What is the surface area of the solid created by revolving #f(x) =2x^3-2x , x in [2,4]# around the x axis?

Answer 1

#A=216pi#

The area is given by the following integral :

#A=int_2^4 2pif(x)dx=2piint_2^4(2x^3-2x)dx=#
#2pi[x^4/2-x^2]_2^4=2pi(4^4/2-4^2-2^4/2+2^2)=#
#2pi*108=216pi#
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Answer 2

To find the surface area of the solid created by revolving the function ( f(x) = 2x^3 - 2x ) around the x-axis over the interval ([2, 4]), you can use the formula for surface area of revolution, which is given by:

[ S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + \left( f'(x) \right)^2} , dx ]

First, find the derivative of ( f(x) ) which is ( f'(x) = 6x^2 - 2 ). Then, plug the function and its derivative into the formula and integrate over the given interval ([2, 4]) to find the surface area.

[ S = \int_{2}^{4} 2\pi (2x^3 - 2x) \sqrt{1 + (6x^2 - 2)^2} , dx ]

Compute this integral to find the surface area of the solid.

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Answer from HIX Tutor

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.

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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