How do you find the volume of the solid formed by rotating the region enclosed by ?

enter image source here
about the x-axis.

3 Answers
Apr 4, 2016

#V=pi(e^(0.6)/2+4e^0.3-3.3)~~9.4577...#

Explanation:

We'll use a method call the disk method to do this.
enter image source here
Take a look at the above picture. Pretend it's an infinitely small cylindrical disk, with radius #r# and height #dx#. When we revolve #y=e^x+2# about the #x#-axis, we can break the resulting solid down to infinitely many infinitely small cylindrical disks like the one above (woah, calculus huh!). All of these disks have a radius equal to the value of the #y# and height equal to an infinitely small change in #x#, or #dx#.

We know volume of a cylinder is given by:
#V=pir^2h#
And we know that the radius of each cylinder is the value of #y#, and the height is #dx#; that means the volume of an infinitely small disk is:
#dV=pi(y^2)dx#
Integrating both sides to find total volume, for #x=0# to #x=0.3#, gives
#V=int_0^0.3pi(y^2)dx#
But #pi# is a constant and #y=e^x+2#, so
#V=piint_0^0.3(e^x+2)^2dx#

Expanding the #(e^x+2)^2# binomial:
#V=piint_0^0.3e^(2x)+4e^x+4dx#
And using the sum rule:
#V=piint_0^0.3e^(2x)dx+piint_0^0.3 4e^xdx+piint_0^0.3 4dx#
Finally, evaluating these one by one:
#V=pi[e^(2x)/2]_0^0.3+4pi[e^x]_0^0.3+4pi[x]_0^0.3#
#V=pi((e^(0.6)/2-e^0/2)+4(e^0.3-e^0)+4(0.3-0))#
#V=pi(e^(0.6)/2-1/2+4e^0.3-4+1.2)#
#V=pi(e^(0.6)/2+4e^0.3-3.3)~~9.4577...#

Apr 11, 2016

Awesome Ken...

Here is a Maple rendering of the rotation!

enter image source here

By the way, you have the coolest name in the world.

May 12, 2016

0.9117

Explanation:

#f(r)=e^x#
#V(R)=2 pi int_0^R f(r)rdr#
#C(h,r)=pi h r^2# now with #h = 2# and #R = 0.3#
#V_t =V(R) + C(2,R) = 0.3462+0.5655 = 0.9117#