Calculating Volume Using Triple Integrals in Spherical Coordinates

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Homework Help Overview

The discussion revolves around calculating the volume of a solid defined by a sphere and a cone using triple integrals in spherical coordinates. The specific solid is bounded by the sphere \(x^2+y^2+z^2=36\), above the xy-plane, and outside the cone defined by \(z=7\sqrt{x^2+y^2}\).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss different methods for setting up the volume integral, including the use of spherical coordinates and potential alternatives like cylindrical coordinates. Questions arise about determining the limits of integration and the boundaries of the solid.

Discussion Status

There is ongoing exploration of the problem, with participants attempting to clarify the setup and boundaries for integration. Some guidance has been offered regarding the limits of integration, but there is no consensus on the correct approach or final setup.

Contextual Notes

Participants are navigating potential misunderstandings regarding the notation and definitions used in the context of spherical coordinates, as well as the specifics of the integration process.

mhs11
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hi all

how can i find the volume of the solid that lies within the sphere x^2+y^2+z^2=36 , above the xy plane, and outside the cone z=7sqrt(x^2+y^2) .

your help is very much appreiated
 
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Welcome to PF!

Hi mhs1! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

Either split it into "vertical" cylinders of thickness dr, or split it into "horizontal" discs-with-holes-in of height dz. :wink:
 
thnx 4 ur reply

but i didn't get it

how can if find the boundaries
 
mhs11 said:
… how can if find the boundaries

Do you mean the limits of integration?

If you integrate over r (= √(x2 + y2)), do it from 0 to the maximum value of r.

If you integrate over z, do it from 0 to the maximum value of z. :smile:
 
i did the following:

0 ≤ σ ≤ 6
σ^2 dσ= σ^3 /3 = 72

0 ≤ q ≤ 2π

dq= q = 2π

arctan 7/√50 ≤ Φ ≤ π

sinΦ dΦ= -cosΦ= 1+cos(arctan 7/√50 )

then i multiply them

(1+cos(arctan 7/√50 ))*2π *72=773.8884482

but when i enter it it gives me that it is wrong
 
mhs11 said:
i did the following:

0 ≤ σ ≤ 6
σ^2 dσ= σ^3 /3 = 72

0 ≤ q ≤ 2π

dq= q = 2π

arctan 7/√50 ≤ Φ ≤ π

sinΦ dΦ= -cosΦ= 1+cos(arctan 7/√50 )

then i multiply them

(1+cos(arctan 7/√50 ))*2π *72=773.8884482

but when i enter it it gives me that it is wrong

I'm not following this at all. :confused:

What is σ ?

What is σ2dσ supposed to be?

What is 7/√50 ?

What are you trying to integrate?
 
i'm trying to find the volume using shperical coordinate
 
mhs11 said:
i'm trying to find the volume using shperical coordinate

(I would have used either cylindrical or Cartesian coordinates.)

(and it's arctan7, not arctan 7/√50, though it is arcsin7/√50)

ok, write it out properly this time … what is the basic formula for volume, using spherical coordinates?

(oh, and have an integral: ∫ and a theta: θ and a phi: φ :wink:)
 

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