So here we're trying to find the surface area of revolution for this this shape i've taken the sine
function from zero to pi just the first hump of the sine function and then i spun it around the
x-axis trying to get the surface area here and again i'm going to take a physical integration
approach to this i'm going to chop it up into a bunch of tiny contributions to surface area
and then add them all up using an integral i think i make that mistake a hundred percent of the time
all right there we go
so i'm gonna try to chop this into ribbons remember it's a surface it's not a solid
alright so we're slicing the surface into infinitesimal little ribbons
and then if i look at this little width of my ribbon that's a little chunk of arc and that's
why these two things are usually put in the same chapter in a calculus book that's a little ds
and the radius of my ribbon so if i go from
i hope this isn't too obnoxious but i'm going to put in a little radius here
so from the center from the axis of symmetry out to the edge of this thing
well that's just the value of f of x itself or sine x
all right and just like we were doing with cylindrical shells
i then slice my ribbon and unroll it
and we'll see what it looks like
whoa i'm getting crazy lag right now
hopefully we don't crash i think i know it's wrong but i can't fix it on the fly so i have
a thickness of ds and then i have a circumference of this guy of two pi times the radius or two pi
sine x and then d s i just used the same thing i was using with arc length d s is the square root
of one plus f prime of x squared dx which means
square root 1 plus well the derivative of this function is just cosine all right and
this little area of the ribbon is what i call d a a differential contribution to the total area
and i'm going to get an expression for d a real quick d a
is just a length times width so it's going to be 2 pi sine x root 1 plus cosine squared x dx
and once you've got an expression for da then the integral just simply adds all those up so i have
the total area is equal to the sum of all the contributions to the area
but now i have an expression for that little differential area contribution
and this is entirely phrased in terms of of the
integration variable x so i'm going to go from the beginning value of x to the ending value you