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Homework Help: Easy triple integral problem

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Use a triple integral to find the volume of the solid.
    The solid in the first octant bounded by the coordinate planes and the plane 3x+6y+4z=12

    2. Relevant equations


    3. The attempt at a solution
    So I'm using the triple integral type of dzdydx.

    At first I set z and x to 0, then z and y to 0 to obtain x and y limits but I think that was incorrect. I'm trying to integrate 3x+6y+4z-12.

    I guess the upper limits of z will be the above equation done in editor, z=3-3x/4-3y/2 and lower limit will be 0?

    That integration gets really dirty really quick. I doubt it is right. Can someone help point me in the right direction?
  2. jcsd
  3. Apr 28, 2010 #2


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    that sound like the way to go, why not try...

    shouldn't get too bad as its just polynomials
    [tex] \int_0^{???}dx \int_0^{???} dx\int_0^{3-\frac{3}{4}x-\frac{3}{2}y} dz[/tex]
  4. Apr 28, 2010 #3


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    lanedance wrote that integral in a very pecular way- especially with two "dx"s! Clearly a typo.

    I would write it as
    [tex]\int_{x= 0}^a \int_{y=0}^{f(x)}\int_{z= 0}^{3- \frac{3}{4}x- \frac{3}{2}y}dz dy dx[/tex]

    That last integral, with respect to x, must give a numeric answer, of course, so the two limits must be numbers. What is the largest value x can take in this figure?

    The second integral, with respect to y, can give a function of x. For every x, what is the largest value y can have? Projecting this figure into the z value is the same as taking z= 0. Since 3x+ 6y+ 4z= 12 becomes the line 3x+ 6y= 12 when z= 0: 6y= 12- 3x so what is y as a function of x?

    Of course, for each (x, y), z goes from 0 up to the plane 3x+ 6y+ 4z= 12 which is what gives the formula for z that you give and the upper limit for the "z" integral.
  5. Apr 28, 2010 #4


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    yep thanks Halls, meant
    [tex] \int_0^{a}dx \int_0^{f(x)} dy\int_0^{3-\frac{3}{4}x-\frac{3}{2}y} dz[/tex]

    nice explanation, apologies if the notation confused things, I find it easier to keep track of the integrals in the operator notation.

    Frozenguy, meant the same thing as Hall's post, but work with whatever you're comfortable with.
  6. Apr 28, 2010 #5
    Ok so y=2-x/2 and that is the upper limit for dy?
    So when it comes to that upper z limit getting squared I distributed the whole thing but now I'm thinking I didn't have to haha.

    So is that the right track for the limits? and dx limits will be [0,4] correct?
  7. Apr 28, 2010 #6


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    sounds good - not sure why you were squaring things, but sounds like you're on the right track now
  8. Apr 28, 2010 #7
    So can someone see any red flags? I'm getting a negative volume which doesn't make sense and plus the book says the answer is 4.
  9. Apr 29, 2010 #8


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    your initial integrand should just be the volume element dV = dx.dy.dz = 1.dx.dy.dz
  10. Apr 29, 2010 #9


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    In the form Hall's wrote:

    [tex]\int \int \int dV
    = \int_{x= 0}^4 \int_{y=0}^{3-x/2}\int_{z= 0}^{3- \frac{3}{4}x- \frac{3}{2}y}1.dz dy dx[/tex]
  11. Apr 29, 2010 #10
    oh... Why is that? lol I'll try to read up on it but why dont we need to integrate a specific formula?

    Thanks so much for your help
  12. Apr 29, 2010 #11


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    no think of the integral as [itex]\int_0^a f(x) dx [/itex] adding up thr contribution from infintesimal itegrand (f(x) dx) over 0<x<a. If f(x) =1, [itex]\int_0^a f(x) dx [/itex] just adds up each incremental length dx to find the total 'x' length.

    For the volume integral its the same, [itex]\int \int \int dV = \int \int \int dx.dy.dz[/itex]. So as we want to calcaute the volume intergal think of it as adding up each infinitesimal block dV = dx.dy.dz, so the "function" to integrate must be identically 1.
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