CAN SOMEONE HELP ME IN THIS INTEGRAND QUESTION PLS PLS PLS PLS ''Find the surface area between the z = 1 and z = 4 planes of z = x ^ 2 + y ^ 2 paraboloid.''

Step-by-step explanation:

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)

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step-by-step explanation: