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I have the following homework problem which I can't seem to get right. The question stated is:

Find the volume of the solid generated by revolving the region bounded by the parabola $y=\frac{x^2}{25}$ and the line y=1 about the line y=1.

So I came up with the following integral and solved to get $\frac{8}{3}\pi$ but the correct answer according to my homework is $\frac{16}{3}\pi$. $$\pi\int_0^5\left(1-\frac{2x^2}{25}+\frac{x^4}{625}\right)dx$$

So is my integral correct and I'm simply not doing the math right when solving it? Or am I just missing something in the integral?

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You have to integrate from -5 to 5 or multiply your integral by two. That does the trick

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  • $\begingroup$ But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct. $\endgroup$ – blizz Jan 13 at 21:46
  • $\begingroup$ What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you? $\endgroup$ – pendermath Jan 13 at 21:48
  • $\begingroup$ I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying! $\endgroup$ – blizz Jan 13 at 21:54
  • $\begingroup$ Exactly! You got it $\endgroup$ – pendermath Jan 13 at 21:55
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You forgot to multiple it by $2$, because you set the lower bound $0$ instead of $-5$.

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