Let's get a volume for this interesting shape by using
shells so this is made out of the region bounded by
this little exponential curve that's the one on the bottom y equals 2e to the negative x
y equals 1 for the upper boundary x equals 3 for the right boundary and i want to use nested
cylindrical shells for this and that requires that i use shells that have the same symmetry axis as
the solid revolution that i'm looking at
so i'm going to do all these nested shells
like this
and they have a little bit of thickness which i can view easiest along their horizontal sides
and i can make some attempt to draw the thickness of the shell
on the end of the shell but it's a little bit tricky to get that done
that's probably good enough and then i noticed that the thickness of this thing is d y and that
immediately tells me i'm going to be doing a y integral when i add all these little shells up
i need to get to get the volume of one of these shells i'm going to need to get
the radius the shell
and that's the distance here from y equals 1 down to whatever the y coordinate is of
the bottom edge of this thing so my plan is to identify each of these shells by that y
coordinate and then add them all up as i go from the minimum value of y all the way up to y equals
one where i have really skinny longer shells so one of the really tricky things about this problem
is i've got to figure out the length of these shells
and that's a horizontal distance between x equals 3 and then the x value corresponding to that y
so we've got to take that y coordinate and we've got to invert and go back to the x coordinate that
it came from so just as a little side note here i had the curve y equals 2e to the negative x
and i need to solve this for x in order to figure out what
that x coordinate is connected to this y coordinate so i'm going to divide by 2
take the natural log of both sides solve for x and i get the x is negative natural log y over 2.
um you can play with this a little you don't have to but if i had the reciprocal
of that interior so this would be 2 over y to the negative 1 and i could bring out the negative 1
using log properties so i could write this as positive natural log 2 over y there's not a lot
of benefit to that but i thought i would mention it all right so this x coordinate is natural log
2 over y and then my y coordinate again is used to identify which shell that
i'm talking about here so the width of the shell is going to be 3 minus
natural log 2 over y okay so now i'm ready to do a cut i'm just going to slice my shell and unroll it
and then i should be able to express the amount of volume in that shell
so here's my unrolled shell
i'm just going to unroll it vertically just the way it was oriented
and it has a little bit of thickness remember the thickness from the diagram was d y
this whole thing has a differential volume of dv and this width here that was 3 minus natural log
2 over y and then the height of this slab well that's the circumference
of the old shell when it was rolled up so that's where i'm going to end
up using my radius which i think i said was 1 minus y but didn't write
so that the circumference of the shell again is equal to this height of the slab when i unroll
the shell so that circumference is 2 pi times the radius 2 pi times the quantity 1 minus y
and now i just use simple geometry to say what the volume of the shell is
length times width times height so i have dv equals 2 pi times the quantity 1 minus y
times 3 minus natural log of 2 over y
times the thickness d y and i'm done setting up my integral basically v is the sum of all the dvs
i have one more complication coming up and i have 2 pi integral 1 minus y
times the quantity 3 minus natural log of 2 over y d y the complication that i just realized is that
my starting y value that's all the way over here has to be determined algebraically and then my
finishing y value is one so i'll put that in let's see what my starting y value is i'll just call it
like y one to make it easy to reference um well that's just the value of this curve two e to the
negative x evaluated at x equals three so that's two e to the negative three and there we have it