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Math Help - need help getting a triple integral problem started

  1. #1
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    need help getting a triple integral problem started

    Find the volume of the region bounded by the paraboloid z = x^2 + y^2 + 4 and the planes x = 0, y = 0, z = 0, x + y = 1

    So I believe that the limits are:

    z : 0 to x^2 + y^2 + 4

    y : 0 to 1-x

    x : 0 to.... here's where I'm lost, if I set x = 1-y I wouldn't be able to solve this integral, right?

    Also I don't understand what I'm supposed to be integrating, is it just
    x^2 + y^2 - z + 4 because it asks for the volume of the paraboloid?
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    Re: need help getting a triple integral problem started

    Hey mrb165.

    If the region you are integrating is not smooth (but continuous), then you will need to split up the integral into multiple parts. Since you have a paraboloid and a plane, there is a good chance of this happening.

    To start you off in the right direction you should find where the x+y=1 plane intersections the paraboloid and look at partitioning the integral based on where this intersection is.

    The analogy is that if you integrate say |x|dx over [-5,5] then you need to calculate [-5,0] and [0,5] when you are integrating.
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    Re: need help getting a triple integral problem started

    Start by drawing a picture. If you are really good draw a three dimensional picture, but, fortunately for me, that isn't really necessary. Yes, you are right that for each (x, y), z goes from 0 (because we are told that z= 0 is one boundary) up to z= x^2+ y^2+ 4.

    Now, we are down to the limits on the x and y integrals so we are in the xy- plane and that is easy to draw. One boundary is, of course, x= 0, another y= 0, and the third is the line x+ y= 1. That is, of course, the line from (0, 1) to (1, 0). For each x, y goes from 0 up to 1- x. Then, of course, with no further restriction on x (such as "for each x" or "for each (x, y)", x must go from 0 to 1. In order to find the integral, you want to integrate "dV= dzdydz"; \int\int\int dV= V.

    So you want to integrate int_0^1\int_0^{1- x}\int_0^{1- x}\int_0^{x^2+ y^2+ 4} dzdydx. I can see where your confusion is, though- look what you get after doing that first integral, \int_0^{x^2+ y^2+ 4} dz.

    Other ways to do this: once you are in the xy-plane, you could also argue that, for each y x goes from 0 to 1- y and then ygoes from 0 to 1. That would give
    \int_0^1\int_0^{1- y}\int_0^{x^2+ y^2+ 4} dzdxdy
    Last edited by HallsofIvy; July 27th 2013 at 05:37 PM.
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    Re: need help getting a triple integral problem started

    ok, here's how I set it up....

    1 1-x x+y+4
    ∫ ∫ ∫ dzdydx
    0 0 0

    then I got this....

    1 1-x
    ∫ ∫ [x^2 + y^3 +4]dydx
    0 0

    and then I kept integrating and now I'm stuck at this...

    1
    ∫ ((1-x)^3)/3 + 4 -4x dx
    0

    I have no idea how to integrate this now, I'm not sure I even did any of this right...
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    Re: need help getting a triple integral problem started

    Quote Originally Posted by mrb165 View Post
    ok, here's how I set it up....

    1 1-x x+y+4
    ∫ ∫ ∫ dzdydx
    0 0 0

    then I got this....

    1 1-x
    ∫ ∫ [x^2 + y^2 +4]dydx
    0 0

    and then I kept integrating and now I'm stuck at this...

    1
    ∫ ((1-x)^3)/3 + 4 -4x dx
    0

    I have no idea how to integrate this now, I'm not sure I even did any of this right...
    Is this a typo or is this what you actually got?
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    Re: need help getting a triple integral problem started

    Quote Originally Posted by Prove It View Post
    Is this a typo or is this what you actually got?
    yeah, thats what I got after the first integration, then it would become:

    1 1-x
    ∫ [x^2 + (y^3)/3 +4y] dx
    0 0

    right? (1-x and 0 should be to the right side)
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    Re: need help getting a triple integral problem started

    ok hold on I forgot you add the variable to constants now I have:


    1
    ∫ x^2(1-x) + (1-x)^3/3 + 4(1-x) dx
    0

    is this right so far? I'm not sure if I know how to integrate this
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    Re: need help getting a triple integral problem started

    Quote Originally Posted by mrb165 View Post
    yeah, thats what I got after the first integration, then it would become:

    1 1-x
    ∫ [x^2 + (y^3)/3 +4y] dx
    0 0

    right? (1-x and 0 should be to the right side)
    \displaystyle \begin{align*} \int_0^{x^2 + y^2 + 4}{dz} = x^2 + y^2 + 4 \end{align*}, NOT \displaystyle \begin{align*} x^2 + y^3 + 4 \end{align*}...
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    Re: need help getting a triple integral problem started

    math help.pdf

    ok, just figured out you can post images... anyway I got -4/3 for the volume so I think it's wrong, not sure where I went wrong though...
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