Answer to Question #291627 in Calculus for harsh

Question

find the volume of the solid generated by revolving region bounded by the curves y = x, y = 2 - x, y = 0, about y = -1

Answer 1

$x=2 - x=>x=1$

$x=0, 2-x=0=>x=2$

$V=\pi\displaystyle\int_{0}^{1}((x+1)^2-(0+1)^2)dx$

$+\pi\displaystyle\int_{1}^{2}((2-x+1)^2-(0+1)^2)dx$

$=\pi[\dfrac{x^3}{3}+x^2]\begin{matrix} 1 \\ 0 \end{matrix}+\pi[8x-3x^2+\dfrac{x^3}{3}]\begin{matrix} 2 \\ 1 \end{matrix}$

$=\pi(\dfrac{1}{3}+1+16-12+\dfrac{8}{3}-8+3-\dfrac{1}{3})$

$=\dfrac{8\pi}{3}({units}^3)$