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A body with volume $V(w)$ is between the curve $y = \frac{1}{x + \sqrt{x}}$ and the $x$ axis, for $1 \leq x \leq w$. I am to rotate the body one turn around the $x$ axis, and determine $\lim_{w \to \infty} V(w)$.

In similar problems, I have used the formula $V = \pi \int_a^b f(x)^2\,dx$, but I have never worked with approaching infinity in similar problems. The textbook answer to the problem only confuses me further.

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In fact you want to solve $\int_1^{w}\pi\frac{dx}{(x+\sqrt{x})^2}$?? and then take the limit?? – Babak S. Jan 19 '13 at 19:44
That's my thinking, yes, but it doesn't seem to be it according to the text book answer. – Quispiam Jan 19 '13 at 19:53
But what you have done seems correct as I thought the same way also. May I ask what is the answer? – Babak S. Jan 19 '13 at 20:00
The answer is $2\pi\left[ \ln \frac{t}{t+1} + \frac{1}{t+1} \right ]_1^\sqrt{w}$ – Quispiam Jan 19 '13 at 20:16
up vote 1 down vote accepted

You have the correct expression for volume, in this case $$\int_1^w \pi\frac{dx}{(x+\sqrt{x})^2}.$$ There is a natural substitution $u=\sqrt{x}$. Then $dx=2u\,du$, and after some mild cancellation we arrive at $$\int_1^\sqrt{w} 2\pi\frac{du}{u(u+1)^2}.$$ The integrand is a natural candidate for partial fractions. Forget about the $2\pi$ for a while. We have $$\frac{1}{u(u+1)^2}=\frac{1}{u}-\frac{1}{u+1}-\frac{1}{(u+1)^2}.$$ An antiderivative of this is $\log u-\log (u+1)+\frac{1}{u+1}$. It is convenient to express $\log u-\log (u+1)$ as $\log\left(\frac{u}{u+1}\right)$ since that makes the later limit process more obvious. So our volume is $$2\pi \left. \left(\log\left(\frac{u}{u+1}\right)+\frac{1}{u+1}\right)\right|_1^{\sqrt{w}}.$$

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