# What is the surface area of the solid created by revolving #f(x) = 6x^2-3x+22 , x in [2,3]# around the x axis?

around the x axis, the volume of revolution is

that looks horrendous and might be made simpler by a shift in axis. completing the square should ultimately make the algebra a little bit easier to bear

and the integral becomes

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To find the surface area of the solid created by revolving the function ( f(x) = 6x^2 - 3x + 22 ) around the x-axis over the interval ([2,3]), we can use the formula for surface area 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 ).

First, find the derivative of ( f(x) ): [ f'(x) = 12x - 3 ]

Then, substitute the function and its derivative into the formula: [ S = 2\pi \int_{2}^{3} (6x^2 - 3x + 22) \sqrt{1 + (12x - 3)^2} , dx ]

Now, integrate this expression to find the surface area.

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