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 a0 about the x axis. If the surface area is 4pi, find the value of a

Question

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

marigirl · Answer

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

marigirl · Answer

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

astrophysics · Answer

So you're having trouble with the integration?

marigirl · Answer

i dont quite get what to integrate,

marigirl · Answer

attachment

astrophysics · Answer

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

marigirl · Answer

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

astrophysics · Answer

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