What is the surface area of the solid created by revolving f(x) = x^2-2x+15 , x in [2,3] around the x axis?

May 14, 2018

${S}_{A} = 2 \pi {\int}_{2}^{3} \left({x}^{2} - 2 x + 15\right) \cdot \sqrt{1 + {\left(2 x - 2\right)}^{2}} \cdot \mathrm{dx} = 52.2146$

Explanation:

To determine the surface area created by revolving y around the
x-axis

${S}_{A} = 2 \pi {\int}_{a}^{b} y \cdot \sqrt{1 + {\left(y '\right)}^{2}} \cdot \mathrm{dx}$

$y = {x}^{2} - 2 x + 15$

$y ' = 2 x - 2$

${S}_{A} = 2 \pi {\int}_{2}^{3} \left({x}^{2} - 2 x + 15\right) \cdot \sqrt{1 + {\left(2 x - 2\right)}^{2}} \cdot \mathrm{dx}$

$= 2 \pi {\int}_{2}^{3} \left({x}^{2} - 2 x + 15\right) \cdot \sqrt{1 + \left(4 {x}^{2} - 8 x + 4\right)} \cdot \mathrm{dx}$

$= 2 \pi {\int}_{2}^{3} \left({x}^{2} - 2 x + 15\right) \cdot \sqrt{4 {x}^{2} - 8 x + 5} \cdot \mathrm{dx}$

$= {\left[\frac{223 \cdot a r \sinh \left(\frac{8 \cdot x - 8}{4}\right)}{64} + \frac{x \cdot {\left(4 \cdot {x}^{2} - 8 \cdot x + 5\right)}^{\frac{3}{2}}}{16} - {\left(4 \cdot {x}^{2} - 8 \cdot x + 5\right)}^{\frac{3}{2}} / 16 + \frac{223 \cdot x \cdot \sqrt{4 \cdot {x}^{2} - 8 \cdot x + 5}}{32} - \frac{223 \cdot \sqrt{4 \cdot {x}^{2} - 8 \cdot x + 5}}{32}\right]}_{2}^{3}$

$= {\left[\frac{223 a r \sinh \left(\frac{8 x - 8}{4}\right) + \left(4 x - 4\right) {\left(4 {x}^{2} - 8 x + 5\right)}^{\frac{3}{2}} + \left(446 x - 446\right) \sqrt{4 {x}^{2} - 8 x + 5}}{64}\right]}_{2}^{3}$

$= \left[\frac{223 \cdot a r \sinh \left(4\right) + 1028 \cdot \sqrt{17}}{64} - \frac{223 \cdot a r \sinh \left(2\right) + 466 \cdot \sqrt{5}}{64}\right]$

$\left[\frac{223 \cdot a r \sinh \left(4\right) - 223 \cdot a r \sinh \left(2\right) + 1028 \cdot \sqrt{17} - 466 \cdot \sqrt{5}}{64}\right]$

$= 52.2146$