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The region R is bound by the curve $y=\frac{1}{(1+4x^2)^2}$, the line $x=\frac{\sqrt{3}}{2}$, the $x$-axis and the $y$-axis.

Find the volume of the solid formed when R is rotated $2\pi$ radians about the $y$-axis.

How do you tackle this question? I tried using

$$V=\pi\int_0^1 \frac{\sqrt{\frac{1}{y}}-1}{4} dy$$ since $$x^2=\frac{\sqrt{\frac{1}{y}}-1}{4}$$ but that's not the answer.

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I find more "natural" the cylinder method $$V=2 \pi \int_0^{\frac{\sqrt{3}}{2}} x\cdot \frac{1}{\left(4 x^2+1\right)^2} \, dx$$ Substitute $4x^2+1=t\to 8x\,dx=dt\to x\,dx=\frac{dt}{8}$ integration limits become

$x=0\to t=1;\;x=\dfrac{\sqrt 3}{2}\to t=4$

so we have

$$V=2\pi\int_1^4 \,\frac{1}{8}\frac{dt}{t^2}=\frac{\pi}{4}\left[-\frac{1}{t}\right]_1^4=\frac{\pi}{4}\left(-\frac{1}{4}+1\right)=\color{red}{\frac{3}{16}\pi}$$

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It should be $$\pi\int_\frac{1}{16}^1 \frac{\sqrt{\frac{1}{y}}-1}{4} dy+\pi\left(\frac{\sqrt3}{2}\right)^2\cdot\frac{1}{16}$$

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This is probably better done as a "volume by shells" problem. Only part of the time is the radius of your disk equal to your last displayed equation. Sometimes the radius is $\sqrt{3}/2.$

Note, for instance, that your integral is improper at $y=0.$

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  • $\begingroup$ I am not understanding your comment. My teacher says that the integral is only valid from $V=\pi\int_\frac{1}{16}^1 \frac{\sqrt{\frac{1}{y}}-1}{4} dy$ since the graph never touches the $x$-axis but I can't understand how to add the last part of the graph to the total volume. $\endgroup$ – user373534 Nov 12 '17 at 13:31
  • $\begingroup$ You have to do a separate integral for $y=0$ to $y=1/16$. Or just compute the volume of that cylinder with the formula. But really, this is a "shells" problem. $\endgroup$ – B. Goddard Nov 12 '17 at 13:36

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