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

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Answer 1 · 2017-07-22T12:28:35

$A = 216 π$ $A=216pi$

The area is given by the following integral :

$A = \int_{2}^{4} 2 π f (x) d x = 2 π \int_{2}^{4} (2 x^{3} - 2 x) d x =$ $A=int_2^4 2pif(x)dx=2piint_2^4(2x^3-2x)dx=$

$2 π {[\frac{x^{4}}{2} - x^{2}]}_{2}^{2} = 2 π (\frac{4^{4}}{2} - 4^{2} - \frac{2^{4}}{2} + 2^{2}) =$ $2pi[x^4/2-x^2]_2^4=2pi(4^4/2-4^2-2^4/2+2^2)=$

$2 π \cdot 108 = 216 π$ $2pi*108=216pi$

What is the surface area of the solid created by revolving f(x)=2x3−2x,x∈[2,4]f(x) =2x^3-2x , x in [2,4] around the x axis?