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?

Question

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?

Emily Ammerman · Accepted Answer

approximately #100.896#

The surface area #S# created by revolving the function #f(x)# around the #x#-axis on the interval #x in[a,b]# can be found through:

#S=2piint_a^bf(x)sqrt(1+[f'(x)]^2)dx#

Using #f(x)=2x^2-4x+8# and #f'(x)=4x-4# on the interval #x in[1,2]# gives:

#S=2piint_1^2(2x^2-4x+8)sqrt(1+(4x-4)^2)dx#

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

The answer you receive (don't forget to multiply by #2pi#) should be

#Sapprox100.896#

Isaiah Allen · Answer

undefined 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.