Triple integration w/spherical coordinates

In summary, the problem involves finding the mass of a part of a solid sphere with a radius of 5 in the first octant, where the mass density is given by the function (x^2 + y^2 + z^2)^3/2. This is solved by dividing the sphere into spherical shells and integrating the function (x^2 + y^2 + z^2)^3/2 using spherical coordinates with limits of integration of 0 to 5 for rho, 0 to 2pi for theta, and 0 to pi/2 for phi. This results in one-eighth of the mass of the full sphere.
  • #1
MAins
18
0
1.

"Find the mass of part of the solid sphere x^2 + y^2 + z^ 2 ≤ 25 in the 1st octant x ≥ 0, y ≥ 0, z ≥ 0 where mass density is f (x, y, z ) = (x^2 + y^2 + z^2 )^3/2 ."

3.

These problems are really stumping me! I need somebody to work it out/explain it to me! What will the limits of integration be for the following question? What do i integrate? I know I need to transform it to spherical coordinates... but beyond that I'm lost.

I know it's a triple integral:
m = ∫∫∫(x^2 + y^2 + z^2)^3/2 dzdydx
transforming to spherical co-ordinates:
0 ≤ rho ≤ 5
0 ≤ theta ≤ ? (how do I figure this out?)
0 ≤ phi ≤ ? (ditto)

dzdydx = rho^2 sinphi drho dphi dtheta
What does f (x, y,z) transform to and how do I figure it out?
 
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  • #2
x^2+y^2+z^2=rho^2. What does that make f(x,y,z)? phi is the polar angle and theta is the equatorial angle. What range of these keeps you in the first octant? Refer to a picture of your preferred system of spherical coordinates.
 
  • #3
… keep it simple …

Hi MAins! :smile:

You're making this too complicated:cry:

It's one-eight of the mass of the whole sphere.

Divide it into spherical shells, of radius r, from r = 0 to 5.

Then the mass of each shell is … ? :smile:
 
  • #4
But just in case, you would like to learn how to do these problems!

Since [itex]\rho= \sqrt{x^2+ y^2+ z^2}[/itex], [itex](x^2+ y^2+ z^2)^{3/2}= \rho^3[/itex]. Thats the function you want to integrate.

Yes, you are correct that the differential of volume in spherical coordinates is [itex]\rho sin^2(\phi)d\rho d\theta d\phi[/itex]

Now, to determine what the limits of integration should be, think about what the variables in spherical coordinates mean. [itex]\rho[/itex] measures the distance from the origin, (0,0,0) to a point. Since your sphere is centered at (0,0,0) and has radius 5, [itex]\rho[/itex] must go from 0 to 5. [itex]\theta[/itex] measures the angle around the "equator" (think of it as "longitude"). For the full sphere, it goes from 0 to [itex]2\pi[/itex]. Here, you have just 1/4 of a full circle: so [itex]\theta[/itex] goes from 0 to ?? [itex]\phi[/itex] measures the angle down from the z-axis to the point (think of it as "co-latitude". For the full sphere, it goes from 0 to [itex]\pi[/itex]. You only want to go from the z-axis to the xy-plane, 1/2 way: so [itex]\phi[/itex] goes from 0 to ??
 

1. What is triple integration with spherical coordinates?

Triple integration with spherical coordinates is a method used to solve three-dimensional integrals over a spherical region in space. It involves breaking down the integral into three separate integrals, each representing a different coordinate axis, and converting the coordinates from Cartesian to spherical form.

2. When is triple integration with spherical coordinates used?

Triple integration with spherical coordinates is commonly used in physics and engineering applications, such as calculating the mass, center of mass, or moment of inertia of a three-dimensional object with spherical symmetry.

3. How do you convert coordinates from Cartesian to spherical form?

To convert coordinates from Cartesian to spherical form, you can use the following equations:
x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ
Where r is the distance from the origin, θ is the polar angle, and φ is the azimuthal angle.

4. Can triple integration with spherical coordinates be used for any shape?

No, triple integration with spherical coordinates is only suitable for solving integrals over a spherical region in space. For other shapes, different coordinate systems and integration techniques may be more appropriate.

5. Are there any advantages to using spherical coordinates for triple integration?

Yes, using spherical coordinates can often simplify the integrals and make them easier to solve, especially for problems with spherical symmetry. It can also help in visualizing and understanding the three-dimensional region being integrated over.

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