Find Volume of Revolved Region: [-pi/2, pi/2] & y= cos x & y = -cos x

[shunae95]

Find the volume of the solid generated by revolving the region enclosed by y= cos x and y = -cos x for [-pi/2, pi/2] about the line y=2pi.

 

[Greg]

Hello shunae95 and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[shunae95]

2π∫π/2 0 (2π+cosx)2−(2π−cosx)2dx
2π∫π/2 (2π+cos⁡x)2−(2π−cos⁡x)2dx16π2∫π/20cosxdx
16π2∫0π/2cos⁡xdx16π2[sinx]π/20=16π2

 

[Greg]

2π∫π/2 0 (2π+cosx)2−(2π−cosx)2dx
2π∫π/2 (2π+cos⁡x)2−(2π−cos⁡x)2dx16π2∫π/20cosxdx
16π2∫0π/2cos⁡xdx16π2[sinx]π/20=16π2

Do you have to use shells? Disk integration seems much more efficient:

\(\displaystyle \pi\int_{-\pi/2}^{\pi/2}\left(2\pi+\cos(x)\right)^2-\left(2\pi-\cos(x)\right)^2\,dx=16\pi^2\)

My apologies for the late reply.

 

[MarkFL]

If I use the shell method, I get:

\(\displaystyle V=4\pi\int_{-1}^1 (2\pi-y)\arccos(|y|)\,dy=16\pi^2\)

Greg, when I evaluate the integral you correctly set up using the washer method, I also get $V=16\pi^2$. D

 

[Greg]

I forgot the coefficients of $2\pi$! Thanks Mark! (I've edited my post).