Triple Integral Help in Spherical Coordinates

In summary, the conversation discusses putting boundaries in triple integrals using spherical coordinates. The figure attached is an ice cream cone and the question is how to arrange the boundaries in the order of dr dz dθ. It is recommended to use cylindrical coordinates instead and integrate in two separate steps for the cone and paraboloid. The final answer is the sum of two triple integrals. The conversation also briefly touches on the projection of the cone and paraboloid onto the xy-plane.
  • #1
sevag00
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1
Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.
 

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  • #2
Hi sevag00
sevag00 said:
Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.

Your question contradicts itself. You want to express the integrals in spherical coordinates, but the boundaries in terms of drdθdz are cylindrical coordinates! :smile:

What is the whole question? Are you required to find the volume of the cone including the section of paraboloid on top?
 
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  • #3
lol. I meant the volume in cylindrical coordinates.
 
  • #4
sevag00 said:
Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.

There's no reason to use spherical coordinates on this problem. You want cylindrical coordinates. It is best not to integrate the dr variable first; better would be ##r\, dz dr d\theta##. The dz limits will be z on the lower surface and z on the upper surface, both expressed in terms of r. Once you have those z values in terms of r, you can set them equal to see what r is at their intersection.
 
  • #5
Yeah, i know it's easy integrating in order dzdrdθ . But the question in the book is asking drdzdθ.
 
  • #6
Well, "[itex]drdzd\theta[/itex]" is the hard way! I recommend doing it as two separate integrals. For z from 0 to 1, we have the cone, z= r, so, for each z, r goes from 0 to z. For z= 1 to 2, we have the paraboloid, [itex]z= 2- r^2[/itex] so, for each z, [itex]r^2=2-z[/itex] and r, being positive, goes from 0 to [itex]\sqrt{2- z}[/itex].
 
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  • #7
The answer will be the sum of two triple integrals; volume of the cone + volume of paraboloid.
 
  • #8
Is this in order dzdrdθ?
 
  • #9
HallsofIvy said:
I recommend that you do these as two separate integrals- one over the cone from z= 0 up to z= 1 and the second over the paraboloid from z= 1 up to 2.
For the second triple integral, i think it should be from z=1 to ##z=2-r^2##
HallsofIvy said:
The projection of both cone and paraboloid to the xy-plane is the circle with center at (0, 0) and radius 2. To cover that, in polar coordinates, [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex] and r from 0 to 1.
The radius of the projection is r=1.
 
  • #10
sevag00 said:
Is this in order dzdrdθ?

Is what in that order? Use the quote button to reply to a message so we know what you are replying to.
 
  • #11
Actually, i know the answer. But i didn't get how r is from 0 to z + 0 to sqrt(2-z). Answer is attached.
 

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  • #12
sevag00 said:
Actually, i know the answer. But i didn't get how r is from 0 to z + 0 to sqrt(2-z).

Doing dr first, on the top part (the paraboloid) r goes from 0 to the r on the paraboloid. The equation of the paraboloid is ##z = 2-r^2##. Just solve that for ##r## to get ##\sqrt{2-z}##.
 
  • #13
Isn't the dr supposed to be from the x-axis to y-axis?
 
  • #14
sevag00 said:
Isn't the dr supposed to be from the x-axis to y-axis?

That doesn't make any sense. Think of starting on the z axis up at the level of the paraboloid. r is perpendicular to the z axis and it goes from r = 0 out till it hits the paraboloid. What direction it goes depends on ##\theta##.
 
  • #15
Aha. I think i get it.

EDIT: Yes, i get it. Thanks for the help.
 

1. What are spherical coordinates and how are they different from rectangular coordinates?

Spherical coordinates are a system used to locate points in three-dimensional space. They use three variables: radius (r), inclination (θ), and azimuth (φ) to represent a point. In contrast, rectangular coordinates use three variables (x, y, and z) to represent a point. The main difference is that spherical coordinates use a radial distance from the origin, while rectangular coordinates use Cartesian coordinates.

2. How do I convert a triple integral from rectangular coordinates to spherical coordinates?

To convert a triple integral from rectangular coordinates to spherical coordinates, the following substitutions can be made: x = rsinθcosφ, y = rsinθsinφ, z = rcosθ. The limits of integration for each variable should also be converted accordingly based on the new coordinate system.

3. What is the volume element in spherical coordinates?

The volume element in spherical coordinates is r^2sinθdrdθdφ. This takes into account the changes in the radius, inclination, and azimuth as the point moves in three-dimensional space.

4. How do I determine the limits of integration for a triple integral in spherical coordinates?

The limits of integration for a triple integral in spherical coordinates depend on the shape and boundaries of the region being integrated. The radius (r) typically ranges from 0 to some value, the inclination (θ) ranges from 0 to π, and the azimuth (φ) ranges from 0 to 2π. It is important to consider the shape and symmetry of the region to determine the appropriate limits for each variable.

5. How can I use triple integrals in spherical coordinates for real-world applications?

Spherical coordinates are often used in physics and engineering applications, such as calculating the electric field generated by a point charge. Triple integrals in spherical coordinates can be used to determine the volume or mass of a three-dimensional object, calculate the average value of a function over a spherical region, or solve problems involving spherical symmetry. They are also useful in solving problems involving spherical planets and celestial bodies.

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