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

Answer:

The volume of the solid V = 41601.6 π

Step-by-step explanation:

From the given information:

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

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

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

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

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

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

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

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

$V = 2 \pi \begin {bmatrix} \dfrac{1x^6}{6} - \dfrac{6x^6}{5} - \dfrac{27x^4}{4}+ \dfrac{108x^3}{3}+ \dfrac{324x^2}{2} \end {bmatrix}^6_0$

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

V = 2π (20800.8)

V = 41601.6 π