What is the surface area of the solid created by revolving f(x) = 2x^2-4x+8 , x in [1,2] around the x axis?

May 29, 2016

approximately $100.896$

Explanation:

The surface area $S$ created by revolving the function $f \left(x\right)$ around the $x$-axis on the interval $x \in \left[a , b\right]$ can be found through:

$S = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$

Using $f \left(x\right) = 2 {x}^{2} - 4 x + 8$ and $f ' \left(x\right) = 4 x - 4$ on the interval $x \in \left[1 , 2\right]$ gives:

$S = 2 \pi {\int}_{1}^{2} \left(2 {x}^{2} - 4 x + 8\right) \sqrt{1 + {\left(4 x - 4\right)}^{2}} \mathrm{dx}$

Actually integrating this is very complex and far beyond the scope of this problem, so put this into a calculator--be mindful of parentheses.

The answer you receive (don't forget to multiply by $2 \pi$) should be

$S \approx 100.896$