Use spherical coordinates to calculate the triple integral of f(x, y, z)=y over the region x^2+y^2+z^2≤8, x, y, z ≤ 0.
since we're dealing with a bottom quarter of the sphere (y,z <= 0),
and y = r sinØ sinθ, we just have
∫[0,√8] ∫[π,2π] ∫[-π/2,0] r sinØ sinθ dθ dØ dr
= ∫[0,√8] ∫[π,2π] r sinØ (-cosθ [-π/2,0]) dØ dr
= ∫[0,√8] ∫[π,2π] r sinØ dØ dr
= ∫[0,√8] ∫[π,2π] r (-sinØ [π,2π]) dr
= ∫[0,√8] -2r dr
= -r^2 ∫[0,√8]
= -8
actually, I mistakenly used the bottom 1/4 sphere, instead of just the 1/8 sphere.
I'm sure you can fix that.
To calculate the triple integral of f(x, y, z) = y over the given region using spherical coordinates, we need to express the region in terms of spherical coordinates and set up the appropriate integral limits.
In spherical coordinates, a point (x, y, z) is represented as (ρ, θ, φ), where:
- ρ represents the distance from the origin to the point,
- θ represents the angle in the xy-plane, measured from the positive x-axis,
- φ represents the angle between the positive z-axis and the line segment connecting the origin to the point.
Since the region x^2 + y^2 + z^2 ≤ 8 lies completely in the negative octant (x, y, z ≤ 0), the appropriate limits for the spherical coordinates are:
- ρ: 0 to √8
- θ: 0 to π/2
- φ: 0 to π/2
Now, we can set up the triple integral:
∫∫∫_V f(ρ, θ, φ) ρ^2 sin(φ) dρ dθ dφ
Substituting f(ρ, θ, φ) = ρ sin(φ), the integral becomes:
∫∫∫_V (ρ sin(φ))^2 sin(φ) dρ dθ dφ
Simplifying, we have:
∫_0^(π/2) ∫_0^(π/2) ∫_0^(√8) ρ^3 sin^3(φ) dρ dθ dφ
Let's evaluate this integral step-by-step:
Step 1: Evaluate the innermost integral with respect to ρ:
∫_0^(√8) ρ^3 sin^3(φ) dρ
Using the power rule for integration, we have:
= [ρ^4/4]_0^(√8) * sin^3(φ)
= [(√8)^4/4 - 0] * sin^3(φ)
= 8 * sin^3(φ)
Step 2: Evaluate the middle integral with respect to θ:
∫_0^(π/2) 8 * sin^3(φ) dθ
Since θ does not appear in the integral, its limits don't affect the result. Hence, we can simply multiply the integrand by the range of θ:
= 8 * sin^3(φ) * [θ]_0^(π/2)
= 8 * sin^3(φ) * (π/2 - 0)
= 4π * sin^3(φ)
Step 3: Evaluate the outermost integral with respect to φ:
∫_0^(π/2) 4π * sin^3(φ) dφ
Using trigonometric identities and integration by substitution, let's simplify this integral:
= 4π * ∫_0^(π/2) sin^2(φ) sin(φ) dφ
Let u = sin(φ), then du = cos(φ) dφ. When φ = 0, u = sin(0) = 0, and when φ = π/2, u = sin(π/2) = 1. Thus, the integral becomes:
= 4π * ∫_0^1 u^2 du
= 4π * [u^3/3]_0^1
= 4π * (1/3 - 0)
= 4π/3
Therefore, the triple integral of f(x, y, z) = y over the region x^2 + y^2 + z^2 ≤ 8, x, y, z ≤ 0, in spherical coordinates is 4π/3.
To calculate the triple integral of a function using spherical coordinates, we first need to know the limits of integration in terms of the spherical coordinates.
In spherical coordinates, a point (x, y, z) can be represented as (ρ, θ, φ), where:
- ρ is the distance from the origin to the point, ρ ≥ 0
- θ is the angle in the xy-plane measured from the positive x-axis, 0 ≤ θ ≤ 2π
- φ is the angle from the positive z-axis to the line segment connecting the origin to the point, 0 ≤ φ ≤ π
Now let's determine the limits of integration for each coordinate:
1. For ρ:
Since x^2 + y^2 + z^2 ≤ 8, and x, y, and z are all ≤ 0, we have ρ ≤ √8 = 2√2.
2. For θ:
Since x, y, and z are all ≤ 0, the region is symmetric in the xy-plane. Therefore, we can integrate θ from 0 to π.
3. For φ:
Since the region x, y, and z are all ≤ 0, φ varies from 0 to π/2.
Now, let's set up the triple integral using spherical coordinates:
∫∫∫ f(ρ, θ, φ) ρ^2 sin(φ) dρ dθ dφ
In this case, f(ρ, θ, φ) = y = ρ sin(φ) sin(θ).
So, we have:
∫(0 to π) ∫(0 to π/2) ∫(0 to 2√2) ρ sin^2(φ) sin(θ) dρ dθ dφ
Next, we can perform the integration in the given order. Solving this triple integral will give us the desired result.