The region bounded by $y= \frac{x}{\sqrt{x^3+8}}$, the $x$-axis, and the line $x=2$ is revolved about the $y$-axis. Find the volume of the solid generated this way.
[Answer= $\frac{8\pi}{3}(2-\sqrt{2})$]
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For the rotation around $y$ axis the formula is $$V=2\pi \int _{a}^{b}xf(x)\,dx$$ In this case we have $$ V=2 \pi \int_0^2 \frac{x^2}{\sqrt{x^3+8}} \, dx=\frac{2\pi}{\color{red}{3}}\int_0^2 \frac{\color{red}{3}x^2}{\sqrt{x^3+8}}\,dx =\frac{2\pi}{3}\left[2 \sqrt{x^3+8}\right]_0^2=\frac{8}{3} \left(2-\sqrt{2}\right) \pi$$ I have multiplied and divided by $\color{red}{3}$ in order to have $3x^2dx$ so the integral becomes $$\int \frac{d(x^3)}{\sqrt{x^3+8}}=\int (x^3+8)^{-1/2} \,d(x^3)=\frac{1}{1-\frac12}\,(x^3+8)^{1/2}+C=2\sqrt{x^3+8}+C$$ Hope this can be useful