Evaluate the surface integral ∫ ∫ₛ G(x, y, z) dS using a parametric description of the surface. G (x, y, z ) = 2z² , over the hemisphere x² + y² + z² = 36, for z >= 0.

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

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step-by-step explanation: