use the shell method to compute the volume of the solid obtained by rotating the region underneath the graph…

Explanation:

Step1: Recall the Shell Method formula

The Shell Method formula for rotating about the (y -)axis ((x = 0)) is (V=2\pi\int_{a}^{b}x\cdot f(x)dx), where (a = 0), (b = 2), and (f(x)=\frac{1}{\sqrt{x^{2}+6}}). So, (V = 2\pi\int_{0}^{2}\frac{x}{\sqrt{x^{2}+6}}dx).

Step2: Use substitution

Let (u=x^{2}+6), then (du = 2xdx) and (xdx=\frac{1}{2}du). When (x = 0), (u=6); when (x = 2), (u=4 + 6=10). The integral becomes (V=2\pi\int_{6}^{10}\frac{1}{2\sqrt{u}}du).

Step3: Integrate

(\int\frac{1}{\sqrt{u}}du=\int u^{-\frac{1}{2}}du=2u^{\frac{1}{2}}+C). So, (V=2\pi\times\frac{1}{2}\left[2\sqrt{u}\right]_{6}^{10}).

Step4: Evaluate the definite - integral

(V=\pi(2\sqrt{10}-2\sqrt{6}) = 2\pi(\sqrt{10}-\sqrt{6})).

Answer:

(2\pi(\sqrt{10}-\sqrt{6}))