# What is the surface area produced by rotating f(x)=(1-x)/(x^2+6x+9), x in [0,3] around the x-axis?

Surface Area $\textcolor{red}{S = - 0.1637650055 \text{ }}$square unit

#### Explanation:

Given $y = \frac{1 - x}{{x}^{2} + 6 x + 9} \text{ }$ and $x = 0$ to $x = 3$

The formula for finding the surface area or revolution

$S = 2 \pi {\int}_{a}^{b} y \cdot \mathrm{ds}$

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

$S = 2 \pi {\int}_{a}^{b} \frac{1 - x}{{x}^{2} + 6 x + 9} \cdot \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Also $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 5}{x + 3} ^ 3$

$S = 2 \pi {\int}_{a}^{b} \frac{1 - x}{{x}^{2} + 6 x + 9} \cdot \sqrt{1 + {\left(\frac{x - 5}{x + 3} ^ 3\right)}^{2}} \mathrm{dx}$

I suggest Simpson's Rule to calculate the integration and

$S = 2 \pi {\int}_{a}^{b} \frac{1 - x}{{x}^{2} + 6 x + 9} \cdot \sqrt{1 + {\left(\frac{x - 5}{x + 3} ^ 3\right)}^{2}} \mathrm{dx}$

$\textcolor{red}{S = - 0.1637650055 \text{ }}$square unit

God bless...I hope the explanation is useful.