Let R be the region enclosed by #y= e^(2x), y=0, and y=2#. What is the volume of the solid produced by revolving R around the x-axis?

Question

6675 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-18T06:49:13

#3/4pi=2.3562# cubic units. nearly.

y = 2 meets # y = e^(2x)#, at # (1/2ln 2, 2).#

The x-axis y = 0 is the asymptote to #y = e^(2x)>0#.

Now,

volume #= pi int y^2 dx#, for x from #-oo to 1/2 ln 2#

#-pi int e^(4x) dx#, for x from #-oo to 1/2 ln 2#

#=pi/4[e^(4x)]#, between the limits

#=pi/4[e^(2ln2)-e^0]#

#=pi/4[4-1)#, using #e^ln a = a#.

#=3/4pi=2.3562# cubic units. nearly.

The graph shows the area that is revolved.

graph{(y-.1)(e^(2x)-y)(x-.5ln2+.0001y)=0 [-1.5, 1.5, 0, 2,1]}