# Math Help - Help finding surface area of a surface

1. ## Help finding surface area of a surface

Suppose the curve defined by y=4x^(1/2) for x <= x <= 1 is revolved about the x-axis.

1.) Find the surface area of the surface.

My answer:
b
S= 2PI ∫ r(x) [sqrt of 1 + (f'x)]^2
a
1
=2PI∫ 4x^1/2 [sqrt of 1 + 2x^3/2]^2
0

1
=2PI/6∫ 6(4x^1/2)(1+2x^3)^1/2
0

1
=2PI/6∫6(4x^1/2)(1+2x^3)^1/2
0

1
=PI/3[((1+2x^3)^3/2)/(3/2)|
0

=PI/4.5[((3^3/2)-1)]

My answer=4.19PI/4.5

Have I come anywhere close to doing this problem correctly? I'm thinking I might have done something wrong. If I have, can someone show me what I did wrong or how to do this problem correctly? Thanks

2. $y = 4\sqrt{x}$

$y' = \frac{2}{\sqrt{x}}$

$(y')^2 = \frac{4}{x}$

$S = 2\pi \int_0^1 4\sqrt{x} \cdot \sqrt{1 + \frac{4}{x}} \, dx$

$S = 8\pi \int_0^1 \sqrt{x + 4} \, dx$

can you finish?