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Homework Help: Setting up a Triple Integral to Find Volume Enclosed by Two Functions

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the region of space enclosed between the functions: 1=-z+2x+2y and 100=z2+y2+x2.

    3. The attempt at a solution
    I am not sure how to set this problem up. I think it is a triple integral, since there is a z-component. I graphed the equation in an program and I see that it it the former function which is a curved sheet, intersecting a sphere (the latter function) with a radius of 10.

    I tried finding their intersection points so that I could understand what the bounds would be for each component, but can't figure it out after several hours.

    Will somebody show me how to set it up so that I can try the triple integral? This practice assignment is more about evaluating integrals, so that is the main part of the work, but I just need it set up first before I can practice evaluation...

    Thanks in advance!
  2. jcsd
  3. Dec 11, 2012 #2


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    The first thing you want to do is express the functions in terms of z (x or y work too but whatever right?).

    This will give you your limits for z.

    Then fix z=0 so you can find the points of intersection. After you find the points of intersection you can determine the intervals for x and y.

    Then you can solve the intersection for either x or y yielding your limits for either x or y respectively. Then you simply determine your limits for x or y depending on which variable you allowed to vary when you solved the intersection.
  4. Dec 11, 2012 #3
    Thanks for the reply... Let me try this:

    So for the first function: z = 2x+2y-1

    And for the second function: z = √(-x2-y2+100)

    So those are the limits in the triple integral for z?

    I am not quite sure what you mean by "fix z=0 so you can find points of intersection." Would you mind giving an example?

  5. Dec 11, 2012 #4


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    Of course.

    Suppose we have a region bounded by [itex]z=4-x^2-y^2[/itex] and [itex]z=x^2+y^2-4[/itex] calculate the volume of the solid bounded by these regions.

    Clearly you have your limits for z right there.

    Suppose we hold z = 0 fixed. Could you tell me where the two curves intersect?
  6. Dec 11, 2012 #5
    Would you set them equal to each other, then move them to one side of the equals sign to make them zero?

    I did [itex]z=4-x^2-y^2[/itex] and [itex]z=x^2+y^2-4[/itex]
    so 4-x2-y2 = x2+y2-4
    then 2x2+2y2=0...

    not sure what I would do next...

    Also, did I express the functions in terms of z correctly? And when you say "limits of z" do you mean those are the upper and lower bounds on that integral when I set it up?

    Thanks for the help thus far.
  7. Dec 11, 2012 #6


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    Woah woah, I think you may have made an arithmetic error. You do indeed get a circle... but clearly a circle of radius zero makes no sense right?

    Lets deal with this first then the rest will be clear to you.

    Also yes I mean that those are the upper and lower bounds on the integral for z.
  8. Dec 11, 2012 #7
    Sorry, 2x2+2y2=8.

    so x2+y2=4.

    Do I set something equal to zero to find an intersection?
    Is it y=±2 and x=±2?
  9. Dec 11, 2012 #8


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    Okay so the two curves intersect on circle of radius 2 with center (0,0). That means either x or y run from -2 to 2. You can pick EITHER one.

    Now you can solve x2+y2=4 for your limits on either x or y depending on which one you picked to run from -2 to 2.

    Could you tell me what your integral looks like now? ( Don't even solve for f(x,y) just show me your limits ).
  10. Dec 11, 2012 #9
    The limits for x would be:√(4-y2)
    And the limits for y would be : √(4-x2)

    Is that right?
  11. Dec 11, 2012 #10
    Or do I have to plug -2 and 2 in for one of them?

    If I made the limits for y -2 and 2, how would that problem be solved? Just by plugging in..?
  12. Dec 11, 2012 #11


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    Okay so what we have so far :

    The two curve intersect on the circle of radius 2 and center (0,0).

    So lets suppose we let : -2 ≤ y ≤ 2

    So we want to solve x2+y2 = 4 for x.

    Solving this we get : x = ± √4 - y2 ( Notice the ± )

    Could you set up the integral for me know? You know your limits for z, x and y.
  13. Dec 11, 2012 #12
    Okay, would it be ∫(4−x2−y2 to x2+y2−4)∫(-2 to 2)∫(-√4 - y to +√4 - y) ... dxdydz


    that "..." is another place where I'm not sure what belongs...
  14. Dec 11, 2012 #13


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    Not quite. We want to hold y fixed as long as possible and allow x and z to vary. So your integral would look like :

    [itex]\int_{-2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \int_{4 - x^2 - y^2}^{x^2 + y^2 - 4} f(x,y,z) dzdxdy[/itex]

    Notice we hold y fixed as long as possible?

    Okay, as for f(x,y,z), think about which of our two surfaces is bigger than the other and subtract them to get your function to integrate. ( Bigger one - Smaller one ).
  15. Dec 11, 2012 #14
    I just graphed both surfaces and it looks like two equally sized paraboloids, one upside down and the other right side up... How do I know which is larger?

    Also, why do we want that specific order for the integral? Is y supposed to be fixed for as long as possible because we made it have numbers and not functions of another variable for its limits? But then, why is dz on the inside and not dx? Is that because it is the one that has limits in terms of two variables?

    Sorry for all the questions, but I can feel it getting clearer!
  16. Dec 11, 2012 #15


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    First note :

    [itex]\int_{-2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \int_{4 - x^2 - y^2}^{x^2 + y^2 - 4} f(x,y,z) dzdxdy[/itex] = [itex]\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{4 - x^2 - y^2}^{x^2 + y^2 - 4} f(x,y,z) dzdydx[/itex]

    So there is more than one way to represent your integral and get the same answer. Like I said before, we could have fixed x OR y for as long as possible and allowed the other to vary respectively.

    dz is on the inside because we have expressed both surfaces in terms of z. We could have also expressed the surfaces in terms of x and y, but i'll let you try that for yourself and observe why we took the convenient way out by integrating with respect to z first :).

    As for your surfaces, if you're inside the circle x2 + y2 < 4, then x2 + y2 - 4 < 0 which implies that 4 - x2 - y2 > 0.

    This should clear up why the limits of integration on z are the way they are and allow you to find your f(x,y,z) since you know which surface is bigger than the other now right?

    Sorry now though, I have to go pass out for my exam tomorrow. Just apply this same logic to your other problem once you figure this one out and it's smooth sailing.
  17. Dec 12, 2012 #16
    Awesome... Before you go, could you just show how I should subtract the two surfaces for the inner function? I have gotten this far on my own problem now, just need that last step which I am unsure of...
  18. Dec 12, 2012 #17
    Would it be (4-x^2-y^2) - (x^2+y^2-4) which is -2x^2-2y^2 +8?
  19. Dec 12, 2012 #18
    If anybody can still help me here... Does this method work for my problem? It is a sphere being cut by a sheet... If so, how can I set up the intersection? I am stuck at 2^x+2^y-1=sqrt(-x^2-y^2+100)
  20. Dec 12, 2012 #19
    Should this be cone in spherical coordinates? If so, how would it be set up?

    I am absolutely lost trying to find where these functions intersect. Please help!!
  21. Dec 12, 2012 #20


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    This may be all fine and dandy, but it appears to be a much simpler problem than the OP. The OP has 2x and 2y in it, and I fail to see how this approach (or any other!) will handle that.
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