surface area?

Question

Find the area of the surface obtained by rotating the curve
y= \sqrt{6 x}
from x=0 to x = 3 about the x-axis.

formula is the integral of 2pi y sqrt(f'(x))^2).

actually pascal my other question i posted i figured out what i did wrong i forgot to add in the ^(3/2) for both sides...thanks though. oh and your explanation helped me with this problem thanks. :)

Answer 1

Actually, the formula is ∫2π y √(1+(y')²) dx -- you were right about the formula when you posted (and for some reason removed?) your previous question, and I hope my (incorrect) answer didn't throw you off.

Assuming that you actually mean √(6x) and that that backslash doesn't indicate something that got misplaced:

2π ∫√(6x) √(1+(6/(2√(6x))²) dx
2π ∫√(6x) √(1+(3/√(6x))²) dx
2π ∫√(6x) √(1+9/(6x)) dx
2π ∫√(6x+9) dx
u=6x+9, du=6 dx
π/3 ∫√u du
2π/9 u√u + C
2π/9 (6x+9)√(6x+9) + C
Evaluating from 0 to 3:
2π/9 (18+9)√(18+9) - 2π/9 (9)√(9)
6π√27 - 2π√9
18π√3 - 6π
A ≈ 79.09561

Answer 2

When rotating about the x axis, your radius is your y-value for each x. You'll want to integrate from 0 to 3. For the surface area, you want to find the circumference at each x. So your integrand (I think is the word) is 2(pi)r(x), where r(x) is the radius in terms of x.

You should be able to solve this from there.