you are integrating over the whole sphere, so
0 <= p <= 1 (inside-outside)
0 <= φ <= 2π (whole x-y plane)
0 <= θ <= π (top-to-bottom of sphere)
There must be some examples in your text. And there are surely some online.
The outer boundaries are from 0 to 1.
The middle one goes from -sqrt(1-x^2) to sqrt(1-x^2)
The inner one goes from -sqrt(1-x^2-z^) to sqrt(1-x^2-z^)
for 1/sqrt(x^2+y^2+z^2) dydzdx
I don't understand how to get the limits of integration. I know for rho it will be from 0 to 1. I want to know the process to get the boundaries for phi and theta since I have a few other similar problems to do.
0 <= p <= 1 (inside-outside)
0 <= φ <= 2π (whole x-y plane)
0 <= θ <= π (top-to-bottom of sphere)
There must be some examples in your text. And there are surely some online.
In spherical coordinates, a point in three-dimensional space is represented by the radial distance ρ, the polar angle φ, and the azimuthal angle θ.
First, let's visualize the region of integration. The given integral is over a region inside a sphere of radius 1. The outer boundaries are from ρ = 0 to ρ = 1, representing the distance from the origin to the surface of the sphere.
Next, let's consider the limits for the angle φ. The middle boundary of integration is defined within the region -sqrt(1 - x^2) ≤ y ≤ sqrt(1 - x^2). By rewriting in spherical coordinates, this becomes -sqrt(1 - ρ^2 sin^2φ) ≤ ρ sinφ ≤ sqrt(1 - ρ^2 sin^2φ). Simplifying, we get -sqrt(1 - ρ^2 sin^2φ) ≤ ρ sinφ ≤ sqrt(1 - ρ^2 sin^2φ). Since -1 ≤ sinφ ≤ 1, this inequality becomes -sqrt(1 - ρ^2) ≤ ρ sinφ ≤ sqrt(1 - ρ^2). Thus, the limits for φ are given by -arcsin(√(1 - ρ^2)) ≤ φ ≤ arcsin(√(1 - ρ^2)).
Finally, let's determine the limits for the angle θ. The inner boundary of integration is defined within the region -sqrt(1 - x^2 - y^2) ≤ z ≤ sqrt(1 - x^2 - y^2). Rewriting in spherical coordinates, this becomes -sqrt(1 - ρ^2 sin^2φ - ρ^2 cos^2φ) ≤ ρ cosφ ≤ sqrt(1 - ρ^2 sin^2φ - ρ^2 cos^2φ). Simplifying, we get -sqrt(1 - ρ^2) ≤ ρ cosφ ≤ sqrt(1 - ρ^2). Since -1 ≤ cosφ ≤ 1, this inequality becomes -sqrt(1 - ρ^2) ≤ ρ cosφ ≤ sqrt(1 - ρ^2). Thus, the limits for θ are given by -π ≤ θ ≤ π.
Therefore, the limits of integration for the integral in spherical coordinates are:
ρ: 0 ≤ ρ ≤ 1
φ: -arcsin(√(1 - ρ^2)) ≤ φ ≤ arcsin(√(1 - ρ^2))
θ: -π ≤ θ ≤ π
By using these limits, you can now evaluate the integral 1/√(x^2 + y^2 + z^2) dydzdx in spherical coordinates.