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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.

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    I will try u-substition of the expression.2012-09-03

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HINT: $u=1-9x$. $\qquad\qquad\qquad$

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    Okay I get V= pi/9 (2- \sqrt 2) as my final answer. That makes a lot more sense! Thanks2012-09-03