Triple Integral

1. Feb 23, 2014

dwn

1. The problem statement, all variables and given/known data

Image attached of problem.

2. Relevant equations

Not sure how this applies.

3. The attempt at a solution

∫(o to ∞) ∫(0 to 5) ∫(0 to 4) 16xy dydxdz
Below, I have completed the integrations in order (dydxdyz):

∫dy : 8xy2|(0 to 4) = 128 x

∫dx : 64x2 = 64(25) = 1600

∫dz : 1600z| (o to ∞)

Do I set a limit on the integration dz, or set it at infinite?

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2. Feb 23, 2014

tiny-tim

hi dwn!
the limits of z depend on x and y

so you must integrate wrt z first (from 0 to … ? )

3. Feb 23, 2014

Dick

If you want to set this up as a triple integral, then since you are trying to find volume, your integrand should be 1. And I would integrate over dz first. And the upper value of z will correspond to its value on the boundary surface. Then integrate over x and y.

4. Feb 23, 2014

dwn

If I want to set this up as a triple integral? Just out of curiosity...what other method? I am going to calculate this using the triple, but I would like to know for my own selfish reasons.

How do I go about finding ;) ? When you say that it depends on x and y....I can't just make it any constant, can I?

5. Feb 23, 2014

Dick

You can also find the volume by integrating (height)dxdy. It's not really a different method, you just aren't writing the z integration out explicitly. Same thing. And the limit of z isn't any constant. It's the value of z on the bounding surface as a function of x and y. Don't think too hard about this.

6. Feb 23, 2014

dwn

#smackforehead : just set the integral dz from 0 to z....

7. Feb 23, 2014

Dick

.... means just set the integral dz from z=0 to z=16xy, right?

8. Feb 23, 2014

dwn

exactly. So does this "thickheaded" thinking go away at some point? :)

9. Feb 23, 2014

Dick

As far as this topic goes it should be going away already. You seem to be pretty clear on how obvious the answer is.

10. Feb 24, 2014

tiny-tim

hi dwn!

(just got up :zzz:)

when you perform the first integration (in this case, z) in a triple integral to find a volume,

you are essentially finding the volume of a tall thin square slice of the volume, of sides dx and dy …

its volume obviously is (approximately) dxdy times the height,

ie dxdy times [z(x,y) - zo(x,y)]​

where z = z(x,y) is the equation of the top surface, and z = zo(x,y) is the equation of the bottom surface (in this case, the constant z = 0)

btw, you will notice that the limits of integration for all variables except the last in a multiple integral will generally not be constants, but will generally depend on the later variables!