Question

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Answer 1

$V = 864\pi$

Step-by-step explanation:

Since one of the boundaries is y = 0, we need to find the roots of the function $f(x)=-2x^2+6x+36$. Using the quadratic equation, we get

$x = \dfrac{-6 \pm \sqrt{36 - (4)(-2)(36)}}{-4}= -3,\:6$

But since the region is also bounded by $x = 0$, that means that our limits of integration are from $x=0$ (instead of -3) to $x=6$.

Now let's find the volume using the cylindrical shells method. The volume of rotation of the region is given by

$\displaystyle V = \int f(x)2\pi xdx$

$\:\:\:\:\:\:\:= \displaystyle \int_0^6 (-2x^2+6x+36)(2 \pi x)dx$

$\:\:\:\:\:\:\:= \displaystyle 2\pi \int_0^6 (-2x^3+6x^2+36x)dx$

$\:\:\:\:\:\:\:= \displaystyle 2\pi \left(-\frac{1}{2}x^4+2x^3+18x^2 \right)_0^6$

$\:\:\:\:\:\:\:= 864\pi$