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

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

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Answer 1 · 2018-07-11T16:02:00

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

$x = ln (\frac{5}{y}), y \in (\frac{5}{e}, 5)$ $x = ln ( 5 / y ), y in ( 5 / e, 5 )$ , and $x = ln (\frac{y}{5}), y \in (5, 5 e)$ $x = ln (y / 5), y in ( 5, 5 e )$ ,

setting limits for integration with respect to y.

The area ls divided into two parts;

$A_{1}$ $A_1$ = the area from y = 5/e to y = 5.and

$A_{2}$ $A_2$ = .the area from y = 5 to y = 5 e.

Volume V = $V_{1}$ $V_1$ obtained by revolving $A_{1}$ $A_1$ , about y-axis

$+ V_{2}$ $+V_2$ obtained by revolving $A_{2}$ $A_2$ , about y-axis

$V_{1} = π \int (1^{2} - x^{2}) d y$ $V_1 = pi int (1^2 -x^2) dy$ , from $A_{1}$ $A_1$

$= π \int (1^{2} - {(ln (\frac{5}{y}))}^{2}) d y$ $= pi int ( 1^2 - ( ln ( 5 / y ))^2) dy$ , between limits for $A_{1}$ $A_1$

$= pi int (1 - ( ln 5 - ln y )^2) dy, with limits for$ 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 · 2018-07-12T04:36:13

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
.

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

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