use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the…

Explanation:

Step1: Find intersection points

Set $9x - x^{2}=18$, which can be rewritten as $x^{2}-9x + 18 = 0$. Factoring gives $(x - 3)(x - 6)=0$. So the solutions are $x = 3$ and $x = 6$.

Step2: Recall the formula for the method of cylindrical shells

The formula for the volume $V$ using the method of cylindrical shells about the line $x = a$ is $V=2\pi\int_{c}^{d}(x - a)h(x)dx$, where $h(x)$ is the height of the shell and $(x - a)$ is the radius of the shell. Here $a = 3$, $c = 3$, $d = 6$, and $h(x)=(9x - x^{2}-18)$.

Step3: Set up the integral

$V = 2\pi\int_{3}^{6}(x - 3)(9x - x^{2}-18)dx$. Expand the integrand: $(x - 3)(9x - x^{2}-18)=9x^{2}-x^{3}-18x-27x + 3x^{2}+ 54=-x^{3}+12x^{2}-45x + 54$.

Step4: Integrate term - by - term

$\int(-x^{3}+12x^{2}-45x + 54)dx=-\frac{1}{4}x^{4}+4x^{3}-\frac{45}{2}x^{2}+54x+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[-\frac{1}{4}x^{4}+4x^{3}-\frac{45}{2}x^{2}+54x\right]_{3}^{6}$. $V = 2\pi\left[\left(-\frac{1}{4}(6)^{4}+4(6)^{3}-\frac{45}{2}(6)^{2}+54(6)\right)-\left(-\frac{1}{4}(3)^{4}+4(3)^{3}-\frac{45}{2}(3)^{2}+54(3)\right)\right]$. $V = 2\pi\left[\left(-324 + 864-810 + 324\right)-\left(-\frac{81}{4}+108-\frac{405}{2}+162\right)\right]$. $V = 2\pi\left[54-\left(-\frac{81}{4}+108 - 202.5+162\right)\right]$. $V = 2\pi\left[54-\left(-\frac{81}{4}+67.5\right)\right]$. $V = 2\pi\left[54-\left(\frac{-81 + 270}{4}\right)\right]$. $V = 2\pi\left[54-\frac{189}{4}\right]$. $V = 2\pi\left[\frac{216 - 189}{4}\right]$. $V = 2\pi\times\frac{27}{4}=\frac{27\pi}{2}$.

Answer:

$\frac{27\pi}{2}$