Find the volume of the solid obtained by rotating the region in the first quadrant bounded by y = x^2 and y = 9 about the y axis.
My first answer was 54, my second 18. Neither are correct, but I feel at a loss for this problem.
Have you learned the disk method for finding a volume of a solid of revolution? You have a vertical axis of revolution so you'll have to rewrite the function $\displaystyle y = x^2$ in terms of x -> $\displaystyle x = \sqrt{y}$.
The integral to set up is
$\displaystyle V = \pi \displaystyle \int_c^d [R(y)]^2 dy$
Plug in $\displaystyle \sqrt{y}$ for R(y), 0 for c, and 9 for d:
$\displaystyle V = \pi \displaystyle \int_0^9 (\sqrt{y})^2 dy$
Take it from here.