Spherical Coordinates Triple Integral

In summary, the conversation discusses using spherical coordinates to evaluate a triple integral and finding the correct volume element to get the right answer. The conversation reveals that the initial attempt was incorrect due to an error in the volume element, and eventually the correct answer is found.
  • #1
qamptr
10
0
I thought this question was elementary... but I apparently know less than I thought I did.

Homework Statement


Use spherical coordinates to evaluate [tex]\iiint_{E} x^{2}+y^{2}+z^{2}dV[/tex]
Where E is the ball [tex]x^{2}+y^{2}+z^{2}\leq 16 [/tex]


Homework Equations


[tex]x^{2}+y^{2}+z^{2}=\rho^{2}[/tex]



The Attempt at a Solution


[tex]\int^{2\pi}_{0}\int^{\pi}_{0}\int^{4}_{0}\left(\rho^{2}\right)\rho Sin \left( \phi \right) d\rho d\phi d\theta = 256\pi[/tex]

which is apparently incorrect. Where am I going wrong?
 
Physics news on Phys.org
  • #2
qamptr said:
I thought this question was elementary... but I apparently know less than I thought I did.

Homework Statement


Use spherical coordinates to evaluate [tex]\iiint_{E} x^{2}+y^{2}+z^{2}dV[/tex]
Where E is the ball [tex]x^{2}+y^{2}+z^{2}\leq 16 [/tex]


Homework Equations


[tex]x^{2}+y^{2}+z^{2}=\rho^{2}[/tex]



The Attempt at a Solution


[tex]\int^{2\pi}_{0}\int^{\pi}_{0}\int^{4}_{0}\left(\rho^{2}\right)\rho Sin \left( \phi \right) d\rho d\phi d\theta = 256\pi[/tex]

which is apparently incorrect. Where am I going wrong?
You may want to check your volume element :wink:.
 
  • #3
Hootenanny said:
You may want to check your volume element :wink:.

Am I blind? I don't understand.
 
  • #4
Hootenanny said:
You may want to check your volume element :wink:.

[tex]\left(\rho^{3}\right) \rho[/tex] instead of [tex]\left(\rho^{2} \right) \rho [/tex] gets me the right answer... but why?
 
  • #5
qamptr said:
[tex]\left(\rho^{3}\right) \rho[/tex] instead of [tex]\left(\rho^{2} \right) \rho [/tex] gets me the right answer... but why?

Oh... :rolleyes:
 

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

Spherical coordinates are a system of coordinates used to represent points in three-dimensional space. They differ from Cartesian coordinates in that they use a radial distance, an azimuth angle, and a polar angle instead of x, y, and z coordinates. This system is particularly useful for describing points on a sphere or other curved surfaces.

2. 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 must first determine the limits of integration for each variable. The radial distance, r, will typically have limits of 0 to the radius of the sphere or curved surface. The azimuth angle, θ, will have limits of 0 to 2π, and the polar angle, φ, will have limits of 0 to π. Then, substitute the spherical coordinate expressions for x, y, and z into the integrand and evaluate the integral accordingly.

3. What are the advantages of using spherical coordinates for triple integrals?

The main advantage of using spherical coordinates for triple integrals is that they are better suited for dealing with spherical or curved surfaces. In Cartesian coordinates, the equations for these surfaces can be very complex, making integration difficult. However, in spherical coordinates, these surfaces are often described by simpler equations, making integration easier and more efficient.

4. Can I use spherical coordinates for any type of triple integral?

No, spherical coordinates are only suitable for triple integrals that involve spherical or curved surfaces. If the region of integration is a rectangular box or an irregular shape, it is better to use Cartesian coordinates. It is important to choose the appropriate coordinate system based on the shape of the region and the complexity of the integrand.

5. How do I visualize a triple integral in spherical coordinates?

Visualizing a triple integral in spherical coordinates can be challenging, but there are a few helpful strategies. First, you can think of the integral as representing the volume under a curved surface, where the limits of integration correspond to the boundaries of the surface. Alternatively, you can use mathematical software or graphing calculators to plot the integrand and the region of integration in spherical coordinates to get a better visual understanding.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
544
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
952
  • Calculus and Beyond Homework Help
Replies
2
Views
789
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
689
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
885
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Back
Top