use the shell method to find the volume of the solid generated by revolving the shaded region about the y…

Explanation:

Step1: Recall shell - method formula

The shell - method formula for volume $V$ of a solid of revolution about the $y$-axis is $V = 2\pi\int_{a}^{b}x\cdot h(x)dx$, where $a$ and $b$ are the limits of integration, $x$ is the radius of the shell, and $h(x)$ is the height of the shell. Here, $a = 0$, $b=\sqrt{3}$, and $h(x)=3\sqrt{x^{2}+1}-3$.

Step2: Set up the integral

$V = 2\pi\int_{0}^{\sqrt{3}}x(3\sqrt{x^{2}+1}-3)dx=6\pi\int_{0}^{\sqrt{3}}x\sqrt{x^{2}+1}dx - 6\pi\int_{0}^{\sqrt{3}}x dx$.

Step3: Solve the first integral

Let $u = x^{2}+1$, then $du = 2xdx$. When $x = 0$, $u = 1$; when $x=\sqrt{3}$, $u = 4$. So, $\int_{0}^{\sqrt{3}}x\sqrt{x^{2}+1}dx=\frac{1}{2}\int_{1}^{4}\sqrt{u}du=\frac{1}{2}\times\frac{2}{3}u^{\frac{3}{2}}\big|_{1}^{4}=\frac{1}{3}(4^{\frac{3}{2}} - 1^{\frac{3}{2}})=\frac{1}{3}(8 - 1)=\frac{7}{3}$.

Step4: Solve the second integral

$\int_{0}^{\sqrt{3}}x dx=\frac{1}{2}x^{2}\big|_{0}^{\sqrt{3}}=\frac{3}{2}$.

Step5: Calculate the volume

$V = 6\pi\times\frac{7}{3}-6\pi\times\frac{3}{2}=14\pi - 9\pi = 5\pi$.

Answer:

$5\pi$