Answer:
S = ⅙ π (65^³/₂ − 5^³/₂)
Step-by-step explanation:
z = x² + y², 1 < z < 4
Surface area is:
S = ∫∫√(1 + (fₓ)² + (fᵧ)²) dA
where fₓ and fᵧ are the partial derivatives of f(x,y) with respect to x and y, respectively.
fₓ = 2x, fᵧ = 2y
S = ∫∫√(1 + (2x)² + (2y)²) dA
S = ∫∫√(1 + 4x² + 4y²) dA
For ease, convert to polar coordinates.
S = ∫∫√(1 + 4r²) dA
S = ∫∫√(1 + 4r²) r dr dθ
At z = 1, r = 1. At z = 4, r = 4.
So 1 < r < 4, and 0 < θ < 2π. These are the limits of the integral.
S = ∫₀²ᵖⁱ∫₁⁴√(1 + 4r²) r dr dθ
To integrate, use u-substitution.
u = 1 + 4r²
du = 8r dr
⅛ du = r dr
When r = 1, u = 5. When r = 4, u = 65.
S = ∫₀²ᵖⁱ∫₅⁶⁵√u (⅛ du) dθ
S = ∫₀²ᵖⁱ (⅛ ∫₅⁶⁵√u du) dθ
S = ∫₀²ᵖⁱ (¹/₁₂ u^³/₂ |₅⁶⁵) dθ
S = ∫₀²ᵖⁱ (¹/₁₂ (65^³/₂ − 5^³/₂)) dθ
S = (¹/₁₂ (65^³/₂ − 5^³/₂)) θ |₀²ᵖⁱ
S = (¹/₁₂ (65^³/₂ − 5^³/₂)) (2π)
S = ⅙ π (65^³/₂ − 5^³/₂)
