Find the volume of a cone using spherical coordinates

In summary, to find the volume of the portion of cone z^2 = x^2 + y^2 bounded by the planes z = 1 and z = 2, one can use spherical coordinates or cylindrical coordinates. In spherical coordinates, the limits of integration are rho from 1/cos(phi) to 2/cos(phi), phi from 0 to pi/4, and theta from 0 to 2pi. In cylindrical coordinates, the limits are theta from 0 to 2pi, z from 1 to 2, and r from 0 to z.
  • #1
clocksmith
3
0
Find the volume of the portion of cone z^2 = x^2 + y^2 bounded by the planes z = 1 and z = 2 using spherical coordinates

I am having trouble coming up with the limits






Relevant equations
dV = r^2*sin(theta)*dr*d(theta)*d(phi)
r = sqrt(x^2+y^2+z^2)

the problem is actually 2 parts, the 2nd part asks to evaluate by cylindrical coordinates and I obtain 7pi/3 which i know is right, I just can't come up with the limits

Attempt
for r i have from sqrt2 to 2sqrt2
for theta i have from 0 to pi/4
and for phi i have from 0 to 2pi
 
Physics news on Phys.org
  • #2
In spherical coordinates, [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], and [itex]z= \rho cos(\phi)[/itex]
Putting those into [itex]z^2= x^2+ y^2[/itex] becomes [itex]\rho^2 cos^2(\phi)= \rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 cos^2(\theta)sin^2(\phi)[/itex][itex]\rho^2 sin^2(\phi)[/itex] which reduces to [itex]cos^2(\phi)= sin^2(\phi)[/itex]. For [itex]\phi[/itex] between 0 and [itex]\pi/2[/itex], that is true only for [itex]\phi= \pi/4[/itex].

Now, for [itex]z= \rho cos(\phi)= 1[/itex] then [itex]\rho= 1/cos(\phi)[/itex] for [itex]z= \rho cos(\phi)= 2[/itex] then [itex]\rho= 2/cos(\phi)[/itex] the limits of integration are: [itex]rho[/itex] from [itex]1/cos(\phi)[/itex] to [itex]2/cos(\phi)[/itex], [itex]\phi[/itex] from 0 to [itex]\pi/4[/itex], and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex].

It looks to me like cylindrical coordinates should be much easier! Again, [itex]\theta[/itex] runs from 0 to [itex]2\pi[/itex], z obviously runs from 1 to 2, and, since [itex]z^2= x^2+ y^2= r^2[/itex], z= r (z is positive) so r runs from 0 to z.
 

1. What are spherical coordinates?

Spherical coordinates are a system of identifying points in 3D space using three variables: radius, inclination, and azimuth. These coordinates are useful in representing objects with spherical shapes, such as cones.

2. How do you find the volume of a cone using spherical coordinates?

To find the volume of a cone using spherical coordinates, you first need to determine the radius and height of the cone. Then, you can use the formula V = 1/3 * π * r^2 * h to calculate the volume. The radius and height can be expressed in terms of spherical coordinates, with the radius being the distance from the origin to the point on the surface of the cone, and the height being the distance from the base of the cone to the point on the surface.

3. Can you explain the formula for finding the volume of a cone using spherical coordinates?

The formula for finding the volume of a cone using spherical coordinates is derived from the general formula for finding the volume of a cone, V = 1/3 * π * r^2 * h. In spherical coordinates, the radius is represented by the variable ρ, the inclination by the variable θ, and the azimuth by the variable φ. Therefore, the formula becomes V = 1/3 * π * ρ^2 * h, where ρ and h are expressed in terms of θ and φ.

4. Can you provide an example of finding the volume of a cone using spherical coordinates?

Sure, let's say we have a cone with a radius of 5 units and a height of 10 units. Using spherical coordinates, we can express the radius as ρ = 5 and the height as h = 10. Plugging these values into the formula V = 1/3 * π * ρ^2 * h, we get V = 1/3 * π * 5^2 * 10 = 83.78 units^3.

5. Are there any limitations to using spherical coordinates to find the volume of a cone?

One limitation of using spherical coordinates to find the volume of a cone is that it can only be used for objects with a spherical shape. Additionally, the formula may become more complex for cones with non-uniform shapes or sizes, making it more difficult to accurately calculate the volume using spherical coordinates.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
542
  • Calculus and Beyond Homework Help
Replies
7
Views
688
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
966
  • Calculus and Beyond Homework Help
Replies
7
Views
964
  • Calculus and Beyond Homework Help
Replies
4
Views
951
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
4K
  • Calculus and Beyond Homework Help
Replies
9
Views
132
Back
Top