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

May 11, 2018

$\frac{\pi}{6} \left[27 - 5 \sqrt{5}\right]$ units^$2$

#### Explanation:

The formula for surface area of revolution is, SA=2piint_a^bf[x]sqrt[1+[d/dxf[x]]^2dx.

From the question, $S A$=2piint_1^2sqrtxsqrt[1+[d/dxsqrtx]^2 [where $f \left[x\right]$=sqrtx].

Differentiating$\frac{d}{\mathrm{dx}} \sqrt{x}$= 1/[2sqrtx and so${\left[\frac{d}{\mathrm{dx}} \left[\sqrt{x}\right]\right]}^{2}$=1/[4x

Therefore, $S A$=2piint_1^2sqrtxsqrt[1+1/[4x]dx = 2piint_1^2sqrtxsqrt[[4x+1]/[4x]dx.

That is, $2 \pi {\int}_{1}^{2} \frac{\sqrt{x}}{2 \sqrt{x}}$sqrt[4x+1dx = piint_1^2sqrt[4x+1dx..........$\left[1\right]$

Integrating ....$\left[1\right]$ w.r.t. gives us $S A$ =, $\frac{\pi}{6} \sqrt{{\left[4 x + 1\right]}^{3}}$ and when evaluted for $x = 1 , x = 2$ will result inthe above answer when simplified.