Surface area of a curve

Find the area of the surface obtained by rotating the curve
y= sqrt(3x)
from x=0 to x = 8 about the x-axis.

I know the formula.... int 2PI * sqrt(3x) * (1+(dy/dx)^2)^(1/2) But it doesn't seem to be working.

y = √(3x)
y = (3x)^(1/2)

y ' = (1/2) * (3x)^(-1/2) * 3
y ' = (3/2) * ( 1 / √(3x) )
y ' = ( 3 / 2√(3x) )

√(1 + (dy/dx)^2 )

√(1 + (( 3 / 2√(3x) ))^2 )
√(1 + (9 / 4(3x) ) )
√(1 + (9 / 12x ) )
√(1 + (3 / 4x ) )
√(4x/4x + (3/4x) )
√( (4x + 3)/4x )
√(4x + 3)/√(4x)

8
∫ 2 * π * y * √(1 + (dy/dx)^2 )
0

8
∫ 2 * π * √(3x) * √(4x + 3)/√(4x)
0

8
∫ 2 * π * √(3)√(x) * √(4x + 3) * (1/√(4)√(x))
0

8
∫ 2 * π * √(4x + 3) * ( √(3)/2 )
0

8
∫ √(3)π * √(4x + 3)
0

u = 4x + 3
du = 4dx ====> dx = du/4

8
∫ √(3)π * √(u) * du/4
0

8
∫ ( √(3) / 4 ) π * u^(1/2) * du
0

. .. . . .. . . . .. . . . .. . . . .. . . . .. . 8
(√(3)π / 4) * u^(1/2 + 1)/(1/2 + 1) ]
. .. . .. . .. . . . . .. . . . . .. . . .. . .. 0

. .. . . .. . . . .. . . . .. . . . . . . .. 8
(√(3)π / 4) * (4x + 3)^(3/2)/(3/2) ]
. .. . .. . .. . . . . . .. . ... . .. . .. 0

. .. . . .. . . . .. . . . .. . . . . . . .. . . 8
(√(3)π / 4) * (2/3) * (4x + 3)^(3/2) ]
. .. . .. . .. . . . . . .. . ... . .. . .. . . 0

. .. . . .. . .. .. . . . . . . .. . . 8
(√(3)π / 6) * (4x + 3)^(3/2) ]
. .. . .. . .. .. . ... . .. . .. . . 0

(√(3)π / 6) * [ (4 * 8 + 3)^(3/2) - (4 * 0 + 3)^(3/2) ]

(√(3)π / 6) * [ (32 + 3)^(3/2) - (0 + 3)^(3/2) ]

(√(3)π / 6) * [ (35)^(3/2) - (3)^(3/2) ]

(√(3)π / 6) * [ 35(35)^(1/2) - 3(3)^(1/2) ] ≈ 183

You need to integrate the y by dx and compute the integral in the interval from 0 to 8.
integral of sqrt(3x) is 2*x^(3/2)/sqrt(3) + c. For 0, the value is c. For 8 the value is 32 * sqrt(2/3) + c.
Hence, the area is 32 * sqrt(2/3) which is approximately 39.19.