# 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?

Jul 11, 2018

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

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

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

${V}_{1} = \pi \int \left({1}^{2} - {x}^{2}\right) \mathrm{dy}$, from ${A}_{1}$

$= \pi \int \left({1}^{2} - {\left(\ln \left(\frac{5}{y}\right)\right)}^{2}\right) \mathrm{dy}$, between limits for ${A}_{1}$

$= \pi \int \left(1 - {\left(\ln 5 - \ln y\right)}^{2}\right) \mathrm{dy} , w i t h \lim i t s f \mathmr{and}$A_1

$= \pi \int \left(1 - {\left(\ln 5\right)}^{2} + 2 \ln 5 \ln y - {\left(\ln y\right)}^{2}\right) \mathrm{dy}$,

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 

$+ \left(2 \ln 5 \left(\ln y - y\right) - \left({\left(\ln y\right)}^{2} - 2 \left(\ln y - 1\right)\right)\right]$,

between the limits

$= \pi y \left[\left(- 1 - 2 \ln 5 - {\left(\ln 5\right)}^{2}\right) - \ln y \left(\ln y - 2 \ln 5 - 2\right)\right]$,

between the limits.

$= \pi y \left[- {\left(\ln 5 + 1\right)}^{2} + 2 \left(\ln 5 + 1\right) \ln y - {\left(\ln y\right)}^{2}\right]$

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 $\frac{\pi}{3}$..

.

Jul 12, 2018

Continuation, for the 2nd part.
Answer: $V = 20 \left(\frac{\pi}{e}\right)$ $= 23.11455 c u$, nearly

#### Explanation:

$V = \pi y \left[- {\left(\ln 5 + 1\right)}^{2} + 2 \left(\ln 5 + 1\right) \ln y - {\left(\ln y\right)}^{2}\right]$

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

At the upper limit, the value is

$\pi \left(5 e\right) \left[- {\left(\ln 5 + 1\right)}^{2} + 2 {\left(\ln 5 + 1\right)}^{2} - {\left(\ln 5 +\right)}^{2}\right]$ = 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#

$= \pi \left(\frac{5}{e}\right) \left[- 4\right]$. And so, .

$V = 20 \left(\frac{\pi}{e}\right)$

$= 23.11455 c u$, nearly
.