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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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Have you tried anything? If your integrand was $x^{-1/2}$, do you know how would you proceed? –  Matthew Conroy Sep 3 '12 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? –  Gerry Myerson Sep 3 '12 at 2:40
    
Would it be (-2)x^(1/2)? Now what do I do with the expression (1-9x)? –  Jared Sep 3 '12 at 2:40
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I will try u-substition of the expression. –  Jared Sep 3 '12 at 2:46
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1 Answer

up vote 2 down vote accepted

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

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I get pi*((\sqrt 2)-2). Is that correct? If so, is there any way to simplify it? –  Jared Sep 3 '12 at 2:56
    
@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.) –  Brian M. Scott Sep 3 '12 at 3:25
    
Okay I get V= pi/9 (2- \sqrt 2) as my final answer. That makes a lot more sense! Thanks –  Jared Sep 3 '12 at 13:32
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