[tex]V = 864\pi[/tex]
Step-by-step explanation:
Since one of the boundaries is y = 0, we need to find the roots of the function [tex]f(x)=-2x^2+6x+36[/tex]. Using the quadratic equation, we get
[tex]x = \dfrac{-6 \pm \sqrt{36 - (4)(-2)(36)}}{-4}= -3,\:6[/tex]
But since the region is also bounded by [tex]x = 0[/tex], that means that our limits of integration are from [tex]x=0[/tex] (instead of -3) to [tex]x=6[/tex].
Now let's find the volume using the cylindrical shells method. The volume of rotation of the region is given by
[tex]\displaystyle V = \int f(x)2\pi xdx[/tex]
[tex]\:\:\:\:\:\:\:= \displaystyle \int_0^6 (-2x^2+6x+36)(2 \pi x)dx[/tex]
[tex]\:\:\:\:\:\:\:= \displaystyle 2\pi \int_0^6 (-2x^3+6x^2+36x)dx[/tex]
[tex]\:\:\:\:\:\:\:= \displaystyle 2\pi \left(-\frac{1}{2}x^4+2x^3+18x^2 \right)_0^6[/tex]
[tex]\:\:\:\:\:\:\:= 864\pi [/tex]
