More specifically I have a problem that states:
Let $\Omega$ be the region bounded by the function $Y=x^3$ and the x-axis for $0 \le x\le1$ .This region is revolved about the x-axis. Find the surface area of the solid formed.
I got $\frac{\pi}{27}$ but that doesnt seem to be answer choice.
Best Answer
Use the formula:
$$\text{S}_x=2\pi\int_a^bf(x)\sqrt{1+\left(f'(x)\right)^2}\space\text{d}x$$
So we get, when $f(x)=x^3$, $a=0$ and $b=1$:
$$\text{S}_x=2\pi\int_0^1x^3\sqrt{1+\left(\frac{\text{d}}{\text{d}x}\left(x^3\right)\right)^2}\space\text{d}x=2\pi\int_0^1x^3\sqrt{1+9x^4}\space\text{d}x$$
Because $\frac{\text{d}}{\text{d}x}\left(x^3\right)=3x^2$.
Now, use a substitution $u=1+9x^4$ and $\text{d}u=36x^3\space\text{d}x$:
$$\text{S}_x=\frac{\pi}{18}\int_1^{10}\sqrt{u}\space\text{d}u=\frac{\pi\left(10\sqrt{10}-1\right)}{27}\approx3.563121852013749$$