Triple Integral, Volume of a solid

In summary, The conversation was about calculating a triple integral in a region bounded by a sphere, a cone, and a plane. The individual tried to solve it using spherical coordinates, but there was a question about the range of the beta angle. Another approach was suggested, changing to polar coordinates and then evaluating the triple integral, which was recommended to be the better option.
  • #1
abbot
5
0

Homework Statement


Well, first of all, I'm not english spoken, so sorry for the mistakes.
I was trying to calculate the integral below:

[tex]\int \int \int_{V} (xy+z) dxdydz [/tex]

where V is a region in [tex]R^{3}[/tex] bounded by

the sphere [tex]x^2+y^2+z^2<=9[/tex]

the cone [tex]z^2<=x^2+y^2[/tex]

and the plane [tex]z>=0[/tex]


2. Relevant equations

The Attempt at a Solution


I tried to calculate it with spherical coordinates in that way:

x= r sin(β) cos(α)
y= r sin(β) sin(α)
z= r cos(β)

0<=α<=2π
π/4<=β<=π/2
0<=r<=3

but I'm not sure if that it's correct so can anybody help me?Thanks a lot
 
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  • #2
You have written z^2 LESS than or equal to x^2+y^2. If that's what you meant to write, I don't think the range of the beta angle is correct. Is it?
 
  • #3
Dick said:
You have written z^2 LESS than or equal to x^2+y^2. If that's what you meant to write, I don't think the range of the beta angle is correct. Is it?

Well, I'm not sure, the solid bounded by those surfaces it's inside the sphere but outside the cone, so i thought beta angle goes from the straight lines that form cone, to the plane z=0, considering that beta angle starts at the z-axis
 
  • #4
abbot said:
Well, I'm not sure, the solid bounded by those surfaces it's inside the sphere but outside the cone, so i thought beta angle goes from the straight lines that form cone, to the plane z=0, considering that beta angle starts at the z-axis

Ok, I'm just thinking of a different polar angle convention. I'd say that looks fine then.
 
  • #5
I've also tried to calculate it integrating z in that way

[tex]\int \int( \int^{\sqrt{9-x^2-y^2}}_{\sqrt{x^2+y^2}} (xy+z) dz)dxdy [/tex]

and then changing to polar coordinates with x and y
What do you think about it?
 
  • #6
abbot said:
I've also tried to calculate it integrating z in that way

[tex]\int \int( \int^{\sqrt{9-x^2-y^2}}_{\sqrt{x^2+y^2}} (xy+z) dz)dxdy [/tex]

and then changing to polar coordinates with x and y

That's not what i would recommend. Change to spherical coordinates, then evaluate the triple integral. Merry Christmas!
 
  • #7
Char. Limit said:
That's not what i would recommend. Change to spherical coordinates, then evaluate the triple integral. Merry Christmas!

Thank you and Merry Christmas for you too!
 

1. What is a triple integral?

A triple integral is a type of integral that is used to calculate the volume of a three-dimensional shape or solid. It involves integrating a function over a three-dimensional region in space.

2. How is a triple integral different from a regular integral?

A regular integral involves integrating a function over a one-dimensional interval, while a triple integral involves integrating a function over a three-dimensional region in space.

3. What is the process for solving a triple integral?

The process for solving a triple integral involves setting up the integral in terms of three variables (typically x, y, and z), determining the limits of integration for each variable, and then evaluating the integral using techniques such as substitution or integration by parts.

4. Can a triple integral be used to find the volume of any solid shape?

Yes, a triple integral can be used to find the volume of any solid shape as long as the shape can be represented by a three-dimensional region in space.

5. Are there any real-world applications of triple integrals?

Yes, triple integrals have various applications in fields such as physics, engineering, and economics. They can be used to calculate the volume of irregularly shaped objects, determine mass and density distributions, and solve optimization problems.

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