Answer:
Step-by-step explanation:
Given the surface
G(x, y, z) =2z²
Over the hemisphere
x² + y² + z² = 36. For z≥0
Using polar coordinate
x=sin Φ cos θ,
y = sin Φ sin θ,
z = cos Φ
0 ≤ Φ ≤ π/2, 0 ≤ θ ≤ 2π
Therefore
r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k
Also, dS= |rθ×rΦ|= sinΦ
dS=sinΦdΦdθ
Then we want to compute the volume integral of
∫ ∫ₛ G(x, y, z) dS
G(x, y, z) =2z²
Therefore in polar forms
G(x, y, z) =2(cos Φ)²
G(x, y, z) = 2cos²Φ
Given that dS=sinΦdΦdθ
∫ ∫ₛ G(x, y, z) dS
∫ ∫ 2cos²ΦsinΦdΦdθ at 0 ≤ Φ ≤ π/2,
0 ≤ θ ≤ 2π
∫ 2cos²ΦsinΦ •θdΦ from 0 ≤ θ ≤ 2π
2∫cos²ΦsinΦ •(2π-0)dΦ
4π∫ cos²ΦsinΦ dΦ from 0 ≤ Φ ≤ π/2
Let U = cosΦ
dU/dΦ =-sinΦ
-dU/sinΦ =dΦ
4π∫ U²sinΦ(-dU/sinΦ) 0 ≤ Φ ≤ π/2
-4π∫ U² dU
-4π U³/3, then U=cosΦ
[-4πcos³Φ / 3 ] from 0 ≤ Φ ≤ π/2
[-4π cos³(π/2)/3 - [-4π cos³(0)/3]
0+4π/3
4π/3
4π/3 unit²
