    Take your y-z and x-z plots and shade in this regions.

    So now we're after [itex]f_1,g_1[/itex], or, functions such that

    [tex] f_1 (y) \leq x \leq g_1 (y).[/tex]

    So, now we're interested in the x-y plot. Or, crap, maybe that should be the y-x plot. Why? Because we're looking for functions of y, so it would be more natural to look at the y-x plane with the x axis vertical. So we need a function of y that bounds x above. How about [itex] \sqrt{y}[/itex] ?! And a function of y that bounds x below...[itex] -\sqrt{y}[/itex] ! So now then [itex]f_1(y) = -\sqrt{y}[/itex] and [itex]g_2(y)=\sqrt{y}[/itex].

    Take your y-x plot (or x-y plot if you're comfortable) and shade in this region.

    And finally, we need the constants a,b bounding y, or,

    [tex] a\leq y \leq b.[/tex]

    But this is pretty straitforward. Look at your y-z and y-x plots with your shaded regions. We've got way too much "y space." On the y-z plane we've shaded a infinite triangular-looking region. On your y-x plot you've shaded an infinite region between the two arms of a parabola. What constants should y be between to make these regions look like your reference plots? Well,

    [tex] 0 \leq y \leq 1. [/tex]


    [tex] \int_0^1 \int_{-\sqrt{y}}^{\sqrt{y}}\int_0^{1-y} \ dz \ dx \ dy.[/tex]

    EDIT: I must say, this is a trial-and-error process usually. There's not really a solid algorithm that can be given. I'm a grad student in math and I often get myself turned around and confused with setting up triple integrals. Post again if you have any questions.
    Last edited: Aug 1, 2011
  5. Aug 1, 2011 #4
    Wow! You guys said in two posts, what I've been struggling to grasp in class for a few units. I never saw the formal lay out of the triple integral (the one Stringy posted), and I must say, it made setting up bounds much easier. Thank you guys for your help! I'll be working these problems much more slowly using that definition as my frame of reference. I have a feeling the rest of my practice problems will flow much more smoothly than when I was staring at 3D plots scratching my head. Stringy, what's the graduate work in mathematics like? I know that is probably a pretty broad question, but I just added Mathematics as my second major- economics is my first- at UGA. I'm trying to see what opportunities await me when I graduate in 2 years.
  6. Aug 1, 2011 #5
    Haha! Well I told a little fib. Insert "incoming" before "graduate student."

    There are others over in the Academic Guidance section that could give you an indication of what graduate work is like.
  7. Aug 1, 2011 #6
    Oh yes, and don't discount the 3D plots entirely! Like I mentioned in the edit to my original post, those functions [itex]f_2 (x,y),g_2(x,y)[/itex] can often be read off strait from the 3D plot. In our case they were planes.

    There will inevitably be integrals where [itex]f_2(x,y),g_2(x,y)[/itex] are functions where both x and y appear. In that case the 3D plot is very helpful. For instance, one of them may be a hyperbolic paraboloid or something.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook