# What is the surface area of the solid created by revolving #f(x) = (3x-2)^2 , x in [1,2]# around the x axis?

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To find the surface area of the solid created by revolving the function ( f(x) = (3x - 2)^2 ) around the x-axis over the interval ( x ) in [1, 2], you can use the formula for surface area of revolution:

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

where ( f'(x) ) is the derivative of ( f(x) ).

First, find the derivative of ( f(x) ):

[ f'(x) = 2(3x - 2)(3) = 6(3x - 2) ]

Now, substitute the values into the formula and integrate over the interval [1, 2]:

[ S = \int_{1}^{2} 2\pi (3x - 2)^2 \sqrt{1 + (6(3x - 2))^2} , dx ]

After solving this integral, you will get the surface area of the solid of revolution.

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