What is the surface area produced by rotating $f(x)=x^3-x^2+1, x in [0,3]$ around the x-axis?

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Answer 1 · 2017-02-15T16:35:36

$9837/70pi=441.5$ cubic units, nearly.

graph{(x^3-x^2+1-y)(x-3+.001y)y(x-.001y)=0 [0, 4, -1, 20]}

Volume = $pi int y^2 dx$ , with $y = x^3-x^2+1$ and x from 0 to 3

$=pi int( x^3-x^2+1)^2 dx$ , with x from 0 t0 3

$=pi int (x^6+x^4+1-2x^5-2x^2+2x^3) dx$ , for the limits

$=pi[x^7/7+x^5/5+x-1/3x^6-2/3x^3+1/2x^4]$ , between x = 0 and 3

$=pi[(2187/7+243/5-243-18+81/2)-(0)]$

$=9837/70pi$

#(140.529)(3.1416)=441.5 cubic units, nearly.

What is the surface area produced by rotating f(x)=x^3-x^2+1, x in [0,3]f(x)=x^3-x^2+1, x in [0,3] around the x-axis?