$y=e^ {-x}\sqrt{x+2}$. what is the volume of solid formed when shaded region is rotated completely about the x-axis? $e^ {-x}\sqrt{x+2}$. what is the volume of solid formed when shaded region is rotated completely about the x-axis? 
I am doing the chapter further calculus in the sub chapter integration by parts. I know how to do regular questions but not this one  I have never encountered a question like this where the solid shaded region is completely rotated about the x-axis- Any help is much appreciated . Thank you.
 A: The volume of rotation about the $x$ axis is given by the following formula: $$\pi\int_{a}^{b}\left(f\left(x\right)\right)^2\,\mathrm{d}x\text{,}$$ where, in this case, $f\left(x\right)=e^{-x}\sqrt{x+2}$, $b=0$ and $a=-2$ (the $x$-intercept of $f\left(x\right)$). $a$ is found by setting $y$ (in the original equation) equal to $0$: $$0=e^{-x}\sqrt{x+2}$$ Either $e^{-x}=0$ or $\sqrt{x+2}=0$. The former will only $\to{0}$ if $x\to\infty$. However, the latter will equal $0$ when $x=-2$ (since $\sqrt{-2+2}=0$). This means the lower bound $a$ is equal to $-2$.
Substituting these values gives the equation: $$\pi\int_{-2}^{0}\left(e^{-x}\sqrt{x+2}\right)^2\,\mathrm{d}x=\pi\int_{-2}^{0}e^{-2x}\left(x+2\right)\,\mathrm{d}x$$ Then, using the integration by parts formula: $$\int{uv^\prime}=uv-\int{vu^\prime}$$ and substituting $u=x+2$ and $v^\prime=e^{-2x}$, the integral $\int_{-2}^{0}e^{-2x}\left(x+2\right)\,\mathrm{d}x$ can be solved: $$u=x+2\implies{u^\prime}=1$$ $$v^\prime=e^{-2x}\implies{v}=-\frac{e^{-2x}}{2}$$ $$\therefore\int{e^{-2x}}\left(x+2\right)\,\mathrm{d}x=-\frac{\left(x+2\right)e^{-2x}}{2}-\int-\frac{e^{-2x}}{2}\,\mathrm{d}x$$ Solving for $\int-\frac{e^{-2x}}{2}\,\mathrm{d}x$ using $u$ substitution: $$u=-2x\implies\frac{\mathrm{d}u}{\mathrm{d}x}=-2\implies\mathrm{d}x=-\frac{\mathrm{d}u}{2}$$ $$\therefore\int-\frac{e^{-2x}}{2}\,\mathrm{d}x=\int\left(-\frac{1}{2}\times-\frac{1}{2}\right)e^u\,\mathrm{d}u=\frac{1}{4}\int{e^u}\,\mathrm{d}u=\frac{e^u}{4}=\frac{e^{-2x}}{4}$$ Substituting $\frac{e^{-2x}}{4}$ back into the equation gives: $$\int{e^{-2x}}\left(x+2\right)\,\mathrm{d}x=-\frac{\left(x+2\right)e^{-2x}}{2}-\frac{e^{-2x}}{4}=-\frac{2\left(x+2\right)e^{-2x}}{4}-\frac{e^{-2x}}{4}$$ $$=-\frac{\left(2x+4\right)e^{-2x}}{4}-\frac{e^{-2x}}{4}=-\frac{\left(2x+4\right)e^{-2x}-e^{-2x}}{4}=-\frac{\left(2x+5\right)e^{-2x}}{4}$$ Finally, multiply the integral by $\pi$: $$\pi\int_{-2}^{0}e^{-2x}\left(x+2\right)\,\mathrm{d}x=\pi\left(\left.-\frac{\left(2x+5\right)e^{-2x}}{4}\right\vert_{-2}^{0}\right)$$ Evaluating this with the upper and lower bounds and multiplying the result by $\pi$, as shown above, will yield the final answer of $38.95429594394258$, or approximately $39$.
