Find the area of the surface obtained by rotating x=1-y^2 0<=y<=1 about x-axis

Question

I keep getting [4pisqrt(1+4y^2)]/3

I feel like i'm a little off. Is this correct

Answer 1

The formula for the area of a rotated surface over the x-axis is.

S = ∫ 2πy*√(1+(dx/dy)²)dy

The problem states.

x = 1-y²

So

dx/dy = -2y

Inserting this into the integral we get.

S = ∫ 2πy*√(1+(-2y)²)dy

simplified to

S = ∫ 2πy*√(1+4y²)dy The limits of the integral are from 0 to 1

To solve this use a u substitution.

u = 1+4y²

du = 8y dy

The integral now becomes

S = ∫ 2πy/8y*√(u)dy /dy du

S = ∫ π/4*√(u) du

Solving the integral we get.

S = u^(3/2)*π/6 Since u = 1+4y²

S = (1+4y²)^(3/2)*π/6 From 0 to 1

Solving with the boundaries you get.'

S = (5^(3/2) - 1)*π/6

Answer 2

Surface area A = 2π∫₀¹y(1 + y'²)^1/2dx; y = (1 - x)^1/2, y' = - 1/2(1 - x)^-1/2.

So, A = 2π ∫₀¹(1 - x)^1/2(1 + 1/[4(1 - x)]^1/2dx = π∫₀¹(5 - 4x)^1/2dx = - π/4 · 2/3(5 - 4x)^3/2 from 0 to 1 =

π/6(5^3/2 - 1) ≈ 5.33

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