Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 18

This problem can be solved by the Washer method easily,
Given
y = secx , y = 0 , 0 <= x <= pi/3
so,
we need to Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 4
so, by the washer method the volume is given as
V= pi*int_a^b [R(x)^2 -r(x)^2]dx
where R(x) is the outer radius and the r(x) is the inner radius
and now ,
R(x) = 4 and r(x) = 4-secx as the rotation of solid is about y=4
as,
0 <= x <= pi/3
so , a=0 , b= pi/3
V= pi*int_0^(pi/3) [(4)^2 -(4-sec x)^2]dx
= pi*int_0^(pi/3) [(16) -(16+sec^2 x-8secx)]dx
=pi*int_0^(pi/3) [(16) -16-sec^2 x+8secx)]dx
=pi*int_0^(pi/3) [8secx-sec^2 x]dx
=pi*[8ln(secx+tanx)-tanx]_0^(pi/3)
=pi*[[8ln(sec(pi/3)+tan(pi/3))-tan(pi/3)]-[8ln(sec0+tan0)-tan0]]
=pi*[8ln(2+sqrt(3))-sqrt(3)]-[0]]
=pi*[8ln(2+sqrt(3))-sqrt(3)]]
is the volume