# What is the surface area of the solid created by revolving f(x)=xsqrt(x-1) for x in [2,3] around the x-axis?

Mar 12, 2018

Area $\approx 19.363$ square units (to be specified from the units of $x$)

#### Explanation:

Double integration may be applied with the element of summation comprising some notional approximate rectangle (becoming exact in the limit as the length of the sides approaches $0$) of sides $\mathrm{dx}$ and $r d \theta$ where $r$ is the radius of the circle being swept to form the surface, such that, in this case, $r = x \sqrt{x - 1}$ and $\theta$ is the angle through which the radius is swept. It will be necessary to sweep from $\theta = 0$ to $\theta = 2 \pi$, over the interval $x = 2$ to $x = 3$.

Denoting the required area by A, retaining $r$ for the moment to illustrate the point, the integral may be set up as

$A = {\int}_{\theta = 0}^{\theta = 2 \pi} {\int}_{x = 2}^{x = 3} r \mathrm{dx} d \theta$

that is

$A = {\int}_{\theta = 0}^{\theta = 2 \pi} {\int}_{x = 2}^{x = 3} x \sqrt{x - 1} \mathrm{dx} d \theta$

Taking the inner integral first,

${\int}_{2}^{3} x \sqrt{x - 1} \mathrm{dx}$

this may be solved using the substitution

$u \left(x\right) = x - 1$

so that

$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

so that

$\int \mathrm{du} = \int \mathrm{dx}$

Noting
$u = x - 1$
implies
$x = 1 + u$
so that
$x \sqrt{x - 1} = \left(1 + u\right) \sqrt{u} = \sqrt{u} + {u}^{\frac{3}{2}}$

and, for the limits,
u(2) = 1
and
u(3) = 2

the required (inner) integral is

${\int}_{1}^{2} \left({u}^{\frac{1}{2}} + {u}^{\frac{3}{2}}\right) \mathrm{du}$

which, by the sum rule is

${\int}_{1}^{2} {u}^{\frac{1}{2}} \mathrm{du} + {\int}_{1}^{2} {u}^{\frac{3}{2}} \mathrm{du}$

$= \left(\frac{2}{3}\right) {\left[{u}^{\frac{3}{2}}\right]}_{1}^{2} + \left(\frac{2}{5}\right) {\left[{u}^{\frac{5}{2}}\right]}_{1}^{2}$

$= \left(\frac{2}{3}\right) \left({2}^{\frac{3}{2}} - 1\right) + \left(\frac{2}{5}\right) \left({2}^{\frac{5}{2}} - 1\right)$

$\approx 3.0817$ (to 4 decimal places, courtesy of Gnu Octave but defer that collapsing to an approximate value until the outer integral is evaluated ... )

Denoting the first (inner) integral by $K$ (as it evaluates to a constant), the outer integral may now be evaluated

$A = {\int}_{0}^{2 \pi} K d \theta$

$= K {\left[\theta\right]}_{0}^{2 \pi}$

$= 2 \pi K$

So the overall double integral evaluates to

$A = 2 \pi \left(\left(\frac{2}{3}\right) \left({2}^{\frac{3}{2}} - 1\right) + \left(\frac{2}{5}\right) \left({2}^{\frac{5}{2}} - 1\right)\right)$

$\approx 19.363$ square units (to be specified from the units of $x$)
(to 3 decimal places, again courtesy of Gnu Octave)