- #1

stonecoldgen

- 107

- 0

^{2}=x(1-x)

^{2}

The idea is to find the surface area of the volume obtained by rotating around both axes.

So let's start with a rotation around the x-axis, I decided to rewrite the equation as:

y=√(x(1-x

^{2})/3)

I know that the surface area for a parallel rotation is S=∫2∏r√((dr/dx)2+1)

and I know the derivative of the function, so I end up with:

S=∫2∏r√2(1-3x

^{2})/(6√(x-x

^{3})/3))+1)

How the HELL am I supposed to integrate that?