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)]$$\sqrt{1+[f(x)]^2}$ $dx$ $-$ $\int_1^2 2\pi(-1)$ $\sqrt{1+(-1)^2} dx $
Is it correct ? Thanks!
Did you draw a picture for this problem? Well, it certainly seems like the part that's right under the x-axis and above the line $y=-1$ is not going to be counted because as you revolve the region about the x-axis, its volume is going to be absorbed by the volume of the whole thing. Nevertheless, the integrals you get for the volume and the surface are the following (it's the disk method and the formula for the area of a surface of revolution):
$$ 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 $$