Hi everyone.
Welcome back to integralcalc.com.
Today we’re going to be doing a surface area or revolution problem.
We're given the equation y equals x squared minus 1/8 ln or the natural log of x.
y={x^2}-{1/8}lnx And we're asked to evaluate the surface area
of revolution between the points where x equals 1 and x equals 2 and we're revolving around
the y axis. 1=x=2;y-axis
Let’s go ahead and take a look at what this is going to look like.
A=int{a}{b}{2 pi x{sqrt{1+[f’ (x)]^2}}}dx Depending on which axis you're revolving on,
whether it's the y-axis or the x-axis, and the form in which you're giving your equation,
that's going to change you're formula a little but in this case because we're revolving around
the y axis, and we're given the equation in this form, we put our x right here.
A=int{a}{b}{2 pi (x){sqrt{1+[f’ (x)]^2}}}dx Everything else stays the same as the standard
formula for surface area of revolution.
So this x is what's important to pay attention to when we're revolving around the y-axis.
We're going to be using that formula.
The first thing that we want to do is to simplify things for ourselves.
We’re given our equation on the form y equals something but in our formula here, we have
f of x. y={x^2}-{1/8}lnx
f(x)={x^2}-{1/8}lnx It’s the same thing but I just want to go
ahead and change the notation to f of x so that we don't get confused.
So I just swapped out y for f of x instead.
Now what I can go ahead and do is work on this piece right here on the formula.
[f’(x)]^2 I want to first take the derivative because
we're looking at f’ of x and then square it.
I want to get this whole thing in place and then just plug it right into the formula.
So I’m going to go ahead and do that.
The first thing I'm going to do is take the derivative of f of x.
So for f’ of x, of course here, we multiply the 2 out in front and then we get 2x.
f’(x)={x^2}-{1/8}lnx f’(x)=2x
This 1/8 coefficient stays here.
f’(x)={2x}-{1/8} The derivative of the natural log of x is
of course 1/x so we multiply the 1/8 by 1/x.
f’(x)={2x}-{1/8}lnx f’(x)={2x}-{1/8}(1/x)
And then when we simplify, our derivative is going to end up being 2x minus 1/8x, which
what we get when we combine this 1/8 and 1/x.
f’(x)={2x}-{1/{8x}} This here is our derivative.
Now that we have f’ of x, let's go ahead and square it before we plug it in to this
part of the formula.
A=int{a}{b}{2 pi x{sqrt{1+[f’ (x)]^2}}}dx So when we square it, obviously, you can see
we're going to square both sides here.
[f’(x)]^2=({2x}-{1/{8x}})^2 Now we can go ahead and simplify the right-hand
side before we plug it into our formula.
When we do that, we'll square this whole thing so you have 2x minus 1/8 x times another 2x
minus 1/8x and when you multiply everything together, when you FOIL or whatever you want
to call it, you get 4x squared minus 1/2 plus 1/64 x squared.
[f’(x)]^2={{4x^2}-1/2+{1/{64x^2}}} So this is now f’ of x squared.
You can see here this part matches exactly what we have up here in our surface area formula
so now we know we can go ahead and plug it in.
When we do that, this is what we get here.
A=int{1}{2}{2 pi x{sqrt{1+{{4x^2}-1/2+{1/{64x^2}}}}}}dx We’re going to plug in our limits here,
our upper and lower bounds, 1 and 2 for a and b.
so we pluygged in 1 and 2, we leave 2pi x alone because that's part of our formula and
then I plugged in this whole derivative under the square root sign.
You'll notice that we have 1 plus f’ of x squared so I did 1 plus this whole piece
right here.
When we do 1 plus a negative 1/2, we get a positive 1/2.
So I just went ahead and combined those.
So what you end up with under the square root when you take this part and you add 1 is this
4x squared plus 1/64 x squared plus 1/2.
A=int{1}{2}{2 pi x{sqrt{{{4x^2}+{1/{64x^2}}+1/2}}}}dx This is now our integral and now we have to
start simplifying before we can integrate and solve.
So the way that we're going to do that and the easiest way to solve these kinds of problems
is to try to find a common denominator for everything under the square root sign.
In this case, our lowest common denominator is going to be 64x squared.
We’re going to multiply each term by whatever we need to get 64x squared in the denominator.
A=int{1}{2}{2 pi x{sqrt{{{4x^2}({64x^2}/{64x^2})+{1/{64x^2}}+1/2({32x^2}/{32x^2})}}}}dx One thing we do too is move the 2 pi out in
front of the integral, which is a really good thing to do in all these kinds of problems
because this 2 pi is just a constant.
A=2 piint{1}{2}{ x{sqrt{{{4x^2}({64x^2}/{64x^2})+{1/{64x^2}}+1/2({32x^2}/{32x^2})}}}}dx It can be moved out in front of the integral
and it simplifies what you have to integrate.
It’s less for you to look at so to make it easier but we leave the x.
Everything stays the same but now in our square root sign here, we end up with three terms
all with 64 x squared in the denominator which means that we could combine those three terms
into 1 and we just have 64x squared on the denominator and we end up with 256 x to the
4th which is this 4x squared here times 64 x squared here.
This 1 gets pulled out right here and we end up with 32 x squared as well here in the numerator.
A=2 piint{1}{2}{x{sqrt{{{256x^4}+{1/{64x^4}}+1}/{64x^2}}}}dx That’s how we simplify what's inside of
our square root sign and now, we can do a couple things.
Remember that whenever you're dealing with square roots, and you have a fraction, just
one single fraction inside of the square root, you can kind of treat the numerator and the
denominator separately.
So the numerator of our fraction is too complicated to immediately take the square root of but
the denominator is simple enough that we can do that.
The square root of 64 is 8, the square root of x squared is x which means that the square
root of 64x squared is 8x.
sqrt{64x^2}=8x A=2 piint{1}{2}{x{sqrt{{{256x^4}+{1/{64x^4}}+1}/{64x^2}}}}dx
A=2 piint{1}{2}{x{sqrt{{{{256x^4}+{1/{64x^4}}+1}}}/{sqrt{64x^2}}}}dx A=2 piint{1}{2}{x{sqrt{{{{256x^4}+{1/{64x^4}}+1}}}/{8x}}}dx
When we have terms multiplied together like that, we can take the square roots separately
so point being that the square root of the denominator is 8x.
So we can pull that outside of our square root sign and we multiply here by 1/8x.
A=2 piint{1}{2}{x(1/{8x}){sqrt{{{{256x^4}+{1/{64x^4}}+1}}}}}dx That 8x still has to stay in the denominator
because it was in the denominator.
In this case, lucky for us, the numerator can be factored into 16x squared plus 1, that
whole quantity squared.
A=2 piint{1}{2}{x(1/{8x}){sqrt{(16x^2+1)^2}}}dx And that's going to be useful to us because
of course when we take the square root, that square root sign is going to go away.
So we pulled out the denominator, we factored the numerator and now our next simplification
step is two-fold.
First of all we have a 1/8 coefficient here on this whole integral.
We can pull this 1/8 outside of this integral and then when we multiply 1/8 by 2pi, we're
going to end up with pi/4.
A=2 pi(1/{8x})int{1}{2}{x{sqrt{(16x^2+1)^2}}}dx A={pi/{4x}}int{1}{2}{x{sqrt{(16x^2+1)^2}}}dx
So that’s how we simplify there.
We have the pi/4.
Also we have an x in the numerator and x in the denominator, so those two will cancel
and go away completely which means everything here in front of the square root is going
to get simplified out in front of the integral or canceled.
A={pi/{4x}}(x)int{1}{2}{{sqrt{(16x^2+1)^2}}}dx A={pi/4}int{1}{2}{{sqrt{(16x^2+1)^2}}}dx
And we're just going to be left with the square root of the quantity 16 plus x squared plus
1 squared.
So of course when we take the square root of that, this squared term here is going to
go away, the square root sign is going to go away.
A={pi/4}int{1}{2}{{(16x^2+1)^2}}dx We’re just left with 16 x squared plus 1.
And luckily, that's easy enough that we can integrate.
So when we integrate this, of course the pi/4 remains out in front and then remember to
integrate, we're adding 1 to the exponent so 2 plus 1 is 3 so we get the exponent 3
and then we divide the coefficient, 16 by the new exponent which is 3 so we get 16/3
x cubed and then of course the integral of 1 is x and we're going to be evaluating that
on the range 1 to 2.
A={pi/4}{[{16/3}{x^3}+x]}tabular{000}{10}{{2} 1}
So that’s what our integral looks like after we evaluate and now we just need to evaluate
by plugging in 1 and 2.
So here's what that's going to look like.
A={pi/4}{[({16/3}{(2)^3}+2)-({16/3}{(1)^3}+1)]} We’ve got this definite integral here.
Remember when we have a definite integral, we always have to plug in the top number here,
in this case 2 first and then you always subtract here in the center and then always plug in
the bottom number in our case, 1 to the second half here.
So the pi/4 remains out in front, we just plug in 2, we subtract and plug in 1.
Now we're going to start simplifying.
So 2 times 2 is 4 times 2 is 8, 8 times 16 is 128.
A={pi/4}{[{128/3}+2-({16/3}{(1)^3}+1)]} So that's how we end up with 128/3, we leave
the 2 by itself.
16/3 times 1 cubed is just 16/3 so we have a minus 16/3.
A={pi/4}[{128/3}+2-{16/3}+1] And then this negative in here carries over
to this 1 term so we end up with a negative 1.
A={pi/4}({128/3}+2-{16/3}-1) Now we could do this with our calculator but
I want to end up with a clean fraction at the end so we're going to find a lowest common
denominator inside the parentheses here which means that we're going to multiply both this
positive 2 and this negative 1 term by 3/3 to get 3s in the denominator.
A={pi/4}({128/3}+2(3/3)-{16/3}-1(3/3)) So we end up with a positive 6/3 and a negative
3/3.
A={pi/4}({128/3}+{6/3}-{16/3}-{3/3}) Now that we've got a lowest common denominator,
we can go ahead and combine these fractions here.
A={pi/4}({128+6-16-3}/3) What we end up with is 115/3.
A={pi/4}(115/3) Now what we do is multiply these two together
and we end up with 115pi divided by 12.
A={{115 pi}/12} And that's obviously as far as you can simplify.
That’s our final answer.
That is the form that you get for your final answer for surface area of revolution.
I hope that helped you guys and I’ll see you in the next problem.