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

first find dS, note the radius of the rotation is x.

to solve make a substitution of

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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