in this video we set up the integral for the surface area of the ellipsoid found by revolving
this curve x over three quantity squared plus y over two quantity squared equals one about the x
axis once the integral is set up it turns out that it can actually be calculated in exact form by
using a trigonometric substitution so we're going to take it all the way through that calculation
in this video at the very end i'm going to verify my symbolic solution by checking with
a computer algebra system now if you haven't seen trigonometric substitutions yet in your
calculus class just look at the chapters down in the description and you can skip ahead to the
end of the video where i verify my symbolic solution by using a computer algebra system
so to find a surface area of revolution we slice into these thin ribbons and i have this red ribbon
already in the picture and the thickness is given by a little arc length element
that we usually call ds and ds is given by 1 plus the derivative of that curve squared dx
and the radius of the ribbon well that's just the height of the function at this arbitrary location
x so the y value corresponding to some arbitrary x now before we can flesh out these expressions
i've got to solve for y in the original equation of this ellipse so i'm going to start there and
say y over 2 squared equals 1 minus x over 3 squared which i'll write as x squared over
9. then i square root both sides and multiply by 2 to solve for y so y is going to be 2 square root
1 minus x squared
over 9. so that's the y that's going to go in here for the radius of the ribbon
but to get that little arc length element and ultimately the area contribution of this ribbon
i've got to get the first derivative of y so when i differentiate the square root
well square root is really a power of one half so i have one half one minus x squared over nine
to the negative one half then the chain rule says multiply by the derivative of the interior there
which is negative two x over nine cleaning things up a little i have negative two x
over nine square root one minus x squared over nine now i don't particularly like
having fractions inside of fractions so i'm going to take one factor of 3 from this 9
and distribute that into the interior of the square root and it becomes a 9 in there
we get negative 2x over 3 times the square root of 9 minus x squared and that's a little bit nicer
so i may as well just get y prime squared right here before i plug it into ds and i get 4x squared
over 9 times the quantity 9 minus x squared again that was y prime squared so now i can write down
the area of this thin area contribution from the ribbon and i have a little ds for the thickness
and then the length of this ribbon well that's the circumference of it from back when it was
rolled up and so that's going to be 2 pi times the radius but the radius is y as a function of x
and normally i would call that little area contribution d a so we have d a
is equal to the area of this little thing well it's just a rectangle so length times width
2 pi y of x times ds so that's 2 pi and then y of x is given by this expression
times 2 root 1 minus x squared over 9 times ds which is this enormous square root 1 plus
4 x squared over 9 times the quantity 9 minus x squared
dx and this is why it's surprising that it turns out this thing has an exact form solution
ds in practice is usually so ugly that there's no symbolic solution to these integrals
well this is one case where there is i'm going to clean this up just a little bit before i slap an
integral sign on it and the integral will simply add up all the area contributions so i have 4 pi
and then i can multiply the interiors of these two square roots and turn them into one single square
root now i'm going to distribute a 3 into the interior of the square root and the point of that
is that it should clean up this piece so that i recognize that it's really the same factor as
this piece in this denominator so i'm going to divide by 3 out in front
multiply by 3 inside the square root which gives you a 9 in there so i have 9 minus x squared
times the quantity 1 plus 4x squared over 9 times 9 minus x squared dx
now i'm going to distribute the 9 minus x squared into this second piece inside the radical it's
going to cancel a factor in that second term so i distributed it to the one out here in
front when i distribute it to that fraction it just cancels the nine minus x squared and
i'm left with a four x squared over nine i'll go ahead and put in an intermediate step here
x squared is just nine x squared over nine so i'm just trying to get a common denominator
with that second term now i don't particularly like the fraction inside of that square boot
so i'm going to multiply by three inside the square root which becomes a 9
and divide by 3 out in front so i end up with 4 pi over 9 and inside i have 81 minus 5x squared
and now i recognize this as a trig substitution type of integral so let me go ahead and finally
apply the integral and say the total area is going to be the sum of all the
area contributions and i'm going to go ahead and just do the right side of the ellipsoid so that's
as x goes from 0 to 3 and then i'll double the answer just cleans things up to do it that way
and plug in d a and now again if you've never seen a trig substitution before you might want
to skip to the end of the video where we do this integral using a computer algebra system
if you want to see the trig substitution then here we go and the idea here is that we have
inside of that square root something that could be simplified to the square of a single function
and then the square root would simplify now a constant minus the square of a function
that fits the form of a trig identity so our goal is to fit this form and i've got
a couple annoying constants floating around so here's how i'm going to do the substitution
when i square that x what i'm trying to get is a factor of 81 to come out of here so that i can
factor the 81 out of the entire square root so i'm going to let x be 9 so when i square it i get 81
over root 5 so that when i square it it cancels this factor of 5 and then the function piece that
i want here is the sine function so we'll put in a sine theta this means that x squared is 81 over 5
sine squared theta and it also means that dx which also has to be transformed here is 9 over root 5
cosine theta d theta furthermore i'd like to transform my limits of integration so i can
forget about x for the rest of the problem so i look at my substitution and i ask myself when x
is zero that's the lower limit of integration down here when x is zero what is theta well the
sine of zero is zero so theta must be zero what about when x is 3 it's not going to be so clean
when x is 3 i would have 3 equals 9 over root 5 sine theta
cancel a factor of 3 here and i end up with root 5 over 3 equals sine theta
so theta is the inverse sine of root 5 over 3. that becomes my new upper limit for the integral
so we transform the integral and as i do this i'm going to pull the constants out in front that were
already there there was a 4 pi over 9 to pull out so this becomes 8 pi over 9 times the integral
as x goes from zero well that means theta was zero so when x is three that means theta is the inverse
sine of root five over three times the square root of eighty one minus five times x squared which i
already determined is 81 over 5 sine squared theta times dx which we figured out was 9 over root 5
cosine theta d theta well i can cancel a 9 here so that's nice and then i'll move that one over
root 5 out in front for now and then if i look at what's happening inside of the square root
these fives are going to cancel each term has an 81 stuck to it which we sort of arranged
intentionally and i factor in 81 out of the square root which gives me a 9 out there so i'm
just going to put that in front of the integral now inside that square root is a 1 minus sine
squared that's all that's left and that's the trig identity we're trying to take advantage of
that's cosine squared but the square root of that is just cosine
so all of this stuff is just going to reduce to cosine of theta i have another one right here
that came from transforming the differential dx so i end up with the integral of cosine squared
theta d theta now the first time you see that integral it's pretty
tricky you have to use a trig identity for it
i'm going to go ahead and smash the 8 and the 9 out in front as i do this so 72 pi over root 5
and the cosine squared has an identity that's one half 1 plus cosine of twice the angle i can cancel
the two by bringing the one half out in front and i'm back to 36 pi out in front of this integral
and now each piece of the integral is guessable the integral of one is theta
the integral of cosine two theta is essentially sine two theta but then i realize i need a one
half out in front of that now if i differentiate it and use the chain rule i'll get back to
what i started with so i know that's the correct anti-derivative and at this point i can sub in the
limits of integration and get my exact solution but there's one more challenge left to overcome
um before i get to that i'll sub in my lower limit when you put in zero for theta you get zero when
you put in zero for theta and sine 2 theta you get another 0. so both of those contribute nothing
and that's really the payoff of doing what we did back here
saying we're just going to take twice the integral of half the area that gave us that nice zero
now when i'm looking at the upper limit plugging in for theta is not a big deal that's
fine but plugging in for sine two theta that's a problem because the sine of twice an inverse
sine i don't know how to deal with but if i had like the sine of the inverse sine or the cosine
of the inverse sine i could deal with that so i'm going to use another identity here i'm going to
use the identity for sine 2 theta that's 2 times the sine times the cosine and the 2 cancels the
one half and i end up with this and i'm trying to sub in inverse sine root 5 over 3. and again when
i sub in the 0 i get 0. so i'm not worried about that so plugging in for theta that's inverse sine
root 5 over 3 then i get into this expression so the sine of the inverse sine of root 5 over 3
that's just root 5 over 3. the one that's non-trivial is the cosine of the inverse
sine so we end up with the cosine of the inverse sine root 5 over 3 and that needs to be simplified
so how do i get a trig function of an inverse trig function we
use what's sometimes called a reference triangle i'm going to use the symbol theta
for the angle whose sine is root 5 over 3 that's what that inverse sine means
and sine is opposite over hypotenuse so here's one way to do it now the sine of this angle
is root five over three what's the cosine of the angle whose sine is root five over three
that requires that i find this horizontal leg to figure this out you use the pythagorean theorem
question mark squared plus root 5 squared which is 5 is equal to 3 squared which is 9 and question
mark squared turns out to be 4 which means that leg is two now i can find the cosine of the angle
whose sine is root five over three it's just the adjacent over the hypotenuse or two thirds
so things are starting to really come together now i have 36 pi
over
root 5
inverse sine root five over three
plus root five over three times the cosine of the angle whose sine is root five over three
and i just figured out that's two thirds so maybe the last thing i'll do here is
rationalize the denominator so i'm going to multiply the denominator out front by root 5
and that means i have to multiply the numerator by root 5 as well but there's some advantage to
distributing that root 5 into the interior of these parentheses so let's see what that looks
like in the second piece i just hit it with a root five so that becomes five five times
two is ten so we end up with ten ninths in there now that's clean enough to be a final answer and
hopefully we'll be able to see how to compare with the answer produced by a computer algebra system
so here i am in maxima and i went back to the step in the lower left hand corner
and i put in two times the integral so that's integrate in maxima four times percent pi
percent pi is the special symbol for pi over 9 times the square root of 81 minus 5x squared
integrate that as x goes from 0 to 3. now it can sometimes be a little tricky to find agreement
between a computer algebra system and what you did by hand one thing i noticed in this solution is
that there's a factor of 9 inside the parentheses in the numerator and that would cancel with a
factor of 9 in this denominator giving me a 5 in the denominator we might be able to get maxima
to simplify that for us so i used the rat simp command and percent means the prior output rat
simp means simplify a rational expression so it did indeed cancel the factor of 9 out of there now
if i look at the first term in the numerator i see the inverse sine of root five over three that's
what a sine is in maxima times 36 pi times root five all divided by five so that term checks out
it's exactly the same as what we got by hand the second term is a 40 pi over five so we're going to
have to think about if we distributed that 36 pi over 5 to the 10 over 9 you could cancel the nines
leaving you with 4 pi over 5 times 10 that's 40 pi over 5 and that agrees with maxima as well
so we've checked out our solution with the computer algebra system and it agrees with
what we did by hand if you find the math content on zack's lab helpful click on the zaxlab logo on
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