Volume Integral of Cone

In summary, to find the mass of the cone bounded by the surface \theta=\alpha and z=a in spherical polar coordinates, we can use the mass density of p_0\cos(\theta) and evaluate a volume integral. To do this, we need to determine the limits for the integral, which can be done by sketching the cone and using the coordinates (r, \theta, \phi). The angle \theta runs from 0 to \alpha, and to find the limits for \phi, we can draw a right-angled triangle and use the angle alpha to determine the r-limits.
  • #1
implet
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Homework Statement


"A solid cone is bounded by the surface [tex]\theta=\alpha[/tex] in spherical polar coordinates and the surface [tex]z=a[/tex]. Its mass density is [tex]p_0\cos(\theta)[/tex]. By evaluating a volume integral find the mass of the cone.


Homework Equations





The Attempt at a Solution


I can't figure out the correct limits for the volume integral. Is it best to solve in Cartesian or spherical polar coordinates?

Many thanks :)
 
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  • #2
implet said:

Homework Statement


"A solid cone is bounded by the surface [tex]\theta=\alpha[/tex] in spherical polar coordinates and the surface [tex]z=a[/tex]. Its mass density is [tex]p_0\cos(\theta)[/tex]. By evaluating a volume integral find the mass of the cone.
Your description of the cone suggests your interpretation of spherical polar coordinates is [tex](r, \theta, \phi)[/tex] where [tex]\theta[/tex] is the angle from the positive z-axis and [tex]\phi[/tex] is the angle from the positive x-axis.

We look to use these coordinates to calculate the integral for the cone. Sketch the cone: it makes an angle of alpha with the positive z-axis and goes up to z=a. More specifically...

[tex]\theta[/tex] runs from [tex]0[/tex] to [tex]\alpha[/tex].

[tex]\phi[/tex] goes from ... to ... ?

To find the r-limits, draw a right-angled triangle:
Code:
   -----
   |   /
 a |  / r      where A is the angle alpha. 
   |A/
   |/
.
 

1. What is the formula for calculating the volume integral of a cone?

The formula for calculating the volume integral of a cone is V = (1/3)πr²h, where r is the radius of the base and h is the height of the cone.

2. How is the volume integral of a cone related to its surface area?

The volume integral of a cone is directly proportional to its surface area. This means that as the surface area of the cone increases, so does its volume integral.

3. Can the volume integral of a cone be negative?

No, the volume integral of a cone cannot be negative. It represents the total volume enclosed by the cone, which is always a positive value.

4. How does changing the radius or height of a cone affect its volume integral?

Changing the radius or height of a cone will directly affect its volume integral. Increasing the radius or height will result in a larger volume integral, while decreasing them will result in a smaller volume integral.

5. Can the volume integral of a cone be used to find the volume of other shapes?

No, the volume integral of a cone can only be used to find the volume of a cone. Other shapes have different formulas for calculating their volume, such as a sphere (V = (4/3)πr³) or a cylinder (V = πr²h).

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