Evaluate the integral by changing to spherical coordinates.
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.
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 solution says the the boundary for phi is from 0 to pi as well as the one for theta.
I'd have to think about it, but you obviously have both halves of the circle and both halves of the sphere.
To evaluate the given integral by changing to spherical coordinates, let's determine the limits of integration for the angles φ and θ.
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.