Question

Hello. I need a bit of help in integration.
Following is the question:
Find the volume of the solid generated by revolving the area between the curve y=(cos x)/x and the x-axis for π/6≤x≤π/2 about the y-axis.

Answer 1

What specifically is your problem?

Answer 2

\[\pi \int\limits_{\pi/6}^{\pi/2}(\cos x/x)^2dx\]\[= \pi (-x) \frac{(-x Si(2x)+\cos^2(x))}{(x)}|_\frac{\pi}{6}^\frac{\pi}{2}\]\[= Si(\frac{\pi}{3})-Si(\pi)+\frac{9}{2\pi}\]

Answer 3

Now that is weird. Someone suggested me to use Shell formula. \[V = \int\limits_{a}^{b}2\pi xf(x)dx\] Here, a=π/6, b=π/2 Hence \[V = \int\limits_{\pi/6}^{\pi/2} 2\pi x \frac{cosx}{x}dx = 2\pi \int\limits_{\pi/6}^{\pi/2}cosxdx = 2\pi [sinx]_{\pi/6}^{\pi/2} = 2 \pi (1 - 1/2)= \pi\]I get different answer??!

Answer 4

The shell method works. You have the correct answer, too. Keonlo's method is using y as the radius, which means that he rotated the curve about the x-axis.

Answer 5

Are you saying\[= Si(\frac{\pi}{3})-Si(\pi)+\frac{9}{2\pi}=\pi \]

Answer 6

Wolfram alpha say\[= Si(\frac{\pi}{3})-Si(\pi)+\frac{9}{2\pi}=0.5659\]
whereas \[\pi=3.14\]

Answer 7

If you use y as the radius, the shape should be like this: