# What is the surface area of the solid created by revolving f(x) =2x+5 , x in [1,2] around the x axis?

Jan 5, 2017

We need to calculate the area of te curved surface and the two end circles.

#### Explanation:

The end circles have radii $f \left(1\right) = 7$ and $f \left(2\right) = 9$ respectively and so have areas $49 \pi$ and $81 \pi$.

The are of the curved surface may be calculated by considering an infinitesimal ring of radius $f \left(x\right)$ between the values $x$ and $x + \mathrm{dx}$, which will have thickness $\mathrm{dl} = \sqrt{{\mathrm{df}}^{2} \left(x\right) + {\mathrm{dx}}^{2}}$ and therefore $A r e a = 2 \pi f \left(x\right) \mathrm{dl}$. The total curved area will then be $\setminus {\int}_{1}^{2} 2 \pi f \left(x\right) \mathrm{dl}$
Now, $\mathrm{dl} = \sqrt{{\mathrm{df}}^{2} \left(x\right) + {\mathrm{dx}}^{2}} = \mathrm{dx} \sqrt{{\left(\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right)}^{2} + 1} = \mathrm{dx} \sqrt{{2}^{2} + 1}$
$= \mathrm{dx} \sqrt{5}$.
So we have to evaluate the integral
$\setminus {\int}_{1}^{2} \mathrm{dx} 2 \pi f \left(x\right) \sqrt{5} = 2 \sqrt{5} \pi {\left[2 {x}^{2} / 2 + 5 x\right]}_{1}^{2} = 2 \sqrt{5} \pi \left[14 - 6\right]$
$= 16 \sqrt{5} \pi$.

The total area is then $= \left(16 \sqrt{5} + 49 + 81\right) \pi = \left(16 \sqrt{5} + 130\right) \pi$.