How Do You Calculate the Volume of a Volcano Using Triple Integrals?

In summary, the volume of the volcano between the graphs z=0 and z=1/(x^2+y^2)^24, and outside the cylinder x^2+y^2=1 can be found by setting up a triple integral in cylindrical coordinates with limits of integration for z going from 0 to 1/r^48, for r going from 1 to infinity, and for theta going from 0 to 2pi. The integrand should be evaluated as r^2/2 between the limits of integration for r, resulting in the final answer of (2pi/2303)/2.
  • #1
seichan
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Homework Statement


A volcano fills the volume between the graphs z = 0 and z =1/(x^2+y^2)^24, and outside the cylinder x^2+y^2=1 Find the volume of this volcano.


Homework Equations


This is a triple integral to be evaluated in cylindrical coordinates.



The Attempt at a Solution


Alright, so, this is what I've done. I don't understand what I'm doing wrong. I set up a triple integral, the limits of integration as follows:
z goes from 0 to 1
theta goes from 0 to 2pi
and r goes from 1 to z^-1/48 (outside the unit circle, but between the other graphs)

I first integrate with respect to r. I get r^2/2, evaluated between z^-1/48 and 1. This gives the integrand (z^-1/2304)/2-1/2. Then, I integrated with respect to z. This gives the answer ((2304z^(2303/2304)/2303)/2)-z/2. Then, this is evaluated from 0 to 1. This gives (2304/2303)/2-1/2, or, (1/2303)/2. This is integrated with respect to theta, giving (theta/2303)/2. Evaluated at 2pi, this results in (2pi/2303)/2. However, my homework software has told me this is wrong multiple times. Where am I going wrong? Thank you for any assistance.
 
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  • #2
z does NOT go from 0 to 1. It goes from 0 up to 1/r48. r goes from 1 to infinity:
[tex]\int_{\theta= 0}^{2\pi}\int_{r= 0}^\infty\int_{z= 0}^{1/r^{48}} r dz dr d\theta[/tex]
 

FAQ: How Do You Calculate the Volume of a Volcano Using Triple Integrals?

What is a triple integral?

A triple integral is a mathematical concept that involves integrating a three-dimensional function over a three-dimensional region. It is used to calculate the volume of irregularly shaped objects that cannot be easily measured using traditional methods.

How is a triple integral different from a regular integral?

A regular integral involves integrating a one-dimensional function over a one-dimensional interval, while a triple integral involves integrating a three-dimensional function over a three-dimensional region. In other words, a regular integral is used to calculate the area under a curve, while a triple integral is used to calculate the volume of a solid.

What is the process for setting up a triple integral?

The process for setting up a triple integral involves identifying the limits of integration for each variable (x, y, and z) and then determining the appropriate order of integration. This is typically done by visualizing the region in three dimensions and breaking it down into smaller, easier-to-calculate sections.

How is the volume of a region calculated using a triple integral?

The volume of a region can be calculated using a triple integral by integrating the function 1 over the given region. This means that each point within the region will contribute a value of 1 to the total volume, and the resulting integral will give the total volume of the region.

What are some real-life applications of using triple integrals to calculate volume?

Triple integrals are commonly used in engineering, physics, and other scientific fields to calculate the volume of complex three-dimensional objects, such as fluid flow through pipes, heat transfer in a solid, or the mass of an irregularly shaped object. They are also used in computer graphics to create 3D models and animations.

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