# What is the surface area of the solid created by revolving f(x)=-2x^3-3x^2+6x-12 over x in [2,3] around the x-axis?

Jan 20, 2016

Use the surface area of rotation integral:
$S A = {\int}_{a}^{b} 2 \pi f \left(x\right) \sqrt{\left(1 + {\left(\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right)}^{2}\right)} \mathrm{dx}$
Integrate the following:
$S A = 2 \pi {\int}_{2}^{3} \left(- 2 {x}^{3} - 3 {x}^{2} + 6 x - 12\right) \sqrt{1 + \left(- 6 {x}^{2} - 6 x + 6\right)} \mathrm{dx}$

#### Explanation:

$S A = {\int}_{a}^{b} 2 \pi f \left(x\right) \sqrt{\left(1 + {\left(\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right)}^{2}\right)} \mathrm{dx}$
may look bad it is not really. Take the derivative of f(x)
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = - 6 {x}^{2} - 6 x + 6$
now integrate

$S A = 2 \pi {\int}_{2}^{3} \left(- 2 {x}^{3} - 3 {x}^{2} + 6 x - 12\right) \sqrt{1 + \left(- 6 {x}^{2} - 6 x + 6\right)} \mathrm{dx}$