Hi everyone! Welcome back to integralcalc.com. Today we’re going to be doing another Surface
Area of Revolution problem. The function that we’re given is y equals the square root
of four minus x squared and they’ve asked us to rotate that function around the x axis
and then find the surface area of that rotation between x equals negative one and x equals
one. So, a couple things here first with this problem
before we get started, we’re going to be using this equation here to find the surface
area of revolution and notice here that we have this y variable, this changes depending
on which axis we’re rotating around. So, when we rotate around the y axis, this is
an x, in this case, we’re rotating around the x axis so it’s a y, everything else
stays the same. So, what we’re going to do is take this formula. We’re going to
start with what’s inside the square root here because this is the most complicated
part of our formula. We’ll calculate the derivative here, dy over dx, and then square
it, add one in it, and when we get that piece together we’ll plug it in to our integral
instead of having to do it all within our integral.
So, let’s go ahead and find the derivative first. When we find the derivative of the
square root of four minus x squared, the first thing that we’ll do is convert the square
root to an exponent, to the one half. The square root of something is the same thing
as that thing to the one half power so I just gone ahead and converted it to four minus
x squared to the one half. So nothing’s changed, we just change the format. So, now
that we’ve found that, it’s easier to find the square root because using chain rule
we first treat this one half power as part of the outside function, this four minus x
squared is the inside function. So, remember with chain rule that we first take the derivative
of the outside. And, that means that we’ll multiply this one half exponent out in front
to get this one half as coefficient and then subtract one from the exponent so one half
minus one gives us a negative one half exponent here. And, we leave the four minus x squared
completely untouched for now but then chain rule tells us that we have to multiply by
the derivative of the inside function. So the derivative of four minus x squared is
a negative two x, that’s the… the derivative of the inside, so we multiply by the derivative
of the inside function and therefore multiply here by negative two x. So, now we’re going
to simplify everything here to the right of the equal sign to get our derivative. Notice
that we have a two in the numerator and a two in the denominator so that’s going to
cancel and we’ll be left just with negative one here. So we have negative x in the numerator
from this negative x right here and then four minus x squared to the negative one half.
Remember that if we move that whole piece down into the denominator, that exponent goes
from a negative to a positive. You can flip-flop, a term from the numerator to the denominator
of our fraction and back again and all that changes is the sign on the exponent. So, because
we’ve moved it from the numerator to the denominator, we change the sign from a negative
to a positive, so we end up with four minus x squared to the one half, to the positive
one half, in the denominator. And, remember that anything to the one half is the same
as the square of that… of that term so this four minus x squared to the negative one half
is the same as the square root of four minus x squared. So that’s how we end up with
that in the denominator. And then, now that we’ve calculated dy/dx
which is this component here of our formula, we can go ahead and plug it in to our formula
and start simplifying within the integral. So, they’ve asked us to evaluate between
negative one and one so we plug in negative one for a and one for b here as our limits
of integration and then we leave two pie alone. For y, remember our original function is y
equals the square root of four minus x squared so we plug that in for y here. And then, within
this square root sine, we leave the one alone and we just take the derivative dy over dx
that we found here and plug it in to our formula, leaving the square root exponent and dx. The
next thing that we’ll do is simplify. A couple things, first of all we’ll pull up
two pie out in front of the integral because two pie here is a constant, it’s like coefficient
on everything else that comes after it, so it can just pop right out here in front of
the integral to make integration simpler for us down the road. This square root of four
minus x squared will stay the same. As for the derivative here that we’ve squared,
we’ll go ahead and square both top and bottom. So negative x squared will be a positive x
squared so we end up with that in the numerator and then because we are squaring a square
root in the denominator, we’ll just end up with what we have inside of the square
root sine which is four minus x squared and everything else drops away.
So, now that we’ve… now that we’ve simplified that, remember that we had a one plus x squared
minus over four minus x squared, well, we want to find a common denominator within the
square root. So we change the one to four minus x squared over four minus x squared
which of course reduces to one so it’s the same thing but it gives us a common denominator
with the second term under our square root sine. And, once we’ve done that, we can
combine those two fractions. So four minus x squared plus x squared means that the x
squared are going to cancel and we’ll just be left with this four here in the numerator
and then of course we have four minus x squared in our denominator.
Now, this is a… a technique of algebra, you have a square root sine over a fraction,
well, you can break apart that square root and take the square root of the numerator
and the denominator separately. So that’s what we’ve done here, we take the square
root of the numerator, the square root of four is two so we get two here, and then we
leave the square root of the denominator as it is. The reason we do that is because notice
now that when we have the square root of the denominator, it will cancel with what we have
here in the numerator. So, we’re going to go ahead and cancel these two out and we’ll
just be left with two inside of the integral here which we got from our numerator, everything
else cancelled. So, we have two and now we can easily take
the integral of two, of course, the integral of two is two x so we end up with two pie
times two x and then we’ll be evaluating on the range negative one to one. And when
we do that, remember when we are evaluating like this, we plug in the top number first
which is one, so you can see we’ve plugged that in here for x, and then we always subtract
in the middle and then plug in the bottom number which in our case is negative one,
so we’ve plugged in that negative one here. And then, to simplify, inside the parenthesis
here we’ll end up with two plus two because we have a double negative here so that gives
us a positive which will be two pie times four which gives us our final answer of eight
pie. So, eight pie is the surface area of the revolution of this function around the
x axis between x equals negative one and x equals one.
So, I hope that helped you guys and I’ll see you in the next video. Bye!