# Solving an Integral: ∫(o to ∞) ∫(0 to 5) ∫(0 to 4) 16xy dydxdz

• dwn
In summary, the conversation is about setting up a triple integral to find volume, and the method of integrating over dz first before integrating over x and y. The limits of integration for z depend on the value of z on the bounding surface, which is a function of x and y. It is also mentioned that the limits of integration for all variables in a multiple integral will generally depend on the later variables.
dwn

## 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?

#### Attachments

• Screen Shot 2014-02-23 at 6.43.03 PM.jpg
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hi dwn!
dwn said:
∫(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
dwn said:

## 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?

dwn said:
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...

dwn said:
#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? :)

dwn said:
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
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!

## What is the purpose of solving integrals?

The purpose of solving integrals is to find the area under a curve or the volume of a three-dimensional shape. It is a mathematical tool that is commonly used in physics, engineering, and other scientific fields.

## What is the process for solving integrals?

The process for solving integrals involves breaking down the integral into smaller parts, using integration rules and formulas to solve each part, and then combining the solutions to find the final answer. This process may vary depending on the type of integral being solved.

## What are the limits of integration in this specific integral?

The limits of integration in this integral are 0 to ∞ for the z-axis, 0 to 5 for the y-axis, and 0 to 4 for the x-axis. This means that the integral is being evaluated over a three-dimensional region bounded by these limits.

## What is the significance of the numbers 16, x, and y in this integral?

The numbers 16, x, and y represent the coefficients and variables in the integrand. They are used to represent the function being integrated over the given limits. In this case, the function is 16xy, which is being integrated over the three-dimensional region.

## What are some real-world applications of solving integrals?

Solving integrals has many real-world applications, such as calculating the area under a velocity-time graph to determine an object's displacement, finding the center of mass of an object, and calculating the work done by a force. It is also used in fields such as economics, biology, and statistics to analyze data and make predictions.

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