What is the surface area produced by rotating #f(x)=x^3-8, x in [0,2]# around the x-axis?

Question

Cameron Alexander · Accepted Answer

first find dS, note the radius of the rotation is x.
#dS=sqrt(1+9x^4)dx#
#int_0^2 2pixdS#
to solve make a substitution of #w=3x^2# If you make the substitution with w your integral becomes #int_0^12 pi/3*sqrt(1+w^2)dw# with a trigonometric substitution #w=tan(theta)# the integral becomes #pi/3 int_0^arctan(12) sec^3(theta)d(theta)# which you can solve by parts or using a table.

Christopher Agresta · Answer

undefined To find the surface area produced by rotating the function $ f(x) = x^3 - 8 $ around the x-axis over the interval [0, 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) $ is the function to be rotated.
- $ f'(x) $ is the derivative of $ f(x) $.

First, find the derivative of $ f(x) $:
$$ f'(x) = 3x^2 $$

Next, plug the function and its derivative into the formula and integrate over the interval [0, 2]:

$$ S = \int_{0}^{2} 2\pi (x^3 - 8) \sqrt{1 + (3x^2)^2} \, dx $$

After integrating, you'll get the surface area.