ttyme411gl

2022-07-04

What is the surface area of the solid created by revolving

$f(x)=2{x}^{2}-4x+8,x\in [1,2]$ around the x axis?

$f(x)=2{x}^{2}-4x+8,x\in [1,2]$ around the x axis?

Charlize Manning

Beginner2022-07-05Added 12 answers

The surface area S created by revolving the function f(x) around the x-axis on the

interval $x\in [a,b]$ can be found through:

$S=2\pi {\int}_{a}^{b}f(x)\sqrt{1+[{f}^{\prime}(x){]}^{2}}dx$

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

$S=2\pi {\int}_{1}^{2}(2{x}^{2}-4x+8)\sqrt{1+(4x-4{)}^{2}}dx$

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 $2\pi $ ) should be

$S\approx 100.896$

interval $x\in [a,b]$ can be found through:

$S=2\pi {\int}_{a}^{b}f(x)\sqrt{1+[{f}^{\prime}(x){]}^{2}}dx$

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

$S=2\pi {\int}_{1}^{2}(2{x}^{2}-4x+8)\sqrt{1+(4x-4{)}^{2}}dx$

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 $2\pi $ ) should be

$S\approx 100.896$

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