Question

find the area of the surface that is generated by revolving to portion of the curve y=x^3 between x=0 and x=1 about x axis?

Answer 1

ANSWER $\frac{\pi}{27}\cdot (10\sqrt{10}-1)\cong3.5631$

EXPLANTION

If $f'(x)$ is continuous on $[a,b]$ and the curve $y=f(x)$ is rotated about x-axis , the area of the surface of revolution so generated is

$S =2\pi \int_{a}^{b}\left | f(x) \right |\sqrt{1+\left ( f'(x) \right )^{2}}dx$

Let $f(x)=x^3$ , $a=0, b=1$ , then $f'(x)=3x^2$ and

$S =2\pi \int_{0}^{1}x^3\sqrt{1+9x^4}dx$

Calculate the integral using the substitution $1+9x^4=t$ . So, since $36x^3dx=dt$ , we have

$S = \frac{2\pi}{36}\int_{0}^{1}36x^3\sqrt{1+9x^4}dx=\\= \frac{\pi}{18} \int_{1}^{10}\sqrt{t}dt=\frac{2\pi}{18\cdot 3}\cdot[t^{\frac{3}{2}}]_{1} ^{10} =\frac{2\pi}{18\cdot 3}\cdot10\sqrt{10} -\frac{2\pi}{18\cdot 3} = \frac{\pi}{27}\cdot (10\sqrt{10}-1)\cong3.5631$