## marigirl one year ago I don't quite understand this question The Surface area of a solid is given by the formula $S=2\pi \int\limits_{a}^{b}f(x)\sqrt{1+(f'(x))^2}dx$ The graph of $f(x)=x^3$ is rotated between x=0 and x=a, where a>0 about the x axis. If the surface area is 4pi, find the value of a

1. marigirl

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

2. marigirl

ill post the ans on here, and if u could explain it please :)

3. Astrophysics

So you're having trouble with the integration?

4. marigirl

i dont quite get what to integrate,

5. marigirl

6. Astrophysics

Oh you can do a u substitution u = 1+9x^4

7. marigirl

so just integrate the whole thing normally..the f(x) and f'(x) parts were putting me off

8. Astrophysics

Yeah, that's just showing you what you plug into the integral