Calculating Volume Using Spherical Coordinates | Sphere & Paraboloid

In summary, the region bounded by x2 + y2 + z2=2 and a sphere is p=Sqrt.(2). A parabloid can be converted to cot(phi)csc(phi)=p and the upper limit of rho is min(sphere, parabloid). The region can be graphed and rho and p can be found when z=1.
  • #1
robierob
11
0
I need to use spherical coordinates to try and find the volume of the region bounded by

x2 + y2 + z2 = 2 which converts to p=Sqrt.(2) a sphere

and

z = x2 + y2, a parabloid which I converted to cot(phi)csc(phi)=p

I hope the greek letters for these are comonly used
x=psin(phi)cos(theta)

so far I have

V=SSS (p^2)sin(phi) dp d(theta) d(phi)

i know that theta goes from 0 to 2pi but how do I find the other limits?
 
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  • #2
Well, think a little about what this shape looks like. It's all in the "upper" half of the universe, so it's not surprising that you can use limits of 0 and pi for phi. As for rho... you'll actually have to divide the integral into two parts: one that starts at 0 and terminates on the sphere, and one that starts at 0 and terminates on the parabloid.

Make sense?
 
  • #3
The part about spliting it into two intergrals doesn't really make sense to me. I have seen the solution to a similar one with a cone instead of a parabloid and it was done with one triple intergal not two added togreather. That's what you mean right?
 
  • #4
Some, you can do with a single triple integral... this one, you have to divide into two pieces, by phi. Well, if you want to say that the upper limit of rho is min(sphere, parabloid), I suppose that's just one integral. But offhand, it seems to me that you need to find the value of phi where this minimum changes from one surface to the other.

I'll ask again... you've thought about what this region looks like, right?
 
  • #5
In the case of a cone instead, you never have a case where an upper limit for rho of min(sphere, cone) is ever the cone, so you have a single integral. :smile:

(Also, I want to make clear that I'm talking about the rho component of the sphere, cone and parabloid here, and I'm too lazy to use proper tex notation and subscripts.)
 
  • #6
When I started out on this problem I attempted to use cylindrical coordinates.

The set up looked nice but integrated, no so nice

V=SSS rdzdrd(theta)

z from r^2 to rt. (2-r^2)
r from 0 to 1
and theta from 0 2pi

but when I got to integrating with respect to r it got weird, mabey a trig sub could be used. I don't know.
anyways, i have graphed it and know what it looks like.
I know that they intersect when z = 1

I just don't see p or rho coming into the picture. I know rho is cot(phi)csc(phi)=rho
 
  • #7
should I stick to cylindrical coordinate for this one?
 
  • #8
No, I think the spherical coordinates are correct.

So, right, the surfaces intersect at z=1. What phi does this correspond to? (Do you see that for smaller phi, you integrate p (or rho) out to the sphere, but for phi larger than this, you go out to the parabloid?)

The integral isn't bad, even when the limit is the parabloid. Don't use cot and csc though, just write it as sin and cos. Do theta first, then rho, then phi. The phi integral can be tackled by converting all but one of the cosines into sines, and then substituting u=sin phi.
 
  • #9
Ok, I am going to try and see how that works out.
Thanks
 

1. How do you calculate the volume of a sphere using spherical coordinates?

To calculate the volume of a sphere using spherical coordinates, you can use the formula V = (4/3)πr^3, where r is the radius of the sphere. This formula can also be rewritten in terms of spherical coordinates as V = ∫∫∫ρ^2sinφdρdφdθ, where ρ is the radial distance, φ is the polar angle, and θ is the azimuthal angle.

2. What is the difference between a sphere and a paraboloid?

A sphere is a three-dimensional shape that is perfectly round in all directions, while a paraboloid is a three-dimensional shape that is curved in one direction and flat in the other. In terms of volume, a sphere has a constant radius throughout, while a paraboloid has a changing radius as it moves away from the center.

3. Can you use spherical coordinates to calculate the volume of any three-dimensional shape?

No, spherical coordinates are only useful for calculating the volume of shapes that have a spherical or near-spherical shape. For other shapes, such as a cube or pyramid, other coordinate systems, such as Cartesian coordinates, would be more appropriate.

4. How do you convert between spherical and Cartesian coordinates?

To convert from spherical coordinates (ρ,φ,θ) to Cartesian coordinates (x,y,z), you can use the following formulas: x = ρsinφcosθ, y = ρsinφsinθ, and z = ρcosφ. Conversely, to convert from Cartesian to spherical coordinates, you can use the following formulas: ρ = √(x^2 + y^2 + z^2), φ = arccos(z/√(x^2 + y^2 + z^2)), and θ = arctan(y/x).

5. How do you use spherical coordinates to integrate a function over a sphere or paraboloid?

To integrate a function f(ρ,φ,θ) over a sphere or paraboloid using spherical coordinates, you can use the triple integral ∫∫∫f(ρ,φ,θ)ρ^2sinφdρdφdθ. This represents the volume element multiplied by the function evaluated at each point within the shape. The limits of integration for ρ, φ, and θ will depend on the specific shape and the region being integrated over.

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