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
Comments
Post a Comment