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I'm pretty confused by how we should determine Integration limits.

$f_p(x)=x^3-px^2$

$V$ is the surface area enclosed by the graph of $f_2$ and the x-axis.

$V$ is revolved around the x-axis. Analytically calculate the volume of the resulting solid of revolution

I know the formula for figuring out the volume of revolutions around the $x$-achse is to just integrate $\pi(f(x)^2$, but I would still need to figure out the integration limits and I'm really lost on how to do that.

Thanks in advance for any help!

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  • $\begingroup$ Where does $f_2$ cross the $x-$axis? $\endgroup$
    – Doug M
    May 10 '18 at 16:09
  • $\begingroup$ But why would the x intercept be the point of upper limit? Doesn't make sense to me. $\endgroup$
    – user472288
    May 10 '18 at 16:11
  • $\begingroup$ you have been asked to find the region enclosed by the curve and the x-axis. $\endgroup$
    – Doug M
    May 10 '18 at 16:11
  • $\begingroup$ So if the graph intercepts the x achse twice? How would I know which one is meant? $\endgroup$
    – user472288
    May 10 '18 at 16:12
  • $\begingroup$ Both of them! One is the lower limit, and one is the upper limit. $\endgroup$
    – Doug M
    May 10 '18 at 16:13
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To find the integration limits, you must find the places where $f_2 (x)$ and the $x$-axis cross. You do this by setting $$x^3 - 2x^2 = 0$$ $$x^2 (x-2) = 0$$ $$x = 0, 2$$ Because $f_2 (x) \le 0$ for $x \in [0, 2]$, $V$ would be equal to the opposite of the integral of $f_2 (x)$. $$V = -\int_{0}^{2} f_2 (x)~ {\rm d}x = -\int_{0}^{2} \left(x^3 - 2x^2\right) {\rm d}x$$ I hope this is of help!

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