Question

How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by `y=5e^(x)` and `y=5e^(-x)`, x = 1, about the y axis?

Answer 1

Slicing to cylindrical-shell elements for integration gives approximation only. Circular-annular elements are used. To be continued, in the 2nd answer.

Explanation:

See graph to see the area that revolves about y-axis ( x = 0 ).
graph{ (y-5(2.718)^x)(y - 5(2.718)^(-x))(x-1+0y)=0[0 1.1 0 13.6]}
The curves meet at A(5, 0).

They meet x = 1 at B( 1, 5 / e ) and C(1, 5e ).

Inversely, the equations are

`x = ln ( 5 / y ), y in ( 5 / e, 5 )`, and `x = ln (y / 5), y in ( 5, 5 e )`,

setting limits for integration with respect to y.

The area ls divided into two parts;

`A_1` = the area from y = 5/e to y = 5.and

`A_2` = .the area from y = 5 to y = 5 e.

Volume V = `V_1` obtained by revolving `A_1`, about y-axis

`+V_2` obtained by revolving `A_2`, about y-axis

`V_1 = pi int (1^2 -x^2) dy`, from `A_1`

`= pi int ( 1^2 - ( ln ( 5 / y ))^2) dy`, between limits for `A_1`

`= pi int (1 - ( ln 5 - ln y )^2) dy, with limits for`A_1#

`= pi int ( 1 - ( ln 5 )^2+ 2 ln 5 ln y - ( ln y )^2 ) dy`,

y from 5 / e to 5

Likewise,

`V_2 = pi int ( 1 - ( ln 5 )^2 + 2 ln y ln 5 - ( ln y )^2 ) ) dy`,

with y from 5 to 5 e.

Note that the integrand is the same function of y, for both.So,

`V = pi int ( 1 - ( ln 5 )^2+ 2 ln y ln 5 - ( ln y )^2 ) ) dy`,

with y from 5/e to 5e. Use integration by parts method.

`V = pi [ (1 - ( ln 5 )^2) y`

`+ 2 ln 5 int ln y dy - int ( ln y )^2 dy ]`,

between 5 / e and 5 e

`= pi [ (1 - ( ln 5 )^2)y`

`+ ( 2 ln 5 ( ln y -y ) - ( (ln y )^2 -2 ( ln y - 1 ))]`,

between the limits

`= pi y [ (-1 -2 ln 5 - ( ln 5 )^2 ) - ln y ( ln y - 2 ln 5 - 2 )]`,

between the limits.

`= pi y [ -( ln 5 + 1 )^2 +2 (ln 5 +1 ) ln y - ( ln y )^2]`

I would review my answer for corrections, if any.

The easier cylindrical-shell elements for integration,

applied to right circular cone of height 1 and base radius 1, gives

volume as `pi` against `pi/3`..

.

Answer 2

Continuation, for the 2nd part.
Answer: `V = 20(pi/e)` `= 23.11455 cu`, nearly

Explanation:

`V = pi y [ -( ln 5 + 1 )^2 +2 (ln 5 +1 ) ln y - ( ln y )^2]`

between the limits y between 5 / e and 5 e. ( Use ln e = 1. )

At the upper limit, the value is

`pi ( 5 e ) [ -( ln 5 + 1 )^2 +2 (ln 5 +1 )^2 - ( ln 5 + )^2]` = 0.

At the lower limit, this becomes

`pi (5 / e ) [ -( ln 5 + 1 )^2 +2 (ln 5 +1 ) ( ln 5 - 1 ) - ( ln 5 - 1 )^2`

`= pi (5 / e ) [- 4 ]`. And so, .

`V = 20(pi/e)`

`= 23.11455 cu`, nearly
.