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

The question seems rather odd.

The formula for computing the surface area is

This integral yields a very complicate solution, which I don't believe your homework could ever ask for. For completeness sake, you can check it from WolframAlpha

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

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

Where ( f'(x) ) is the derivative of ( f(x) ) with respect to ( x ), and ( a ) and ( b ) are the limits of integration.

First, find ( f'(x) ):

[ f'(x) = 2(3x-1) \cdot 3 = 6(3x-1) ]

Then, substitute into the surface area formula and integrate:

[ S = 2\pi \int_1^3 (3x-1)^2 \sqrt{1 + (6(3x-1))^2} , dx ]

After integrating, you'll have the surface area of the solid of revolution.

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