Help with what I think it's an impossible integral

[stonecoldgen]

So they give me the equation 3y²=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-x²)/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(1-3x²)/(6√(x-x³)/3))+1)

How the HELL am I supposed to integrate that?

 

[elvishatcher]

I think you made a mistake in determining the integrand. What you have for the dy/dx is not what I got when I differentiated the function (you also write the function as containing x(1-x)^2 initially and then when you rewrite it in terms of y you have an x(1-x^2) but I believe I got something different when I used either). Also, it doesn't look like you squared the dy/dx in the integrand. Finally, keep in mind that the r is equal to the value of the function, so you have to plug that in (You give the function in terms of y, but then you have r and dr/dx in the formula - those should be y and dy/dx... in this case, the radius is the height of the function)

 

[christoff]

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(1-3x²)/(6√(x-x³)/3))+1)

How the HELL am I supposed to integrate that?

I can't even read that. Please consider typing up those integrals in LaTeX.