HELP WITH CALCULUS II ARC LENGTH AND SURFACE AREA

1) L = ∫(x = 1 to 8) √(1 + [(d/dx) x^(2/3)]^2) dx
.......= ∫(x = 1 to 8) √(1 + [(2/3) x^(-1/3)]^2) dx
.......= ∫(x = 1 to 8) √(1 + (4/9) x^(-2/3)) dx
.......= ∫(x = 1 to 8) √[(1/9) x^(-2/3) * (9x^(2/3) + 4)] dx
.......= ∫(x = 1 to 8) (1/3) x^(-1/3) * √(9x^(2/3) + 4) dx
.......= ∫(w = 13 to 40) (1/3) √w * dw/6, letting w = 9x^(2/3) + 4, dw = 6x^(-1/3) dx
.......= (1/18) * (2/3)w^(3/2) {for w = 13 to 40}
.......= (1/27) [40^(3/2) - 13^(3/2)].
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2) L = ∫(y = 0 to 1) √(1 + [(d/dy) (3y^(3/2) - 1)]^2) dy
......= ∫(y = 0 to 1) √(1 + [(9/2) y^(1/2)]^2) dy
......= ∫(y = 0 to 1) √(1 + (81/4)y) dy
......= ∫(y = 0 to 1) √((1/4)(4 + 81y)) dy
......= (1/2) ∫(y = 0 to 1) (4 + 81y)^(1/2) dy
......= (1/2) * (2/3)(1/81)(4 + 81y)^(3/2) {for y = 0 to 1}
......= (1/243) [85^(3/2) - 8].
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3) Since this region is being revolved about the x-axis,
A = ∫ 2πy ds
...= ∫ 2πy √(1 + (dy/dx)^2) dx
...= ∫(x = 0 to 1) 2πx^3 √(1 + (3x^2)^2) dx
...= ∫(x = 0 to 1) 2πx^3 (1 + 9x^4)^(1/2) dx
...= 2π * (1/36)(2/3)(1 + 9x^4)^(3/2) {for x = 0 to 1}
...= (π/27)[10^(3/2) - 1].
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4) Since this region is being revolved about the y-axis,
A = ∫ 2πx ds
...= ∫ 2πx √(1 + (dx/dy)^2) dy
...= ∫(y = -2 to 2) 2π * √(9 - y^2) * √(1 + (-y/√(9 - y^2))^2) dy
...= ∫(y = -2 to 2) 2π * √(9 - y^2) * √(1 + y^2/(9 - y^2)) dy
...= ∫(y = -2 to 2) 2π * √((9 - y^2) + y^2) dy
...= ∫(y = -2 to 2) 2π * 3
...= 24π.
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I hope this helps!