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

1 Answer 1

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

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  • $\begingroup$ I get pi*((\sqrt 2)-2). Is that correct? If so, is there any way to simplify it? $\endgroup$
    – Jared
    Commented Sep 3, 2012 at 2:56
  • $\begingroup$ @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.) $\endgroup$
    – Brian M. Scott
    Commented Sep 3, 2012 at 3:25
  • $\begingroup$ Okay I get V= pi/9 (2- \sqrt 2) as my final answer. That makes a lot more sense! Thanks $\endgroup$
    – Jared
    Commented Sep 3, 2012 at 13:32

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