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

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

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Answer 1 · 2016-05-29T02:04:01

approximately $100.896$ $100.896$

Explanation:

The surface area $S$ $S$ created by revolving the function $f (x)$ $f(x)$ around the $x$ $x$ -axis on the interval $x \in [a, b]$ $x in[a,b]$ can be found through:

$S=2piint_a^bf(x)sqrt(1+[f'(x)]^2)dx$

Using $f(x)=2x^2-4x+8$ and $f'(x)=4x-4$ on the interval $x in[1,2]$ gives:

$S=2piint_1^2(2x^2-4x+8)sqrt(1+(4x-4)^2)dx$

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 $2pi$ ) should be

$Sapprox100.896$

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

1 Answer

Explanation:

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What is the surface area of the solid created by revolving f(x) = 2x^2-4x+8 , x in [1,2]f(x)=2x2−4x+8,x∈[1,2]f(x) = 2x^2-4x+8 , x in [1,2] around the x axis?

1 Answer

Explanation:

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