use the shell method to find the volume of the solid generated by revolving the regions bounded by the…

Explanation:

Step1: Find intersection points

Set $\sqrt{x}=\frac{x - 9}{8}$. Let $t=\sqrt{x}(t\geq0)$, then $t=\frac{t^{2}-9}{8}$, which simplifies to $8t=t^{2}-9$ or $t^{2}-8t - 9 = 0$. Factoring gives $(t - 9)(t+1)=0$. Since $t\geq0$, $t = 9$ and $x=t^{2}=81$. Also, $\sqrt{x}=0$ when $x = 0$ and $\frac{x - 9}{8}=0$ when $x = 9$.

Step2: Use the shell - method formula for volume about x - axis

The shell - method formula for volume $V$ of a solid of revolution about the $x$ - axis using horizontal shells is $V = 2\pi\int_{c}^{d}y\left(x_{right}-x_{left}\right)dy$. We need to express $x$ in terms of $y$. For $y=\sqrt{x}$, $x = y^{2}$; for $y=\frac{x - 9}{8}$, $x=8y + 9$. The limits of integration for $y$ are from $y = 0$ to $y = 9$. $x_{right}=8y + 9$ and $x_{left}=y^{2}$. So $V=2\pi\int_{0}^{9}y\left((8y + 9)-y^{2}\right)dy$.

Step3: Expand the integrand

$y\left((8y + 9)-y^{2}\right)=8y^{2}+9y - y^{3}$.

Step4: Integrate term - by - term

$\int(8y^{2}+9y - y^{3})dy=8\times\frac{y^{3}}{3}+9\times\frac{y^{2}}{2}-\frac{y^{4}}{4}+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[\frac{8y^{3}}{3}+\frac{9y^{2}}{2}-\frac{y^{4}}{4}\right]_{0}^{9}$ $V = 2\pi\left(\frac{8\times9^{3}}{3}+\frac{9\times9^{2}}{2}-\frac{9^{4}}{4}\right)$ $V = 2\pi\left(8\times\frac{729}{3}+\frac{9\times81}{2}-\frac{6561}{4}\right)$ $V = 2\pi\left(1944+\frac{729}{2}-\frac{6561}{4}\right)$ $V = 2\pi\left(\frac{7776 + 1458-6561}{4}\right)$ $V = 2\pi\times\frac{2673}{4}=\frac{2673\pi}{2}$.

Answer:

$\frac{2673\pi}{2}$