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?

Question

Luke Alvarez · Accepted Answer

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

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#.. .

Isabella Ackerman · Answer

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

#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 .

Scarlett Allison · Answer

undefined To use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $ y = 5e^x $, $ y = 5e^{-x} $, and $ x = 1 $ about the y-axis, follow these steps:

1. Determine the interval of integration. In this case, it's from $ x = 0 $ to $ x = 1 $ since the intersection point of $ y = 5e^x $ and $ y = 5e^{-x} $ is at $ x = 0 $.

2. Set up the integral for the volume using cylindrical shells formula:
   $$ V = 2\pi \int_{0}^{1} x \cdot h(x) \, dx $$

3. Find the height function, $ h(x) $, which represents the height of each cylindrical shell. In this case, it's the difference between the top curve and the bottom curve:
   $$ h(x) = (5e^x) - (5e^{-x}) $$

4. Now, substitute $ h(x) $ into the integral:
   $$ V = 2\pi \int_{0}^{1} x \cdot ((5e^x) - (5e^{-x})) \, dx $$

5. Integrate with respect to $ x $ over the interval $[0, 1]$ to find the volume.