All right we're going to compute a surface area of revolution the curve that we started with is
y equals 7x and we've revolved it around the x-axis and so i've already drawn into the
picture a little ribbon of surface and the length of or the the thickness the length of this little
segment right here is d s a small section of arc so remember how we write that down ds
is the square root of 1 plus
f prime of x all squared dx so we're going to have to get f prime of x and we might as
well do it right now so that's y prime for this y uh the square root of x is
just x to the one half and i use the power rule on that so i get seven times one half
x to the negative one-half which cleans up nicely as seven over two square root of x
um so i think i'll go ahead and plug that into ds a little bit later
the other part that i need on this ribbon is the radius
so this ribbon is located at some horizontal value of x the y coordinate corresponding
to that is 7 root x so there's my radius and then i'm going to slice and unroll this ribbon
so i get this long skinny approximately rectangular thing
the thickness is ds the length of it is 2 pi times the radius
of the original ribbon that we unrolled we could clean that up and call it 14 pi
root x so now i'm ready to write down an expression for the differential contribution
to the area we call that d a and it's just the area of a rectangle i get 14 pi root x d s
plug in our d s from over here including the f prime over here and i get 14 pi root x
times the square root of 1 plus f prime that's this thing
squared so i'm going to go ahead and square it while i'm at it so i get 49 in
the numerator and then 4 in the denominator and then square root of x squared is just x
dx and we'll just clean this up a little i'm going to distribute the root x
into the interior of that square root
that just puts an x into the interior
and i end up with x plus well the x's cancel in this term x plus 49 fourths
now that we have an expression for d a entirely in terms of one variable i use an integral
to add up all the area contributions plug in the expression for da and then i'll start
oh something i forgot to write here is that this interval was 2 to 4.6 so i'll start at 2
end at 4.6
and i get 14 pi square roots x plus 49 4
dx now nothing special has to be done to perform this integral i have a function of x in here but
the derivative of it is one so the chain rule backwards is not is not necessary so
this is just this expression to the one-half power it might be good to look at it that way
let's get 14 pi
x plus 49 fourths
to the one half dx then i just use the power rule on this so i add one to the exponent and divide
by the results that gives me three halves for the exponent dividing by three halves
is the same as multiplying by two thirds so let's do all that in the next step so i have
maybe i'll just move constants out in front i have 14 pi
times two thirds and then i have
x plus 49 fourths
to the three halves evaluated from 2 to 4.6 and we're about to wrap it up i have 28 pi over 3
times 4.6 plus 49 fourths to three halves
minus two plus 49 fourths to the three halves and now it's time to just plug that into a calculator
and when i did this i got approximately 128.
oh excuse me let's see if i can i should redo it i was remembering a different calculation
so excuse me while i type for a while
so i have 4.6 49 fourths
three halves and then minus
right so i got 450.8 approximately
the units on that are just square units for whatever
units these represent and the calculus classes usually aren't labeled as like
inches or centimeters or something so it's okay to leave it without units on your final answer
you