# Find the Surface Area of Revolution

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?

• the horizontal line is $y=-1$ ? below the rotation axis? – G Cab Mar 28 '19 at 22:17

$$S = \int_{1}^{2}2\pi (1 - 2x)(x - 3)\sqrt{(-4x+ 7)^2 + 1}dx$$ $$= -16\pi \int_{1}^{2}((x - \frac{7}{4})^2 - (\frac{5}{4})^2)\sqrt{(x - \frac{7}{4})^2 + (\frac{1}{4})^2}dx$$ 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 $$u = tant \implies du = sec^2tdt$$
$$\int cos^{-3}txt = \int \sqrt{u^2 + 1}du$$ $$= \frac{u}{2}\sqrt{u^2 + 1} + \frac{1}{2}ln(u + \sqrt{u^2 + 1})$$ $$= \frac{1}{2}tant sect + \frac{1}{2}ln(tant + sect)$$ $$I_1 = \frac{1}{1024}sintcos^{-4}t - \frac{1}{1024}(\frac{1}{2}tant sect + \frac{1}{2}ln(tant + sect))$$ $$= \frac{1}{1024}(4x - 7)((4x - 7)^2 + 1)^{\frac{3}{2}} - \frac{1}{1024}(\frac{4x - 7}{2}\sqrt{(4x - 7)^2 + 1} + \frac{1}{2}ln(4x - 7 + \sqrt{(4x - 7)^2 + 1}))$$
• I tried two time adding in the rest of the answer but the system did not take it. Anyway you have to calculate the second part of I with $(\frac{5}{4})^2$ which is in standard form. After that just substitute x = 2 and x = 1. – KY Tang Mar 28 '19 at 21:29