`y = secx , y = 0 , 0 <= x <= pi/3` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 4

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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...

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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

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