What is the Triple Integral for the Given Solid in Spherical Coordinates?

In summary: the volume is the same as in the picture, but in terms of the coordinates used, the sphere is at the origin and the cone is at the point (0,0)
  • #1
MozAngeles
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0

Homework Statement



Set up the triple integral for the volume of the given solid using spherical coordinates:

The solid bounded below by the sphere ρ=6cosθ and above by the cone z=sqrt(x2+y2)

Homework Equations





The Attempt at a Solution



I thought i had this set up right where ρ goes from 6cosθ to sqrt(3)/2 ( how i got sqrt(3)/2 was by z=sqrt(x2+y2 is th esame as ρcos∅=ρsin which means ∅=pi/4 plugging that into ρ=6cosθ, gives me sqrt3/2) i don't know if this is right but for ∅ it goes from 0 to pi/4 and for θ it goes from 0 to 2pi

thanks in advance
 
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  • #2
What is ρ, and what is θ?

How can be ρ=6cosθ the equation of a sphere? ehild
 
  • #3
ρ is the distance from P to the origin

∅ is the angle that the line makes with the positive x axis

θ is the angle from the cylindrical coordinates (0<θ<2pi)

i know that the equation for a sphere is when ρ= constant, in this case she told us it was cos∅
 
  • #4
Could you please show that sphere and cone and the volume you need to find, placed into the system of coordinates you use? Is that the same as in the picture? (From http://mathworld.wolfram.com/SphericalCoordinates.html)


SphericalCoordinates_1201.gif


ehild
 
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  • #5
is that the volume you nee to find?

http://img811.imageshack.us/img811/4920/conesphere.jpg

ehild
 
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  • #6
i couldn't get a picture but that is how the sphere looks with a radius of six and the cone is just the top half as looking as any cone would look
 
  • #7
Is the volume as shown in the picture in my previous post? If so, you have to integrate the volume element dV=r^2sin(∅)dr d∅ dθ from 0 to 2pi for θ,
from 0 to pi/4 for ∅ (it is the angle of the cone it makes with the positive y axis) and for r from the top of the cone to the radius of the sphere r=ρ=6.

ehild
 

FAQ: What is the Triple Integral for the Given Solid in Spherical Coordinates?

1. What are spherical coordinates?

Spherical coordinates are a system of coordinates used to specify the position of a point in three-dimensional space. They are based on a sphere, with the origin at the center of the sphere and three coordinates that represent the distance from the origin, the angle from the positive z-axis, and the angle from the positive x-axis.

2. How are spherical coordinates different from Cartesian coordinates?

Spherical coordinates use a different set of coordinates than Cartesian coordinates. In spherical coordinates, the coordinates are based on a sphere rather than a plane, and they use distance, polar angle, and azimuthal angle instead of x, y, and z coordinates. This makes them useful for describing points on a sphere or in spherical objects.

3. What are the advantages of using spherical coordinates?

Spherical coordinates are particularly useful for describing points on a sphere or in spherical objects, such as planets, stars, and other celestial bodies. They also have advantages in certain mathematical calculations, such as integrating over spherical volumes or solving differential equations with spherical symmetry.

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

To convert from spherical to Cartesian coordinates, 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. To convert from Cartesian to spherical coordinates, you can use the following equations:
r = √(x² + y² + z²)
θ = arccos(z/√(x² + y² + z²))
φ = arctan(y/x)

5. In what fields are spherical coordinates commonly used?

Spherical coordinates are commonly used in fields such as physics, astronomy, and geology to describe points on a sphere or in spherical objects. They are also used in computer graphics and navigation systems, as well as in certain types of mathematical calculations and simulations.

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