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Let $g(x) = (1-2x)(x-3)$, B be the region enclosed by $g(x)$, $x = 1$, $x = 2$ and $y = -1$

after revolve the region B about x-axis, we have a solid. The volume of solid and the total surface area is?

I am wondering the correctness of my approach...

Volume = $\int_1^2 \pi[f(x)]^2 - \pi (-1)^2 dx $

TSA = $\int_1^2 2\pi [f(x)]$$\sqrt{1+[f(x)]^2}$ $dx$ $-$ $\int_1^2 2\pi(-1)$ $\sqrt{1+(-1)^2} dx $

Is it correct ? Thanks!

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Did you draw a picture for this problem? Well, it certainly seems like the part that's right under the x-axis and above the line $y=-1$ is not going to be counted because as you revolve the region about the x-axis, its volume is going to be absorbed by the volume of the whole thing. Nevertheless, the integrals you get for the volume and the surface are the following (it's the disk method and the formula for the area of a surface of revolution):

$$ V=\pi\int_{1}^{2}(-2x^2+7x-3)^2\,dx $$

$$ S=2\pi\int_{1}^{2}(-2x^2+7x-3)\sqrt{1+\left[\frac{d}{dx}\left(-2x^2+7x-3\right)\right]^2}\,dx $$

enter image description here

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  • $\begingroup$ That mean when I use the above formula to calculate V, I also calculate the part which below x axis and above y = - 1? $\endgroup$ – eulerisgod Mar 3 '19 at 5:49
  • $\begingroup$ Well, think about what's going on there for a minute. The axis of revolution is the x-axis. As you rotate that shape (the entire region) around the x-axis, the volume of that part below the x-axis is going to be entirely inside the volume of that whole thing that you're rotating. It contributes nothing to the total volume because its volume is completely inside the volume of the solid that you get. You might as well say that the region is enclosed from below by the line $y=0$ or even the x-axis. It will make no difference. Do you see what I'm trying to say? $\endgroup$ – Michael Rybkin Mar 3 '19 at 5:57
  • $\begingroup$ I think i got it. There are only one part ( above / below ) of rotation axis will be counted. Thank so much ! $\endgroup$ – eulerisgod Mar 3 '19 at 6:57
  • $\begingroup$ Sorry about annoying.... I really can't compute the integral of the Surface area ... $\endgroup$ – eulerisgod Mar 3 '19 at 13:17

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