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The question is : Find the volume of the solid obtained by rotating about the $y$-axis the region bounded by the curves $y= e^{-2x^2}$, $y=0$, $x=0$, $x=1$.

Should the bounds for the problem be taken from the $y$-axis or the $x$-axis?

I think that the integral for this problem would be:

$$\prod\int(e^{-2x^2})^2\,dx,$$

or do i have to rewrite the equation in terms of x?

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For a solid of revolution about the $y$ axis, the integral looks like

$$\pi \int_{e^{-2}}^1 dy \: (x(y))^2 = \frac{\pi}{2} \int_{e^{-2}}^1 dy \: (-\log{y})$$

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  • $\begingroup$ where does the x come from? $\endgroup$ – user71317 Apr 8 '13 at 2:56
  • $\begingroup$ The $x$ comes from the fact that we are integrating thin disks of radius $x$ that make up our volume. $\endgroup$ – Ron Gordon Apr 8 '13 at 2:58
  • $\begingroup$ You dont have to rewrite the equation to make it x = ...? $\endgroup$ – user71317 Apr 8 '13 at 2:59
  • $\begingroup$ the bound is e^-2 and not 0? $\endgroup$ – user71317 Apr 8 '13 at 3:00
  • $\begingroup$ do the bound stay the same? $\endgroup$ – user71317 Apr 8 '13 at 3:03

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