# Compute the volume of a solid by revolving a region about the $y$-axis

How to compute the volume of the solid generated by revolving the region between the curve $$y=\dfrac{\cos x}{x}$$ and the $$x$$-axis for $$\pi/6\leq x\leq \pi/2$$ about the $$y$$-axis?

I think we need to compute a definite integral of the form $$\int_a^b \pi x^2dy$$, however it's impossible to express $$x$$ in terms of $$y$$ explicitly.

Can someone help me? Thanks a lot!

The shell method allows you to calculate the volume of a solid of revolution that rotates to an axis that is perpendicular to the one where we're integrating - in this case, we can rotate a shape around the $$y$$-axis by integrating over $$x$$. So you don't need to express the function in terms of $$y$$.
The volume of the solid will be an integral of the form $$V=2\pi \int_a^b x|f(x)|dx,$$ where in this case $$f(x)=\frac{cos(x)}{x}$$, $$a=\pi/6$$ and $$b=\pi/2$$. Note that $$f(x)\geq 0$$ in the interval $$[\frac \pi 6, \frac \pi 2]$$ (check this yourself). Thus we get:
$$V=2\pi \int_{\pi/6}^{\pi/2} x\left|\frac{cos(x)}{x}\right|dx=2\pi \int_{\pi/6}^{\pi/2} x\frac{cos(x)}{x}dx=2\pi\int_{\pi/6}^{\pi/2} cos(x)dx=2\pi(sin(x))^{\pi/2}_{\pi/6}=2\pi(1-\frac 1 2)=\pi.$$
You may use shell method to compute the volume. Then volume is $$\int_{\pi/6}^{\pi/2}2\pi x\frac{\cos x}{x}dx=2\pi\int_{\pi/6}^{\pi/2}\cos x dx=\pi$$
There is another formula, namely, $$\int 2\pi x f(x) dx$$ which could be used to find the volume of revolution about the y-axis.
You will get $$\int_{\pi/6}^{\pi/2} 2\pi \cos x dx =\pi$$