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The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.

Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about $x-axis$ is given as $$V=2\pi\int_0^{\infty}(f(x)^2-0)dx$$

Which gives: $2\pi\int_0^\infty\dfrac{a^6}{(a^2+x^2)^2}dx$

Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form $\dfrac{1}{x^4+2x^2+1}$ for $a=1$, which is transformed into $\dfrac{\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}=2[\dfrac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}-\dfrac{(1-\frac{1}{x^2})}{x^2+2+\frac{1}{x^2}}]$ which can be integrated, but this integral is tending to infinity. Can anyone help ? IS there a simpler way of doing this problem ?

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  • $\begingroup$ The integral definitely converges, by comparing the integrand to $\frac{a^6}{x^4}$. $\endgroup$ Jun 28, 2017 at 16:19
  • $\begingroup$ It does, but I don't know why, the way I am integrating, it yields $\infty$. @MatthewLeingang $\endgroup$
    – User9523
    Jun 28, 2017 at 16:21

3 Answers 3

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This seems like a trig substitution problem to me, albeit a not-so-straightforward one.

Let $x = a \tan\theta$. Then $dx = a \sec^2\theta\,d\theta$, and $a^2 + x^2 = a^2 \sec^2 \theta$. Since $\theta = \tan^{-1}\left(\frac{x}{a}\right)$, $\theta = 0$ when $x=0$, and $\theta\to\frac{\pi}{2}$ as $x\to\infty$. Therefore \begin{align*} 2\pi \int_0^\infty \frac{a^6}{(a^2 + x^2)^2}\,d\theta &= 2\pi \int_0^{\pi/2} \frac{a^6}{a^4\sec^4\theta}\,a \sec^2\theta \,d\theta\\ &= 2\pi a^3 \int_0^{\pi/2} \cos^2\theta\,d\theta \\ &= \pi a^3 \int_0^{\pi/2}(1+\cos2\theta)\,d\theta \\ &= \pi a^3 \left[\theta + \frac{1}{2}\sin2\theta\right]^{\pi/2}_0 \\ &= \frac{\pi^2 a^3}{2} \end{align*} This agrees with the Wolfram Alpha answer.

I can't quite figure out how you intended to integrate $\int\frac{1 + \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx$, but it could be that you were trying to represent a convergent improper integral as the difference of two divergent improper integrals. That leads to trouble, since $\infty - \infty$ is an indeterminate limit form.

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Your integral is equal $\frac{\pi^2}{2}( \frac{1}{a^2})^{\frac{3}{2}}a^6$ according to wolfram alpha. Your algebra must be wrong somewhere. I recommend trying trigonometric substitutions.

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  • $\begingroup$ That's just $\frac{\pi^2}{2}|a|^3$ right? (nice username btw) $\endgroup$
    – user541686
    Jun 28, 2017 at 21:23
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Neither of your two integrals in the last step are divergent. The first one is $$ I_1 = \int_0^\infty\frac{1 + \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \int_0^\infty \frac{x^2 + 1}{x^4 + 2x^2 + 1} \, dx = \int_0^\infty \frac{dx}{x^2 + 1} = \left[\arctan x\right]_0^\infty = \frac{\pi}{2}. $$ So that's nice and convergent. The other one is a bit trickier, but we can find it with a quick jaunt into the complex plane: $$ \int \frac{1 - \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \int \frac{x^2 - 1}{x^4 + 2x^2 + 1} \, dx = \frac{1}{2} \left[ \int \frac{1}{(x+i)^2} + \int \frac{1}{(x-i)^2} \right] \\= -\frac{1}{2} \left[ \frac{1}{x + i} + \frac{1}{x - i} \right] = -\frac{1}{2} \frac{2x}{x^2 + 1} = - \frac{x}{x^2 + 1} $$ and so $$ \int_0^\infty \frac{1 - \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \left[ - \frac{x}{x^2 + 1} \right]_0^\infty = 0. $$

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  • $\begingroup$ Good work. I figured out the first integral but not the second. $\endgroup$ Jun 28, 2017 at 19:03

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