Vind average value of a solid..

[hank]

Consider the solid generated by revolving the region enclosed by y = sec x, x 0, x = pi/3, and y = 0 about the x-axis. Find the average value of the area of a cross section of this solid taken perpendicular to the x-axis.

Ok, here's my setup:

V = 3/pi S[0 - pi/3] secx dx
= 3/pi[ln|secx + tanx] [pi/3 to 0]
= 3/pi * ln(2 + sqrt(3))

Book says 3 * sqrt(3).

What am I doing wrong?

 

[soroban]

Hello, Hank!

Your setup is wrong . . .

Consider the solid generated by revolving the region enclosed by:
\(\displaystyle y\,=\,\sec x,\;x\,=\,0,\,x\,=\,\frac{\pi}{3},\;y\,=\,0\) about the x-axis.
Find the average value of the area of a cross-section of this solid taken perpendicular to the x-axis.

Don't know how you got that volume equation . . .

\(\displaystyle \L V \;=\;\pi\int^{\;\;\;\;\frac{\pi}{3}}_0 \sec^2x\,dx\;=\;\pi \tan x\,\bigg]^{\frac{\pi}{3}}_0\;=\;\pi\cdot\tan\frac{\pi}{3}\;=\;\sqrt{3}\pi\)


Divide by the width of the interval: \(\displaystyle \L\:\frac{\sqrt{3}\pi}{\frac{\pi}{3}} \:=\:3\sqrt{3}\)