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Help with what I think it's an impossible integral!

  • #1
107
0
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?
 

Answers and Replies

  • #2
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)
 
  • #3
123
0
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.
 

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