Question

The region bounded by y = − 1x^2 + 3x + 18 , x = 0 and y = 0 is rotated about the y -axis. Find the volume of this solid of revolution:

Answer 1

The volume of the solid V = 41601.6 π

Explanation:

From the given information:

The volume of the shaded region bounded by rotating the shaded region about the y-axis is: dV

`dV = \pi (-1x^2 +3x+18)^2 ((x+dx)^2 -(x)^2)`

`dV =2 \pi x (-1x^2 +3x+18)^2 \ . dx`

`dV =2 \pi x (-1x^2 +3x+18) ( -1x^2 +3x + 18). dx`

`dV =2 \pi x (1x^4 -3x^3 -18x^2 -3x^3 +9x^2 +54x -18x^2 +54x + 324 ) \ .dx`

`dV =2 \pi x (1x^4 -3x^3-3x^3 -18x^2 +9x^2 -18x^2 +54x +54x +324) \ .dx`

`dV =2 \pi x (1x^4 -6x^3 -27x^2 +9x^2+ 108x +324) \ .dx`

`\implies \int ^V_o \ dv = 2 \pi \int ^6_0 (1x^5 -6x^4 -27x^3+108x^2 +324x) \ dx`

`V = 2 \pi \begin {bmatrix} (1x^6)/(6) - (6x^6)/(5) - (27x^4)/(4)+ (108x^3)/(3)+ (324x^2)/(2) \end {bmatrix}^6_0`

`V = 2 \pi \begin {bmatrix} (1(6)^6)/(6) - (6(6)^6)/(5) - (27(6)^4)/(4)+ (108(6)^3)/(3)+ (324(6)^2)/(2) -0\end {bmatrix}`

V = 2π (20800.8)

V = 41601.6 π