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?

Question

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?

Emma Aldridge · Accepted Answer

#14086/5 pi#

around the x axis, the volume of revolution is

#V = pi int_{x=2}^3 y^2 \ dx#

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

so #y(x) = 6 (x^2 - 1/2 x + 22/6)# #= 6 ((x - 1/4)^2 - 1/16 + 22/6)# #= 6 ((x - 1/4)^2 + 173/48)# so #y(u) = 6 u^2 + 173/8# where #u = x - 1/4, du = dx#

and the integral becomes

#pi int_{u=7/4}^{11/4} y^2 \ du# #= pi int_{u=7/4}^{11/4} ( 6 u^2 + 173/8)^2 \ du# #= pi int_{u=7/4}^{11/4} 36 u^4 + 519/2 u^2+ 29929/64 \ du# #= pi ( 36/5 u^5 + 519/6 u^3+ 29929/64 u)_{u=7/4}^{11/4} # #= 14086/5 pi#

Cameron Alexander · Answer

undefined 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.