integration problem?

Question

how do I find the surface area of a shape swept out by the curve x^2 rotated through 360 degrees about the x axis. Between x=1 and x=2. Imagine it like a lampshade on its side. I want the surface area of it.

This is not a homework, I was just messing about and realised that I couldn't do it and now it is bugging me.

I know that I need to find the length of the curve by Integrating sqrt(4x^2+1)dx. So my answer is found by Integrating 2Pi*x^2*sqrt(4x^2+1)dx using the limits x=1 and x=2.

Now for this I have used trigonometric substitution and end up needing to integate
(Pi/4)[(tanT)^2] [(secT)^3]dx

Note: theta=T

This is where I get stuck how do I integrate this? Have I gone about it the right way?

On the computer I get 49.419 using the original integration and also when changing the limits to fit my trigonometric substitute.

Answer 1

I'll just denote tanT by T and secT by S:
disregard the dT anyway.

∫T²S³ = ∫(S²-1)S³ = ∫S^5 - S³

Now, for ∫S^5 use integration by parts
u = S³ , dv = S² , then du = 3S³T , v = T
∫S^5 = S³T - 3∫S³T²
=S³T - 3(∫S^5 - S³)
Thus, 4∫S^5 = S³T + 3∫S³
∫S^5 = ¼S³T + ¾∫S³

Then, ∫T²S³ = ∫S^5 - ∫S³ = ¼S³T - ¼∫S³

Now, for ∫S³ use another IBP again
u = S , dv = S² , then du = ST , v = T
∫S³ = ST - ∫ST² = ST - ∫S(S²-1)
Thus ∫S³ = ST - ∫S³ + ∫S
2∫S³ = ST + ln|S + T|
∫S³ = ½ST + ½ln|S + T|

Finally, ∫T²S³ = ¼S³T - ¼∫S³
=¼S³T - (1/8)ST - (1/8)ln|S+T| + constant.

d:

Answer 2

I'm not convinced that we need the length of the curve. That's one way to do it, but not the only way. And since the length of the curve is turning out to be a pain, I want to try something else.

We should be able to integrate the circumference of circles of varying radii from x=1 to x=2. (Since dx has negligible width, the surface area on dx has negligible width, and so the surface area on dx is just a circle). The radius of the circle at each point x is given by x^2. The circumference is then given by 2πx^2.

We need to integrate 2πx^2 dx from x=1 to x=2. That gives (2/3)πx^3 from 1 to 2, so (2/3)π(7) = (14/3)π. This would be my answer. Anyone get something different?

Edit: Fixed a silly arithmetic error.

Answer 3

Hi

The curved surface area of revolution is ∫2πy √[1 + (dy/dy)²] dx

You are absolutely correct up to and including your last integral.

CSA = ∫2π x² √[1 + 4x² ] dx

let 4x² = tan²θ then 8xdx = 2tanθsec²θdθ

CSA = ∫2π .tan²θ /4 . secθ. 2tanθsec²θdθ / (4tanθ)

......... = ∫(π/4) tan²θ sec³ θdθ

......... = ∫(π/4) ( 1 + sec²θ) sec³ θdθ

......... = (π/4)∫(sec³θ + (secθ)^5) dθ

now the only way that I know to integrate this is by reduction formulae or repeated integartion by parts.

The reduction formula for the integral of powers of secθ is
(n - 1)S(n) = tanθ(secθ)^(n - 2)) + (n - 2)S(n - 2)

I see that you have some completed solutions...I had better take this a bit further.

From above 4S(5) = tanθ sec³ θ + 3∫sec³θdθ

S(5) = tanθ sec³ θ / 4 + (3/4)∫sec³θdθ

CSA = (π/4)tanθ sec³ θ / 4 + (π/4)∫(7/4)sec³θ dθ

2S(3) = tanθ secθ + ∫secθdθ

so ∫sec³θ dθ = ½tanθ secθ + ½∫secθdθ

CSA = (π/16)tanθ sec³ θ + (7π/16)[½tanθ secθ + ½∫secθdθ]

I will leave you to finish it from here...hope that is OK let me know if not

What a question...looked harmless

Answer 4

Let's get started on your last integral, which really is
messy:
∫ tan² θ sec³ θ dθ
I would write tan² θ = sec² θ -1
and then you have to integrate
sec^5 θ - sec³ θ.
Now use integration by parts to get
∫ sec^5 θ dθ = 1/4 sec³ θ tan θ + 3/4 ∫ sec³ θ dθ.
From here it's just a matter of doing the integral
of sec³ θ. I'll let you carry on from here!

Answer 5

You sure its the surface area you have to find orthe volume?
The volume is 6.2pie
Pie *integration of y=(x^2)^2 with limit 2 to 1

Answer 6

You know, you need to use the calculator to integrate a lot trigonometric expressions. I can't think of any rules to apply by hand. This, of course, depends on your highest level of calculus.