# Triple integral (spherical coordinate)

• naspek
In summary, the problem is finding the volume of solid D which lies above the cone z = (x^2 + y^2)^1/2 and below the sphere z = (x^2 + y^2 + z^2). The integration has been done until substitution of cos phi = u, but the range needs to be changed. The correct u limit is 1/sqrt(2) and the lower limit is not 0. The integral may not be set up correctly and it appears to involve working with two parts, the hemisphere and the cone with a flat base. For the cone, the range of ρ is from 0 to z=1/2 which is ρcos(φ)=1/
naspek
find the volume of the solid D that lies above the cone z = (x^2 + y^2)^1/2
and below the sphere z = (x^2 + y^2 + z^2)

i've done the integration until i need to substitute cos phi = u..
however.. i don't know to change the range..

http://imageshack.us/photo/my-images/839/spherical.jpg/"

Last edited by a moderator:
 Never mind the below paragraph. Your work is correct; I misread the equation at first. All you have to do is put in the correct u limit in your last equation. Since u = cos(φ) and φ goes from 0 to pi/4, your upper u limit is 1/sqrt(2). Be careful with your lower limit for u, it isn't 0. Just finish your integration.I don't think your integral is set up correctly. I see a sphere centered at (0,0,1/2) with radius 1/2 intersecting the cone at z = 1/2. So you apparently want the volume under the top half of the sphere but above the cone. Are you working this in two parts, the hemisphere and the cone with a flat base? If so, for the cone, ρ goes from 0 to z=1/2 which is ρcos(φ)=1/2 or ρ = (1/2)sec(φ). And you have to add the hemisphere.

Last edited:

## What is a triple integral in spherical coordinates?

A triple integral in spherical coordinates is a method of calculating the volume of a three-dimensional region in space using a coordinate system that is based on a sphere. It involves integrating a function over three variables: r (radius), θ (polar angle), and φ (azimuthal angle).

## How is a triple integral expressed in spherical coordinates?

A triple integral in spherical coordinates is expressed as ∫∫∫f(r, θ, φ) dV, where f(r, θ, φ) is the function being integrated and dV is the volume element in spherical coordinates, given by dV = r²sinθ dr dθ dφ.

## What are 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 of the region being integrated. The limits for r typically range from 0 to the radius of the region, θ ranges from 0 to π, and φ ranges from 0 to 2π.

## How is a triple integral in spherical coordinates related to a triple integral in Cartesian coordinates?

A triple integral in spherical coordinates can be converted to a triple integral in Cartesian coordinates using the transformation equations: x = r sinθ cosφ, y = r sinθ sinφ, and z = r cosθ. This allows for easier calculation of the integral in some cases.

## What are some practical applications of triple integrals in spherical coordinates?

Triple integrals in spherical coordinates are commonly used in physics and engineering, particularly in fields such as electromagnetism, fluid mechanics, and thermodynamics. They can be used to calculate the electric field around a charged sphere, the flow of a fluid through a spherical pipe, or the volume of a spherical tank, among other applications.

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