# Triple Integral

## Homework Statement

Image attached of problem.

## Homework Equations

Not sure how this applies.

## 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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tiny-tim
Homework Helper
hi dwn! ∫(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?

the limits of z depend on x and y

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

• 1 person
Dick
Homework Helper

## Homework Statement

Image attached of problem.

## Homework Equations

Not sure how this applies.

## 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?

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.

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?

Dick
Homework Helper
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?

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.

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

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

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

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

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

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

• 1 person
tiny-tim
Homework Helper
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!