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!
On
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 $$\int_{\pi/6}^{\pi/2} 2\pi \cos x dx =\pi$$
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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 $$\int_{\pi/6}^{\pi/2}2\pi x\frac{\cos x}{x}dx=2\pi\int_{\pi/6}^{\pi/2}\cos x dx=\pi$$