# What is the surface area of the solid created by revolving f(x)=sqrt(4x) for x in [0,1] around the x-axis?

Jun 26, 2016

$= 2 \pi$

#### Explanation:

the thin vertical strip width $\Delta x$ and between $y = 0$ and $y = \sqrt{4 x}$ will, when revolved round the x-axis, have volume

$\Delta V = \pi {y}^{2} \Delta x$

So

$V = \pi {\int}_{0}^{1} \setminus {y}^{2} \mathrm{dx}$

$= \pi {\int}_{0}^{1} \setminus 4 x \setminus \mathrm{dx}$ ........ as $y = \sqrt{4 x}$

$= 4 \pi {\int}_{0}^{1} \setminus x \setminus \mathrm{dx}$

$= 4 \pi {\left[{x}^{2} / 2\right]}_{0}^{1}$

$= 2 \pi$