in this video we're looking at a surface of revolution formed by revolving the graph of cosh x
on zero to one about the y axis once we set up the surface integral for this surface of revolution it
turns out that the resulting integral requires integration by parts so you can see i've already
drawn an increment of area into this surface and that's a thin ribbon with a thickness given by an
increment of arc ds in addition i can say that the right end of this ribbon is at a horizontal
position of x well that means the radius of this ribbon is given by x so the next thing we do is
cut and unroll this ribbon and we'll get a thin rectangle the thickness of that rectangle is ds
and the length of it well that corresponds to the circumference of it when it was rolled up
into a ring so that's going to be 2 pi times the radius which here is just 2 pi times x we call the
area of this thing d a and that's just given by the product of length and width so 2pi xds
now we need a reminder of how to write the arc length element ds and this is where the hyperbolic
cosine curve comes into the setup the arc length increment is given by the square root of 1 plus
the derivative of the curve squared times dx now the derivative of cosh is just cinch the
hyperbolic sine function and i'll post a link to the video where i derived that well this means i
can set up my expression for the area increment as 2 pi x d s which is square root 1 plus cinch
squared x dx and because this is phrased entirely in terms of one variable now i can go ahead and
slap an integral sign on it and add up all the contributions so a is the sum of all the
da's that's what integration does so i have the integral as x goes from 0 to 1 2 pi x square root
1 plus cinch squared x dx so here i have to remember an identity for cinch and cosh 1 plus
cinch squared turns out to be cosh squared and if i square root that i just get the cosh function
all right this is a classic integration by parts situation
i have this simple function x that i know will get simpler if i differentiate it so
i'm going to let that be u in my integration by parts and then i have a second function
cosh of x that shouldn't be hard to integrate so i'm going to let cos x dx be dv in my substitution
i'm also going to move the 2 pi out of this thing as i do my integration by parts
so to find v out of this i just have to guess the anti-derivative
of this dv just recalling that the derivative of cinch is cosh so i'm going to pull my 2 pi
out of there as i rewrite this thing using integration by parts and the integration by
parts formula tells me that i get a uv out of this and that's going to be x cinch x
evaluated from zero to one minus the integral of v d u so it's the integral from zero to one
of cinch x that's v and then d u is just dx
so i'll go ahead and evaluate that first piece across the limits of integration
when i plug in the upper limit i get one times cinch of one so i get cinch one
minus zero times cinch of zero so that's zero then i have this other integral piece and i have
to guess the anti-derivative of cinch which again is just cosh and i get cosh x evaluated from zero
to one so this is one way to leave the answer but it's not ideal i can write cinch one minus cosh1
plus cosh of zero i'm pretty confident we can simplify this thing better than that by using
the definitions of cinch and kosh and i can see that cinch of one that would require me to replace
both those x's with ones is going to give me a one-half e to the one minus e to the negative one
then i look at my cosh of one term and that's one half e to the one plus e to the negative one
remember i'm subtracting cosh of one so minus one half
e to the 1 plus e to the negative 1 which is 1 over e
and then i have cosh of 0 at the end of this expression
and replacing the x's with zeros here in the definition of cosh i would get 1 plus
1 in the parentheses and half of that is just 1. so when i look at simplifying these terms
i notice there's a one-half e that's going to pop out of this first piece and a negative
one-half e out of the second piece so those cancel and i'm left with two pi
distributing the one half in here
minus one over two e
minus another one over two e plus one and negative one over two e minus another one over two e
that just gives me negative one over e and i'm gonna go ahead and write that after the one to
make it look a little nicer and there's our exact surface of revolution using integration by parts
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