Disk Method for Volume of Solid with negative exponent.

Use the disk method to find the volume of the solid generated when the region bounded by $y=(1-9x)^{-1/4}$, $y=0$, $x=0$, and $x=1/18$ is revolved about the x-axis.

I know that to set this problem up, I have to use the equation $$V=\pi \int_0^{1/18} (1-9x)^{-1/2} \,dx$$ I get the exponent -1/2 because you must square the original equation to get the volume using the disk method. I do not remember exactly what to do when integrating the problem from here.

• Have you tried anything? If your integrand was $x^{-1/2}$, do you know how would you proceed? Commented Sep 3, 2012 at 2:37
• Are you able to do $\int(1-9x)^{1/2}\,dx$? How does the negativity of the exponent make the question harder for you? Commented Sep 3, 2012 at 2:40
• Would it be (-2)x^(1/2)? Now what do I do with the expression (1-9x)? Commented Sep 3, 2012 at 2:40
• I will try u-substition of the expression. Commented Sep 3, 2012 at 2:46

HINT: $u=1-9x$. $\qquad\qquad\qquad$
• @Izzy: I think that you forgot to take care of $dx$ when you did the substitution: you have $u=1-9x$, so $du=-9dx$, and $dx=-\frac19du$. You’re missing that factor of $-\frac19$ in the final answer. (And no, the final answer doesn’t simplify significantly.) Commented Sep 3, 2012 at 3:25