# What is the surface area of the solid created by revolving #f(x) = 2x^2-4x+8 , x in [1,2]# around the x axis?

approximately

Actually integrating this is very complex and far beyond the scope of this problem, so put this into a calculator--be mindful of parentheses.

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To find the surface area of the solid created by revolving the function ( f(x) = 2x^2 - 4x + 8 ) around the x-axis over the interval ([1, 2]), you can use the formula for the surface area of a solid of revolution:

[ S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

- ( f(x) = 2x^2 - 4x + 8 )
- ( f'(x) ) is the derivative of ( f(x) )

First, calculate the derivative of ( f(x) ): [ f'(x) = 4x - 4 ]

Now, substitute ( f(x) ) and ( f'(x) ) into the formula and integrate over the interval ([1, 2]):

[ S = \int_{1}^{2} 2\pi (2x^2 - 4x + 8) \sqrt{1 + (4x - 4)^2} , dx ]

Once you integrate this expression, you'll obtain the surface area of the solid.

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