in this video we're looking at a surface of revolution formed by revolving the graph of sine
x on zero to pi about the x-axis once we set it up the resulting integral turns out
to be really interesting it requires multiple substitutions and as you can see in the hints
it requires the formula for the integral of secant cubed as well i'm going to leave the proof of the
secant cubed formula to another video because this one is already long enough so you can see in the
picture i've already drawn an increment of area into the surface and that's a thin ribbon with a
width given by a small increment of arc called ds and the ribbon is located at a position of
x which means the radius of this thing is given by the height of the function corresponding to x
in other words the radius is sine x next we cut and unroll this ribbon and that looks something
like this and there's my width d s and then the length of this very skinny rectangle is going
to correspond to the circumference of the ribbon when it was rolled up so that's 2 pi
times the radius or 2 pi sine x then we'll refer to this small increment of
area as an area contribution d a so here's where we're at we have d a is equal to 2 pi sine x
d s here's where we have to be reminded how to write down the arc length increment d s
and i'll post a link to the video where i derive this so d s is given by the square root of one
plus the derivative of the curve squared and the derivative of sine x is cosine x
so there's my arc length d s now i can plug this back into the formula for d a and i've got 2
pi sine x times the square root of 1 plus cosine squared x since everything is phrased in terms of
x now we're ready to set this up as an integral we say that the total surface area is the sum
of all the contributions that's what an integral does it just adds up these continuously changing
infinitesimal contributions and i've phrased it entirely in terms of x now so i can put
in my x limits of integration that's 0 to pi 2 pi sine x square root 1 plus cosine squared x
dx now we get into this non-trivial integral so when you look at one plus cosine squared
you might be tempted to say there must be some identity that's going to help me out of this jam
and it just doesn't work that way there's no identity for one plus cosine squared x that's
going to give you anything useful but what i do notice here is that the derivative of cosine is
sitting in this integrand the derivative of cosine is negative sine so all i have to do is throw a
minus sign in there and i've got the derivative of cosine so this looks like a u substitution to me i
can replace cosine with u and then the derivative of u is going to be basically the sine function
and i should be able to transform the integral so here's my first substitution let u equal cosine x
then d u is negative sine x dx i'm also going to transform the limits of integration so i don't
have to go back and express my answer in terms of x at the end of the problem so when x is equal to
zero i look at my substitution here to figure this out u is going to be cosine 0 which is 1
and when x is equal to pi u is going to be cosine pi which is negative 1. so because i'm tied on
space here i'm going to go ahead and manipulate this integral in the previous line i need to
multiply by negative 1 to find negative sign x in this interior so i'm just going to jam it in here
negative that means i need to multiply the outside of the integral by negative 1 as well
so i put that outside and now this negative sine x dx can be replaced with d u and the square root
part can be replaced with one plus u squared again my new limits of integration are 1 to negative 1.
this 2 pi is a spectator i'll take care of that in a second there's a couple things i can do to
clean this up before i get into how to find the integral of a square root of one plus u squared
um first i have a minus sign out in front and if i take the negative of an integral
it reverses the limits of integration and i'm just a little uneasy having this sort
of backwards integral going from one to negative one so let's just take care of that real quick
and i'm also going to move the 2 pi out in front so i reverse my limits of integration to
something that looks a little more ordinary then i notice that i'm dealing with an even function
in the integrand 1 plus u squared if i replace u with negative u i just get back the original
function that means it has even symmetry and given this interval symmetric about the origin
i can just say that i get twice so i'm going to move a factor of 2 out in front make that a 4 pi
i get twice the integral of just the right half of this thing from zero to one and it's going to
turn out that having that zero as one of my limits of integration actually saves us a lot of work in
this problem all right next step i have a square root of one plus u squared this is a classic
trig substitution integral i need to think of an identity that involves 1 plus something squared
and in this case it's the tangent function 1 plus tangent squared gives me secant squared
so i'm going to replace u with tangent theta that means d u is secant squared theta d theta
again i'm going to transform the limits of integration so when u is zero
i look up here at my substitution when u is zero i have tangent theta equals
zero well that happens at theta equals zero and when u is one i have one equals tangent
theta that happens at theta equals pi over four so i modify my integral again and i have four pi
integral from 0 to pi over 4 square root 1 plus tangent squared theta
and then d u is secant squared theta d theta and 1 plus tangent squared is secant squared but when i
square root that i just get another copy of the secant function and now i can see how i end up
with a secant cubed integral again that integral itself is very non-trivial it requires multiple
applications of integration by parts that's a really interesting integral in its own right
here we're just going to quote the formula for secant cubed so i get one half secant x tangent x
plus a one half natural log of the absolute value of secant x plus tangent x and i'm evaluating
this from zero to pi over four first off i can pull the one half out of this thing so i end up
with a two pi out in front then when i look at my lower limit i'm plugging in zero for x and
in this first expression that would give me a tangent of zero which is zero so that vanishes
in my second expression again the tangent would come out to zero so that's gone
and the secant of zero is one over the cosine of zero which gives me one and the natural log of
one is zero so that vanishes all this left is what we get from the upper limit so it really
paid off to recognize that even symmetry it gave us a really useful zero in this case all right
i've already pulled out the one-half and i'm going to plug in pi over 4 here i get secant pi over 4
tangent pi over 4 plus the natural log absolute value secant pi over 4 plus tangent pi over four
secant again is one over cosine then the cosine of pi over four can be written as one over root two
so one over one over root two and then the tangent of pi over four is just one
in my natural log piece i have the natural log of the absolute value well both of these are going to
give me positive numbers so i'm not worried about the absolute value signs and again secant pi over
4 is root 2. tangent pi over 4 is 1. so there's our exact area for the surface of revolution
using the sine function if you find the math content on zack's lab helpful click on the zac's
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