What is the surface area of the solid created by revolving f ( x ) = 2 x <m

ttyme411gl 2022-07-04 Answered
What is the surface area of the solid created by revolving
f ( x ) = 2 x 2 4 x + 8 , x [ 1 , 2 ] around the x axis?
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Answers (1)

Charlize Manning
Answered 2022-07-05 Author has 12 answers
The surface area S created by revolving the function f(x) around the x-axis on the
interval x [ a , b ] can be found through:
S = 2 π a b f ( x ) 1 + [ f ( x ) ] 2 d x
Using f ( x ) = 2 x 2 4 x + 8 and f'(x)=4x-4 on the interval x [ 1 , 2 ] gives:
S = 2 π 1 2 ( 2 x 2 4 x + 8 ) 1 + ( 4 x 4 ) 2 d x
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 2 π ) should be
S 100.896

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