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What is the surface area of the solid created by revolving #f(x) = 3x^2-4x+8 , x in [1,2]# around the x axis?

Answer 1

Evelyn Agard

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Answer 2

Cameron Alexander

To find the surface area of the solid created by revolving the function ( f(x) = 3x^2 - 4x + 8 ) around the x-axis over the interval ( x ) in [1,2], you would use the formula:

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

where ( f'(x) ) is the derivative of ( f(x) ). Then integrate over the given interval [1,2].

First, find ( f'(x) ), which is ( f'(x) = 6x - 4 ).

Now, substitute ( f(x) ) and ( f'(x) ) into the formula and integrate over the interval [1,2]:

[ S = 2\pi \int_{1}^{2} (3x^2 - 4x + 8) \sqrt{1 + (6x - 4)^2} , dx ]

Calculate this integral to find the surface area.

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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.