# How do you find the area of the surface generated by rotating the curve about the x-axis y=1/3x^3, 0<=x<=1?

Mar 12, 2018

$4 \frac{\pi}{324} \left[10 \sqrt{10} - 1\right]$$u n i t {s}^{2}$

#### Explanation:

The standard formula for the surface area of revolution is ,

SA="2piintf[x]sqrt[1+[[f'x]]^2dx, Let $y = f \left[x\right] = {x}^{3}$.....so in this case having taken the $\frac{1}{3}$ outside the integrand we obtain the following,

SA=2pi/3intx^3sqrt[1+[d/dx[x^3]]^2dx

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] = 3 {x}^{2}$ and squaring..we have $9 {x}^{4}$

Therefore $S A = 2 \frac{\pi}{3} \int {x}^{3} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$.... Now let $u = 1 + 9 {x}^{4}$..........$\left[1\right]$

Differentiating ......$\left[1\right]$, $\frac{\mathrm{du}}{36 {x}^{4}} = \mathrm{dx}$ and so the integral now becomes , $2 \frac{\pi}{3} \int {x}^{3} \frac{\sqrt{u \mathrm{du}}}{36 {x}^{3}}$ The terms in $x$ will now cancel and $\frac{1}{36}$can also be taken outside the integrand leaving

SA=[2pi]/[108]intsqrt[udu= $\frac{2 \pi}{108} \int {u}^{\frac{1}{2}} \mathrm{du}$

After integration, $S A = \frac{4 \pi}{324} \sqrt{{u}^{3}}$............ From......$\left[1\right]$, $u = 1 + 9 {x}^{4}$ and changing the bounds of integration

When $x = 1$, $u = 10$, and when $x = 0$, $u = 1$

Evaluated this becomes $S A$= $\frac{4 \pi}{324}$$\left[10 \sqrt{10} - 1\right]$units^2. Hope this is helpful.