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)

step-by-step explanation:

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.