Integrating in Spherical Co-Ordinates.

In summary, the conversation discusses converting an integral to spherical coordinates and finding the correct limits for integration. The final integrand is determined to be \frac{\rho^{2}\sin\phi}{1+\rho^{2}} and a possible method of integration is suggested.
  • #1
Chaz706
13
0
I have the following Integral

[tex] \int ^1 _0 [/tex] [tex] \int _0 ^\sqrt{1-x^2} [/tex] [tex] \int _0 ^\sqrt{1-x^2-y^2} [/tex] [tex] \frac{1}{1+(x^2)+(y^2)+(z^2)} dzdydx [/tex]

(With the limits working properly!)

Converted to spherical Cor-ordinates, I have

[tex] \int ^\frac{\pi}{2} _0 [/tex] [tex] \int _0 ^\frac{\pi}{2} [/tex] [tex] \int _0 ^1 [/tex] [tex] \frac{1}{1+\rho} \rho^2 sin(\phi) d\rho dr d\phi [/tex]

I've converted the function, but how would I start integrating?
 
Last edited:
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  • #2
Alright.Transform it in spherical coordinates.But u need the limits...

Daniel.
 
  • #3
I have the limits... it's just that I can't get them right on Latex (stupid coding! I'm getting it right, it's just not displaying it that way!)

Hang on...

EDIT: Problem above now has working limits, and my original question as intended.
 
Last edited:
  • #4
A substitution MIGHT work, but i would probably go for integration by parts. Remember sin(phi) is constant for the first integral.
 
  • #5
[tex] \frac{r^{2}}{1+r}=r-1+\frac{1}{1+r} [/tex]

is all u need.

Daniel.
 
  • #6
Your integrand in spherical coordinates should be: [tex]\frac{\rho^{2}\sin\phi}{1+\rho^{2}}[/tex]
 
  • #7
I've got this thanks. Thanks to a form of integration in the back of my book.
 

Related to Integrating in Spherical Co-Ordinates.

1. What are spherical coordinates and how are they different from Cartesian coordinates?

Spherical coordinates are a system of representing points in three-dimensional space using two angles (azimuth and elevation) and a distance from the origin. This is different from Cartesian coordinates, which use three distances from perpendicular axes to represent a point.

2. What is the purpose of integrating in spherical coordinates?

Integrating in spherical coordinates allows us to solve problems involving three-dimensional objects with spherical symmetry. This can be useful in physics, engineering, and other fields where spherical objects are commonly encountered.

3. How do I convert between spherical and Cartesian coordinates?

To convert from spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z), you can use the following formulas:

x = r sinφ cosθ

y = r sinφ sinθ

z = r cosφ

Converting from Cartesian to spherical coordinates involves solving for r, θ, and φ using the equations x = r sinφ cosθ, y = r sinφ sinθ, and z = r cosφ.

4. What types of integrals can be solved using spherical coordinates?

Spherical coordinates can be used to solve triple integrals, which involve integrating over three variables (usually x, y, and z). These integrals are commonly used to calculate volumes, moments of inertia, and other physical quantities.

5. Can I use spherical coordinates for any three-dimensional object?

Spherical coordinates are best suited for objects with spherical symmetry, such as spheres, cones, and spherical shells. They may also be useful for objects with cylindrical symmetry, but they are not as effective for objects with other shapes or symmetries.

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