let $r$ be the region in the first quadrant enclosed by the $x$-axis, the $y$-axis, the line $y = 2$, and…

Explanation:

Step1: Find the intersection point of $y = 2$ and $y=\sqrt{9 - x^{2}}$

Set $2=\sqrt{9 - x^{2}}$, then square both sides: $4 = 9 - x^{2}$, so $x^{2}=5$ and $x=\sqrt{5}$ (since we are in the first - quadrant).

Step2: Use the method of cylindrical shells

The volume $V$ of the solid of revolution about the $y$ - axis using the shell method is given by $V = 2\pi\int_{a}^{b}x\cdot h(x)dx$, where $a = 0$, $b=\sqrt{5}$, and $h(x)=\sqrt{9 - x^{2}}-2$. [ \begin{align*} V&=2\pi\int_{0}^{\sqrt{5}}x(\sqrt{9 - x^{2}}-2)dx\ &=2\pi\left(\int_{0}^{\sqrt{5}}x\sqrt{9 - x^{2}}dx-\int_{0}^{\sqrt{5}}2xdx\right) \end{align*} ]

Step3: Evaluate $\int_{0}^{\sqrt{5}}x\sqrt{9 - x^{2}}dx$

Let $u = 9 - x^{2}$, then $du=-2xdx$. When $x = 0$, $u = 9$; when $x=\sqrt{5}$, $u = 4$. So $\int_{0}^{\sqrt{5}}x\sqrt{9 - x^{2}}dx=-\frac{1}{2}\int_{9}^{4}\sqrt{u}du=\frac{1}{2}\int_{4}^{9}u^{\frac{1}{2}}du$. [ \begin{align*} \frac{1}{2}\int_{4}^{9}u^{\frac{1}{2}}du&=\frac{1}{2}\times\frac{2}{3}u^{\frac{3}{2}}\big|_{4}^{9}\ &=\frac{1}{3}(27 - 8)\ &=\frac{19}{3} \end{align*} ]

Step4: Evaluate $\int_{0}^{\sqrt{5}}2xdx$

$\int_{0}^{\sqrt{5}}2xdx=x^{2}\big|_{0}^{\sqrt{5}} = 5$

Step5: Calculate the volume

[ \begin{align*} V&=2\pi\left(\frac{19}{3}-5\right)\ &=2\pi\times\frac{19 - 15}{3}\ &=\frac{8\pi}{3} \end{align*} ]

Answer:

$\frac{8\pi}{3}$