## Calculus 8th Edition

$\displaystyle{V=16\pi}$
Using the quadratic formula $\displaystyle{y = \frac{12 \pm \sqrt{\left(-12\right)^2-4(3)(9)}}{2(3)}}\\ y=1\qquad y=3$ $\displaystyle{V=\int_{1}^{3}(2\pi y)\left(-3y^2+12y-9\right)\ dy}\\ \displaystyle{V=2\pi\int_{1}^{3}12y^2-9y-3y^3\ dy}\\ \displaystyle{V=2\pi\left[4y^3-\frac{9}{2}y^2-\frac{3}{4}y^4\right]_{1}^{3}}\\ \displaystyle{V=2\pi\left(\left(4(3)^3-\frac{9}{2}(3)^2-\frac{3}{4}(3)^4\right)-\left(4(1)^3-\frac{9}{2}(1)^2-\frac{3}{4}(1)^4\right)\right)}\\ \displaystyle{V=16\pi}$