$y=\sqrt{\cot 2x} \;$ revolved around x-axis between $x = \frac{\pi }{12} \;$ and $x=\frac{\pi }{4} \;$ find volume

Question

The region between the curve $y=\sqrt{\cot 2x} \;\;$ and the x-axis, from $x=\frac{\pi }{12} \;\;$ to $x=\frac{\pi }{4} \;\;$ is revolved around the x-axis to generate a solid. Find the volume of the solid.

I tried finding critical points, by taking derivative, but it seems y' ≠ 0 so I just plugged in $\frac{\pi }{12}$ and $\frac{\pi }{4}$ into $\sqrt{\cot 2x}$ and it seems that at $\frac{\pi }{12}$ function has the maximum y-value of $3^\frac{1}{4}$.

My radius will vary, r = $(3^\frac{1}{4} - \sqrt{\cot2x})$

Area = $\pi \cdot r^2$

Volume = $\pi \cdot r^2 \cdot dx$

Integration:

$$ \int^{3^\frac{1}{4}}_{0} \pi (\sqrt{3} - 2\cdot3^\frac{1}{4} \cdot \sqrt{\cot2x} + \cot2x) $$

but wolfram can't integrate it...

is it even correct?

Answer 1

Both your bounds and integrand are incorrect. $x$ goes from $\dfrac{\pi}{12}$ to $\dfrac{\pi}{4}$. These are our bounds of integration; there is no need to calculate $y$ at these values because we are integrating with respect to $x$. The radius is given by the function itself. $y=r.$ So $$V=\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \pi \cdot \left(\sqrt{\cot 2x}\right)^2\ \mathrm{d}x = \pi \cdot\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \cot 2x\ \mathrm{d}x$$ This is easy to integrate.

Answer 2

HINT: The volume of the solid generated by rotating the curve $y=f(x)$ about the x-axis between $x=a$ & $x=b$ is $$=\color{red}{\int_{a}^{b}\pi y^2dx}$$