can you finish?
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