Find the area obtained from obtained by rotating the curve about the axis. \[9x=y^2+18\]. \[2 \le x \le 6\]

 Write the curve as x = y^2/9 + 2. 
If x = 2, then y = 0 
If x = 6, then y = 6 (the rotation of this region takes care of the negative counterpart.) 

A = integral 2 pi y sqrt(1 + [f '(y)]^2) dy 
= (2 pi) * integral(y = 0 to 6) y sqrt(1 + [2y/9]^2) dy 
= (2 pi/9) * integral(y = 0 to 6) y sqrt(81 + 4y^2) dy 
= (2 pi/9) * [(1/8) * (2/3)(81 + 4y^2)^(3/2) for y = 0 to 6] 
= 98 pi / 3. 

I hope that helps!