HELP Setting up triple integral in spherical coordinate

In summary, the conversation discusses setting up a triple integral in spherical coordinates, specifically for two different problems. The first problem involves encountering an error in the calculation due to a zero value for the denominator, and the suggestion is made to use cylindrical coordinates instead. The second problem involves an error in using the int() command and manually evaluating the integrals in Maple.
  • #1
kyva1929
18
0
HELP! Setting up triple integral in spherical coordinate

Homework Statement



http://img517.imageshack.us/img517/9139/83291277.jpg

Homework Equations



I set up the bound for this problem as following:

r=0..2/cos(phi), phi=pi/2..3pi/4, theta=0..2pi, but maple always return an error in calculating such integral. I've included the Jcobian transformation in the integral, and convert x^2+y^2 to rho^2 sin(phi). I found that the problem was due to the denominator cos(phi) become zero at the lower bound. But the volume representing the respective volume integral makes sense, so how to correct for this? Am I setting up the integral wrong?

second problem:

http://img695.imageshack.us/img695/3585/38264414.jpg
The same thing happens, please help!

Thank you!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2


Use cylindrical coordinates instead.
 
  • #3


Thank you in advance. But is there a way to do it in spherical coordinate?

And for question 2, I did the command int(function,[x=..,y..,z..]) and it returns error

Error, (in assuming) when calling '`property/ConvertProperty`'. Received: 'real is an invalid property'

but I can evaluate it either by hand or by evaluating the integrals one by one in maple and that gives me [2*Pi*(3^21-1)]/21.

What's wrong?

Thank you
 

1. What is a triple integral in spherical coordinates?

A triple integral in spherical coordinates is a type of integral used to calculate the volume of a three-dimensional object with a curved surface. It involves integrating a function over a region in space defined by spherical coordinates, which are based on a sphere with a center point, a radius, and two angles.

2. How do I set up a triple integral in spherical coordinates?

To set up a triple integral in spherical coordinates, you need to first identify the limits of integration for each of the three variables: r, θ, and φ. These limits will depend on the shape and orientation of the object you are trying to find the volume of. You can then use the formula ∫∫∫ f(r,θ,φ) r²sinθ dr dθ dφ to set up the integral.

3. What is the difference between a triple integral in spherical coordinates and Cartesian coordinates?

The main difference between a triple integral in spherical coordinates and Cartesian coordinates is the coordinate system used. In spherical coordinates, the variables r, θ, and φ represent distance from the origin, angle from the positive z-axis, and angle from the positive x-axis, respectively. In Cartesian coordinates, the variables x, y, and z represent distance from the origin in the x, y, and z directions.

4. How do I convert a triple integral from Cartesian coordinates to spherical coordinates?

To convert a triple integral from Cartesian coordinates to spherical coordinates, you can use the following equations: x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ. These equations can be used to express the Cartesian variables in terms of the spherical variables, allowing you to set up the integral in terms of r, θ, and φ.

5. What are some applications of triple integrals in spherical coordinates?

Triple integrals in spherical coordinates are commonly used in physics, engineering, and other fields to find the volume of objects with curved surfaces, such as spheres, cones, and cylinders. They can also be used to calculate the mass, center of mass, and moment of inertia of three-dimensional objects. Additionally, they have applications in calculus and vector calculus, particularly in the calculation of flux and surface area.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
994
  • Calculus and Beyond Homework Help
Replies
3
Views
557
  • Calculus and Beyond Homework Help
Replies
4
Views
752
  • Calculus and Beyond Homework Help
Replies
7
Views
699
  • Calculus and Beyond Homework Help
Replies
4
Views
958
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
958
  • Calculus and Beyond Homework Help
Replies
9
Views
163
  • Calculus and Beyond Homework Help
Replies
4
Views
896
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top