Question: Let g(x) = (1 - 2x)(x - 3), and let A be the region enclosed by y = g(x), x = 1, x = 2, and y = -1.
If we revolve the region A about x-axis we obtain a solid. i. Find the volume of this solid. ii. Find its total surface area.
My Attempt: I have already drawn a graph and calculated the volume. For the volume, although a bit tedious, I expanded the function so that it's easier to integrate. But for the surface area, I have absolutely no clue on how to do it. I know its formula, but apparently there's no way of simplifying it. I think it may have something to do with trigonometric substitution? Can anyone please give me clues on how to approach it?
Your integral setup is correct. Though it is complicated, below is what I can do.
$$S = \int_{1}^{2}2\pi (1 - 2x)(x - 3)\sqrt{(-4x+ 7)^2 + 1}dx$$ $$= -16\pi \int_{1}^{2}((x - \frac{7}{4})^2 - (\frac{5}{4})^2)\sqrt{(x - \frac{7}{4})^2 + (\frac{1}{4})^2}dx$$ Let $ tan t = 4x - 7 \implies dx = \frac{1}{4}sec^2tdt$
Then $$ I_1 = \int(x - \frac{7}{4})^2\sqrt{(x - \frac{7}{4})^2 + (\frac{1}{4})^2}dx$$ $$= \int \frac{1}{256}\frac{sin^2tdt}{cos^5t}$$ $$= \frac{1}{1024}\int sintd(cos^{-4}t)$$ $$= \frac{1}{1024}sintcos^{-4}t -\frac{1}{1024}\int cos^{-3}tdt$$
Let $u = tant \implies du = sec^2tdt$
$$\int cos^{-3}txt = \int \sqrt{u^2 + 1}du$$ $$= \frac{u}{2}\sqrt{u^2 + 1} + \frac{1}{2}ln(u + \sqrt{u^2 + 1})$$ $$= \frac{1}{2}tant sect + \frac{1}{2}ln(tant + sect)$$ $$I_1 = \frac{1}{1024}sintcos^{-4}t - \frac{1}{1024}(\frac{1}{2}tant sect + \frac{1}{2}ln(tant + sect))$$ $$= \frac{1}{1024}(4x - 7)((4x - 7)^2 + 1)^{\frac{3}{2}} - \frac{1}{1024}(\frac{4x - 7}{2}\sqrt{(4x - 7)^2 + 1} + \frac{1}{2}ln(4x - 7 + \sqrt{(4x - 7)^2 + 1}))$$