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So given this curve: $$y=\sqrt{9x-18},\ \ 2 \le x \le 6$$

And using this lovely formula: $$\int2πy\sqrt{1+(\frac{dy}{dx})^2} dx$$

This is what I get for a set up:

$$\int_2^62π \sqrt{9x-18}\sqrt{(\frac{2x-3}{2x-24})} dx$$

I don't know, this set up is looks pretty weird. I think I screwed up when I was squaring (dy/dx).

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2 Answers 2

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Our function is $3\sqrt{x-2}$. We have $\frac{dy}{dx}=\frac{3}{2\sqrt{x-2}}$. Square, add $1$, and bring to a common denominator. We get $\frac{4x+1}{4(x-2)}$.

Take the square root, multiply by $(2\pi)3\sqrt{x-2}$. There is very nice cancellation.

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  • $\begingroup$ Sorry the curve I entered was incorrect. It should be: y=(9x-18)^1/2 $\endgroup$
    – user106039
    Sep 28, 2014 at 22:26
  • $\begingroup$ Yeah, problems such as these should simplify nicely... $\endgroup$
    – bjd2385
    Sep 28, 2014 at 22:37
  • $\begingroup$ Anyway, I was able to modify the work accordingly.Now I'm just wondering... Is u substitution required for this question at all? $\endgroup$
    – user106039
    Sep 28, 2014 at 22:38
  • $\begingroup$ With the change in function, we end up integrating a constant times $\sqrt{4x+1}$. Some people can safely just write down the answer. If you are more cautious, you will let $u=4x+1$. $\endgroup$ Sep 28, 2014 at 23:46
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You could also use $x=\frac{1}{9}(y^2+8)$ to get $\frac{dx}{dy}=\frac{2}{9}y$, so

$\displaystyle S=\int_{\sqrt{10}}^{\sqrt{46}}2\pi y\sqrt{1+\frac{4}{81}y^2} dy=\frac{2\pi}{9}\int_{\sqrt{10}}^{\sqrt{46}}y\sqrt{81+4y^2} dy$.

Now let $u=81+4y^2$ $\;$to get $\displaystyle\frac{\pi}{36}\int_{121}^{265}\sqrt{u} \;du$.

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