Finding the Volume of a Revolved Curve: y = (cos x)/x from pi/6 to pi/2

[sparsh]

Hi

Could someone please give me an idea on how to go about this problem

Find the volume of the curve genereated by revolving the area between the curve y =(cos x)/x and the x-axis in the interval pie/6 to pie/2

Thanks a lot..

 

[Tinaaa]

you know there is a formula that solves your problem perfectly...

https://www.physicsforums.com/attachment.php?attachmentid=6761&stc=1&d=1145703282





try it...

 

[HallsofIvy]

Better than a formula is to think through it: Draw a line from any point on the x-axis up to the curve. As the curve is rotated around the x-axis that line sweeps out a disk of radius y= cos(x)/x. It's area is [itex]\pi y^2[/itex] and if we imagine that as a very shallow cylinder of height dx (the height of the disk is in the x direction) its volume is [itex]\pi y^2 dx[/itex].
The volume of the whole thing is a sum of those volumes (a Riemann sum) and becomes the integral Tinaaa said:
[tex]\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}y^2 dx[/tex]
Since y= cos(x)/x, put that in and integrate.

Was this really for a PRE-Calculus course?

 

[sparsh]

@ Hallofivy

Thanks a lot. Actually I couldn't think up where to put this post so i just dropped it in Pre calculus.