Okay let's integrate this thing by using the shell method
and i put up a video of what it looks like using the washer method and this shell method is going
to be just a lot easier because we're not going to have to invert functions to find out
x values corresponding to y values so we'll see how that all works out
so there's a little cylindrical shell it has a thickness of dx
the edge of the shell is located at a value of x and i'm i'm just going
to try to like map in all the dimensions of this thing so this height of the shell
well that's just 2x for that top y coordinate minus tangent x for the one below it so i have 2x
minus tangent x for that height and then i need to get a circumference for this thing i'm just going
to put the radius in the diagram and that's just x okay so the next step is to cut and
unroll my shell so i'm going to slice it right there and then i'm going to unroll it
and the little thickness the little differential thickness there is just dx
and my height was 2x minus tangent x
and then my my length of this slab is the circumference of
the shell when it's rolled up so that's 2 pi x
so i can write down my differential contribution to volume and it's just length times width times
height for this thing so i end up with 2 pi x times the quantity 2x minus tangent x dx and so
now i'm ready to add these up by using an integral but there's one detail that has has to be worked
out first what's the intersection of that point and for that we need a numerical approximation
so what i'm trying to solve is 2x equals tangent x there's no analytical solution to this thing
so i'm going to use the find root function in wx maxima and that function is for approximating
the value of x for which an expression is equal to zero so i had to rewrite it
this way so let's toggle the maxima i was actually just doing this for another problem
and it's in this top line of maxima i said find root tangent x minus 2x as x goes from 1 to 1.5 so
the important thing there is to choose an interval where you are approximating that you know there's
a root and look at the graph i can tell there's an intersection between one and one and a half
so you have to put that in for maximum and this thing turned out to be at 1.166
so x is approximately 1.166 all right so now we're going to use an integral to add up all the dvs
all the way from x equals zero to one point one six six
two pi x times the quantity two x minus tangent x
d x and um this integral looks to me like the the kind of thing you'd be able to do with
integration by parts but we already had to use an approximation anyway so why don't we go ahead
and just numerically approximate the integral i just need to jot something down real quick okay
i'm trying to reduce my toggling
so we're going to integrate but i'm just going to go ahead
and tell maxima i want to do a numerical integral so that's quad q a g
i have 2 times pi that's percent pi times x multiplied by the quantity two times x
minus tangent of x
as x goes from zero to one point
one six six let's round it there
oops and i forgot to put in which algorithm to use you could choose any number from one to six so
i put in one and i get a volume of 1.787 or i could i could put in 1.788 as my rounded answer
you